Principles of Chemistry
Principles of Chemistry CHE 115
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CHE 115 structure othe Atom V Structure of the Atom A Early Models of the Atom By the begin of the twentieth century experimental evidence indicated that the atom was composed of positive and negative electricity Most of the mass of the atom is associated with the positive electricity The negative electricity was found to be composed of negative charged particles The charge on one particle was determined to be 160103919 coulombs or 480103910 esu electrostatic units and the mass of one particle was found to be 91103928 g These particles were named electrons 1 Thomson Model 1904 A spherical sea of positive electricity in which a suf cient number of electrons are embedded to neutralize the positive charge Thomson suggested that high energy 1 particles HeH upon striking an atom would pass through the atom rather than be de ected 2 GeigerMarsden Experiment 191 l 01 Defleo red 999 quot quotquotquotquotquotquotquot Beam of High A Energy Alpha Metal Foil Particles The 01 of the 01 particles that were de ected through large angles encountered centers of high positive charge 3 Rutherford Planetary Model 1911 39 Most of the atom is empty space 39 Positive electricity is concentrated in a very small volume called the nucleus CHE 115 Structure othe Atom 39 Electrons suf cient in number to equal the positive charge on the nucleus reside around the nucleus and determine the effective volume of the atom 39 The unit of positive charge on the nucleus is called a proton The number of protons on the nucleus is equal to the atomic number of the element Element Atomic N0 N0 of Protons N0 of Electrons H l l l B 5 5 5 C1 l7 l7 17 The protons account for only about halfthe mass of the atom Most of the remaining mass is associated with the neutrons A neutron is a neutrally charged particle on the nucleus The mass number of an atom is the sum of the number of protons and the number of neutrons Mass Number of an Atom 7 Atomic Number of the Element 1 Number of Neutrons Atoms of the same element with diiTerent masses are called isotopes Example Some Isotopes of Neon Mass Number Atomic Number The atomic weights molar masses listed in the periodic table are the weighted average of the masses of the various isotopes Example Atomic Weight of Chlorine Isotope Natural Abundance AW gmole 35Cl 754 3497 37Cl 246 3697 The natural abundance is the percentage of the natural occurring atoms of an element that are this isotope Structure of the Atom CHE 115 8 male 0 L64 3697 g 3545 0 754 A 3497 g male AW i average male B Dual Nature of Light 1 Wave Light is regarded as an electromagnetic wave that travels through space at a constant speed ofc 29981010 cmsec Diffraction is a wave property and light is diffracted Pin Hole quot Light Bulb r H Diffraction Pattern y i i I i i i i i i i i i i i i i i i i i i i 1 i i i i i i J i i i i39 i l i i i i i i i i i i i i i i i i i i i i i 1 i i f i i i i i i i i i i i Wavelength L is the distance between the crests of two consecutive waves Units 1A Angstrom 13910quot1 nm nanometer 13910quot8 cm 13910quot10 m 139102 pm picometer Frequency v is the number of waves that pass a given point in unit time CHE 115 Structure othe Alom Units lsec or 1 Hz hertz one cyclesec Since the speed of light is constant c V7 2998391010 1 sec Visible Spectrum of Light ultraviolet violet blue green yellow orange red I I l I 4000 4500 5000 5500 6000 6500 7000 D A A 2 Particle Monochromatic light is light of a single wavelength k Quantum Theory 39 When light interacts with matter it behaves as if it were composed of discrete units or quanta called photons 39 The energy of one photon is he Ephoton hv T where hPlanck ls constant 662610 27 erg sec Note 1 erg 10 7 J and erg gcm2 sec2 39 Only whole numbers of photons are absorbed or emitted by matter 3 Photoelectric E ect When monochromatic light of an appropriate wavelength strikes a metal surface electrons are ejected KE max Ephoton h CIA r Metal CHEHS Structure ofthe Atom The maximum kinetic energy KEmax of the ejected electron was experimentally found to be independent of the intensity Hm of the in light and dependent on the wavelength it of the monochromatic light Einstein s explanation E photon E KEmax where the threshold energy El is the minimum energy required to remove the electron from the surface of the metal Example When monochromatic light of 4500 A strikes the surface of metallic sodium electrons with KEmax 336103912 erg are ejected What is the threshold energy for sodium Unknown EL for sodium Knowns I 4500 A and KEmax 336103912 erg Concepts Quantum theory photoelectric effect Relationships E 7 and E phatan phaton 7 E 1mmax E iEtiKEmax i 662610 Z7erg sec2998101 7 E i 33610 12erg H 10 cm 4500AT El 7 10510 12erg C Line Spectra of the Elements When an electric current is passed through a discharge tube that contains hydrogen gas 10 torr a white light is emitted Analysis of the white light reveals that only certain wavelengths of light are emitted White Light Hydrogen Lamp Tube containing hydrogen gas 10torr A 486143404101 Spectroscope Line Spectrum Conclusion atoms emit only certain discrete energies CHE 115 Structure othe Atom Spectroscope is an instrument that analyzes white light by separating it into its various wavelengths Line spectrum consists of a limited number of lines and each line corresponds to a diiTerent wavelength of light A mathematical equation that predicts the wavelengths of the bands in the hydrogen line spectrum was developed by J J Balmer v 32891015sec1 1 4 quot02 quot02 where n and n2 are dimensionless integers ie n 1 2 3 4 5 and n2 gt n Example Calculate the wavelength of light in the hydrogen line spectrum that corresponds to n1 2 and n2 6 l l 7 32891015 sec l 2 2 n1 quot2 V E X 29981010 8 7 32891015sec 1l l W 62 1A 7 7 4102A D Bohr Model of the Hydrogen Atom 39 The electron revolves around the nucleus with a xed energy and does not emit or absorb energy spontaneously For the hydrogen atom in its ground state the energy of the electron is E E 2179103912 erg gt Infinity 3951 Em0 E2179103912ergEm E1 oE1 Radius 0 Orblt Orblt 0529A E121791012erg where is the energy of the electron at an in nite distance from the nucleus Structure othe Atom 39 The electron in the hydrogen atom may have other possible energies namely 21793910 12 er g 5 quot2 En where n called the Principal quantum number is a dimensionless integer ie l 2 3 4 Energy Level Diagram E for Hydrogen Radius n n of Orbit 0871012 erg 5 1322 1361012 erg 4 846 2421012 erg O 3 476 0 4861 A Energy of electon 6563 A increasing 5451012 erg 2 212 21791012 erg 1 0529 39 Absorption of energy by the hydrogen atom corresponds to the transition of the electron from a lower energy level En with quantum number n1 to a higher energy level En with quantum number n2 Emission of energy by the hydrogen atom corresponds to the transition of the electron from a higher energy En with quantum number n2 to a lower energy level En with quantum number n1 E energy emitted or absorbed En Equot 6 Z l where n2 gt 711 Note The energy emitted or absorbed by an atom must equal the difference in energies of two energy levels Substituting eq 5 into eq 6 yields the following equation for the hydrogen atom Eenergy emitted or absorbed 217910 12erg 1 2 1 2 7 n2 n1 where n2 gt 711 Example What wavelength of light will be emitted when an electron in a hydrogen atom drops CHE 115 Structure othe Atom from the energy level with n 5 to the level with n 2 nknown Wavelength 7c of light emitted anons Since nZ gt n1 then nZ 5 and n1 2 Concepts Bohr model of the hydrogen atom quantum theory 1 l w Relationships Ephmn 7 if and E 7 217910 12erg l he 39 quot02 n1 2 T 7 217910 12erg 662610 Z7erg sec2998101 A 8 217910 12erg L l E 22 1A L 7 4341A E Matter Waves 1 de Broglie 1924 Proposed that particles such as electrons may have wave character Particles with wave character are called matter waves The momentum mv of the photon or electron is related to the wavelength 9 by the eq 8 x71 8 m v where m mass v velocity p mv momentum and h is Planck s constant Example Calculate the wavelength of an electron m 91103928 g that has a velocity of 59108 cmsec 27 AA 7 1 662610 erg sec 1A 7 12A mV 9110 2 g59108 10 8cm 2 Heisenburg s Uncertainty Principle It is impossible to know simultaneously both the position and momentum of a particle with wave character AxApx z i 9 47 CHE 115 Structure othe Atom where AX is the uncertainty in the position X of the particle and ApX is the uncertainity of the momentum in the X direction of the particle F Quantum Mechanics The electron is considered to be a wave as well as a particle and the energy of the electron is given by the Schrodinger wave equation Schrodinger wave equation for one dimension X 61ij Sitzm39 d 2 hz E V 0 11 x where h is Planck s constant m is the mass of the electron X is the coordinate position of the electron E is the total energy of the electron V is the potential energy is the solution a mathematical equation that depends on the coordinate X to the diiTerential equation and is called the wave function and d2 dX2 is a mathematical operator To illustrate some of the features of the Schrodinger wave equation we consider the following problem Calculate the allowed energies of a particle m 91103928 g constrained to a one dimensional b0X of length a 2510398 cm VCD V0 VCD Particle I I Inside the b0X V 0 and thus 2 2 d87tmE0w0 dxz hz It can be shown that the solution to this second order homogeneous linear diiTerential CHE 115 Structure of the Atom equation has the form V A sin0x where A and 0t are constants The constraint that the particle must be inside the box is used to evaluate the constants and the solutions take the form 22 1n E sin and E n h n a Smut2 where 11 called the quantum number is an integer ie n l 2 3 4 5 6 Energy Level Diagram for Particle in the Box and Plot of 112 versus x its Zen112 sin3nXa 7 E3 32 r12 8ma2 3 E2 22 hz8ma2 a 2 m1 2a391 2 sinuxla E1 12 h2 8ma2 1 O 1e 8 2e 8 x cm Observations 0 For each wn solution to the wave equation there is a corresponding En 0 Only certain discrete values of the energy for the particle are possible namely EIl n2h28ma2 The energy of the particle En is independent of the particle s position X 10 Structure othe Alom CHE 115 The number of diiTerent types of quantum numbers is a consequence of the number of coordinates Coordinates Quantum Numbers X n X y z n 1 m 1 X y ztime n 1m1mS Spherical Polar Coordinates Z x y z Cartesian n re p Spherical Polar Shadow made by the r vector on the xy plane G Hydrogen Atom Quantum Mechanical Model Z v Q2 r Z charge on nucleus 1 for H e charge on electron 48 103910 1 Solutions to the Schrodinger wave equation for the hydrogen atom have the following features a The wave functions are characterized by four quantum numbers n 1 m 1 and ms b Acceptable solutions Ly are obtained only when n I m 1 and ms have the following values Principal Qunatum Number n n l 2 3 4 5 6 Azimuthal Quantum Number 1 1 l 2 3 n 1 m1 1 3210123 1 Magnetic Quantum Number m 1 Spin Quantum Number ms ms 12 12 0 2 Solution wquot m when n l 1 0 m1 CHE 115 Stmctme ofthe Atom lzlitewtel 1 1 E 21791012erg ls where a0O529A R called the radial part is the part of y that depends on r 91 called the angular part is the part of y that depends on 6p Note the angular part of Wm is a constant If the electron is in state WIS then it will have energy E1S irrespective of its position in space ie r may be any value from o to 00 We know the energy of the electron exactly but we do not know its position r 6 p in space 3 Probability Electron density Wm 15m2 is proportional to the probability per unit volume of nding the electron at a point r 6 go in space Units of Wm I m volume391 Graphical Representations of Probability a Plot of q versus r with 6 and p xed Plot of 11J1s2 Versus r 20 Fore 90 and p 0U r3 12 CHE 115 Structure othe Alom Points on the curve represent the probability of nding the electron at distance r from the nucleus b Radial Distribution Function What is the probability of nding the electron in a spherical shell of thickness Ar and radius r Probability Probability per unit volumeXvolume of spherical shell Cross Section of a Spherical Shell ltAI When Ar is very small the volume of the spherical shell is approximately Volume of Spherical Shell 2 4nrzAr and Probability R24nrzAr where R is the radial part of I Since Ar is very small and effectively constant 4rlr2R2 called the radial dism39 button function is plotted as a lnction of r CHE 115 Structure of the Atom Plot of Scaled Radial Distribution Function Versus r 4m2R152 O 01 Most Probable Distance rmp 0529 A Points on the curve represent the probability of nding the electron in spherical shell of very small thickness at distance r from the nucleus c Electron Density Cloud A threedimensional plot of ln Jym2 versus r 6 and p The density of dots at a speci c coordinate r 6 and p is proportional to the probability of nding the electron at that coordinate d Contour Plot A cross section of the electron density cloud The symbols which are printed below represent the relative electron densities wzwmaxf where llmx2 is the maximum value calculated for W Range of Electron Dinsity ED Symbol ED 3 001 001 3 ED 3 002 002 3 ED 3 010 010 3 ED 3 025 0 l4 CHE 115 Structure othe Atom 025 5 ED 5 050 050 5 ED 5 10 Contour Plot for Hydrogen ls Atomic Orbital WIS 53 II0000000I 40 00000000000 27 OOOOOO l3 z000000z 000 z000000z 0l3 z000000z O27 OOOOOO 040 00000000000 053 II0000000I 067 I 093 oooooool I I I I I k0 w The printed area represents the region in space in which there is a 99 probability of the nding the electron e Solid Representation Boundary surface of the electron density cloud for the hydrogen ls atomic orbital WIS The enclosed volume represents the region in which there is a 99 probability of 15 CHE 115 Structure of the Atom finding the electron H Atomic Orbitals The one electron wave lnction ln I m which is identi ed by the three quantum numbers n 1 and m is called an atomic orbital A0 A number and letter are used to designate the values of the n and 1 quantum numbers of an A0 Letter s p d f g 1 O l 2 3 4 Example A 3p A0 is an A0 with n 3 and 1 l ie W311 w3p 1 2s A0 for Hydrogen n 2 I O m O 0 36mm L g nl 0 Graphical Representations Plot of 111252 Versus r For 9 90 and lt1 0 25 Node Zero Probability 0001 t 0 1 2 3 4 5 6 r A 16 CHE 115 OOOOOOOI I H 33 20 07 93 80 67 53 40 27 13 OO 13 27 40 53 67 80 93 07 20 33 Structure of the Atom Plot of Scaled Radial Distribution Function for 23 AO Versus r 10 i 05 Node Zero Probability 00 i i i I i i i 0 1 2 3 4 5 6 7 8 H50 Boundary surface of the electron density cloud for the hydrogen 2s atomic orbital12S 17 CHE 115 Structure othe Atom Note The number of nodes in addition to the node at the nucleus n 1 1 2 2pX A0 for Hydrogen n 21 1 m11 1 1sz 3zntaQ3 Za exp2lt cose Simp Contour Plot for 2px Atomic Orbital 2 1 1 1 1 1 120 107 00000 093 00000000000 00000000000 080 OOOOOOOOOOOOOOO OOOOOOOOOOOOOOOz 067 nnnnnn Annz znnn nnnnnn 053 quot quotquotquotquotquot 040 quot quot quotquot 027 quot 7777777 77 quot quot 013 T T T T T T T T T H quotn m 000 n T T T T T T T T T H quotn m 0 l3 T T T T T T T T T H quotn m 027 quot 7777777 H quot 040 quot quot quotquot 053 quot quotquotquotquotquot 067 nnn nnnnnn O80 000000000000000 093 OOOOOOOOOOO 107 120 133 147 l60 173 l87 CHE 115 Structure othe Atom Boundary surface of the electron density cloud for the 2pX atomic orbital 3 When 1 then In 1 0 1 Thus there are three AOs with the same n and 1 values but di erent m 1 values The electron density clouds for these AOs have the same size and shape but di erent spatial orientation 2p The change from solid cross hatching to dashed cross hatching on the boundary surfaces indicates an algebraic sign change in the wave function 141 m ie positive to negative algebraic sign change 4 When 2 then In 2 l 0 l 2 Thus there are ve AOs with the same n and 1 values but di erent m 1 values The electron density clouds for these AOs have the same size and shape but di erent spatial orientation CHE 115 Structure othe Atom z YL X Summary The size shape and spatial orientation of the electron density wnv L m 2 cloud are related to the quantum numbers n 1 and m 1 that de ne the atomic orbital ym L m 39 n is related to size For the hydrogen atom the size of the electron density cloud increases as n increases 25 35 39 1 is related to the shape 39 m is related to the spatial orientation 20 CHE 115 structure othe Atom 21 5 Polyelectronic Atoms Exact solutions to the Schrodinger wave equation for atoms with two or more electrons are not possible and we must settle for approximate solutions The independent particle model is o en utilized to obtain approximate solutions In this model each electron in the atom is treated independently of the other electrons Each electron is treated as if it were the only electron in an atom that has a nucleus with a charge Zeff For example consider the lithium atom To calculate the energy of Electron 3 we solve the Schrodinger wave equation for a hydrogenlike atom oneelectron atom with a nuclear charge of Zef 13 To calculate the energy of Electron 2 we solve the Schrodinger wave equation for a hydrogenlike atom with a nuclear charge of Zef 27 Thus solutions to the threeelectron problem are approximated by the solutions to three oneelectron problems Actual Charge on Nucleus Effective Nuclear Charge Zeff N I m Screening Constant the part of the nuclear charge Z quotscreenedquot from the quotouterquot electrons by the quotinnerquot ele rons 21 Structure othe Atom With this model the electronic structure of a polyelectronic atom may be described in terms of atomic orbitals In L m that are hydrogenlike and are de ned by the quantum numbers n 1 and m 1 The shape and spatial orientation of the electron density clouds for these AOs are identical to shape and spatial orientation of the clouds for the hydrogen AOs However the size of the cloud will increase as n increases anal Zg alecreases CHEHS I n 1 Rule and Electron Con guration 1 n 1 Rule In the absence of a magnetic eld the energy of an A0 for a polyelectronic atom is determined by the quantum numbers n and 1 In a neutral isolated atom the smaller the sum of the values of n and 1 n I the more stable lower energy the AO described by n and 1 If two AOs have the same n I value the A0 with the smaller n value will be the more stable AO Relative Orbital Energy Level Diagram 4pX 4py 4pZ n I 5 3122 3dXZ 3d 3d 3dx2y2nl5 yz Xy 48 n 4 Orbital Energy Increasing 3px 3Py 3pz n I 4 and Stability 3Sn H 3 Decreasing 2pX 2py 2pZ n3 The set of AOs with the same n value is called a shell Example 2s 2px 2py and 2pZ The set of AOs with the same n value and the same 1 value is called a subshell Example 2px 2py and 2pZ Note The AOs in a subshell have the same energy and are said to be degenerate 2 Pauli Exclusion Principle No two electrons in an atom can have the same values for all four quantum numbers 22 structure ofthe Alom CHE 115 Example He Atom Electron n 1 m 1 ms 0 0 12 2 l 0 0 12 Since both electrons have the same values for the quantum numbers n 1 and mi then the two electrons must have di erent values for the mS quantum number in order to satisfy the Pauli principle Implication Since there are only two possible values for mS 12 12 then only two electrons may have the same values for the quantum numbers n 1 and m 1 Hence only two electrons can occupy the same A0 wquot L m and these electrons must have di erent ms values 3 Electron Con guration Ground State A representation of the occupied AOs in an atom Symbols represents the relative energy of the A0 1 represents an electron with ms 12 l represents an electron with ms 12 Procedure Determine the total number of electrons and distribute the electrons as follows 39 Fill the most stable AO the A0 with the smallest n Ivalue according to the Pauli principle 39 Fill the next most stable AO according to the Pauli principle 39 Continue in this fashion until all the electrons have been distributed Example Write the ground state electron con gurations for Li through Ne Hund 8 Rule r K L l L 2pX 2py 2pZ 2pX 2py 2pZ 2pX 2P 2pZ 2pX 2py 2pZ i l 28 28 28 28 ll ll ll ll 18 18 18 18 Li Be B C Short Form Li 1s2 281 Be 1s22s2 B 13223221 C 1s22s22p2 f Subshell labels Subshell labels 23 CHE 115 Structure othe Awm H and 3 rule states that the most stable con guration is the one in which the electrons are in ah erent orbitals with the SAME energy Y t t L L t i i ii L L 2pX 2py 2pZ 2pX 24B 2pZ 2pX 2py 2pZ 2pX 2py 2pZ 28 f 28 28 i i i i 18 18 18 18 N O F Ne N 1822822p3 O 1822822p4 F 1822822p5 Ne 1822822p6 f Subshell Labels Anomalies are found for Cr Cu and members of their groups 6 and 11 Example Ground State Electron Con guration for Cr Predicted Cr ls22s22p63s23p64sz3d4 Found Cr ls22s22p63s23p64313d5 Reason H alf lleal anal lled subshells are more stable lower energy than partially lled subshells Note The 4s and 3d subshells of Cr are half lled 4 Excited States To a very crude approximation the total electronic energy of the atom is equal to the sum of the energies of the electrons Orbital Energ v 4 835 35 T a L L 4 A no 4 L 29 2pX 2py 2pZ E 2PX 2Py 291 ll gt ll 825 28 E EE 39 E6 28 ll ii 815 15 T N N Ground State Configuration Excited State Con guration E6 2815 2 25 3 2p EE 2815 225 2 2p 835 24 Structure othe Alom When a nitrogen atom in its ground electronic state absorbs monochromatic light of wavelength h it is promoted to an excited electronic state CHEHS The ground state is the lowest energy state of an atom ion or molecule An excited state is any energy state other than the ground state J Periodic Table The properties of the elements are a periodic lnction of their atomic numbers Pen39odic Table 1 2 13 1B Gmup Number R N 39 Nonmatals r 3 12 r m 2 r t Transition Metal 2 V The horizontal diVisions of the periodic table are called periods and the vertical diVisions are called groups The least stable occupied shell in the electron con guration of the atom is called the valence shell Example Valence shell is underlined Group 1 10 13 17 u 1amp21 B 11 F 1amp5 Na 131 Al 15222292391 cl 9292132425 K 1522522p 3s73p 1 Ni 1522822p 3s73p 2ampl Ga 1amp2522p 3523p 3d1 5 gn an 1szzs 2p 3523p 3d sf Visliemrii5 1 2 a s2 1 2 MalMon ns ns n1d n np ns np5 25 CHEHS Structure othe Atom Note Elements in the same group of the periodic table have the same type of valence shell con guration K Size of the Atom 1 Atomic Radii a van der Waals radius distance from the center of the atom to a point where the electron density is effectively zero b Covalent radius one halfthe distance between the centers of two bonded atoms Contour Plot for Bonding Molecular Orbital in H2 37 Covalen 037A 2 Periodic Trends in Atomic Radii a The atomic radii decrease with an increase in atomic number across a period Explanation Zef increases while n remains constant across a period b The atomic radii increase with an increase in atomic number down a group Explanation n increases while Zeff increases only slightly down a group Atomic Radii Decreasing Increasing Periodic Table 26 Structure othe Atom CHE 115 L Ionization Energy The minimum energy required to remove an electron from a gaseous atom or ion in its ground state Example He Atom He 9 gt He g e IE1 2372 Id and AH 2372 kJ we r He g gt H 2 g 9 IE2 5247 H and AH 5247 H Second Ionization Energy energy required to remove second eledmn Periodic Trends in Ionization Energies Ionization Energy Increasing Periodic Table Decreasing Note The periodic trends for ionization energies are the opposite of the trends for atomic radii As Zeff increases across a period more energy is required to remove the electron As n increases while Zef increases only slightly down a group less energy is needed to remove the electron M Electron Af nity The energy released when a gaseous atom or ion in its ground state adds an electron cu e39gt Cl39g EA3486kJ andAH3466kJ Since energy is released AH is negaiive Periodic Trends in Electron Af nities 27 CHEHS Thamochemistry III Thermochemistry Thermodynamics of which thermochemistry is one area is the study of the heat change in a system and work done on the system by the surroundings A Heat Th initial temperature To initial temperature mh mass ofhot block m6 mass ofcold block ch speci c heat of hot block co speci c heat ofcold block When a hot block is placed next to a cold block the temperature of the cold block rises and the temperature of the hot block drops At thermal equilibrium the temperatures of the two block are the same value The change in temperature of each block is related by the equation mh ch Th Tf mccC Tf T where Tf is the temperature of each block at thermal equilibrium and c is the speci c heat The speci c heat is the amount of heat required to raise the temperature of 00 g of a substance one degree The right side of the equation is the change in heat of the hot block and the le side is the change in heat of the cold block heat gained by cold block ac chCTf TC heat lost by hot block ah mhch Tf Th Since qc is a positive number heat absorbed and qh is a negative number heat evolved then 61 q i 0 q is defined as the heat absorbed by a substance and T nal Initial q chT where AT T or q nC AT where n is the number of moles of the substance and C is the molar heat capacity of the substance The molar heat capacity is the amount of heat required to raise the temperature of one mole of substance one degree CHE 115 Thamochemish39y B Work Work is the result of an action against an opposing force over some distance work forceXdistance Although there are many kinds of work we are interested primarily in the work done by a frictionless piston on an ideal gas frictionless cylinder PM is the external pressure and Pnt is the internal pressure of the gas When Pext lt Pint then the gas expands and work is done against Pm when Pext gt Pin then the gas is compressed and work is done against Pint and when Pext Pint then the gas is in equilibrium with the piston Thermodynamic terms System any piece of matter under consideration Examples ideal gas chemical reaction Surroundings any matter that can interact with the system Examples cylinder piston reaction ask w is de ned as the work done on the system by the surroundings For pressurevolume work PV work w is de ned by the equation f VZ w PdV area under PV curve V1 Consider the expansion of an ideal gas from PIV1 to P2V2 Em us unmanned Expansion gt State 1 PV curve Stale M Pressure aim State 2 Since V2 gt V the area under the PV curve gt 0 and w area under the PV curve lt 0 The reverse ofthe de nition ofw occurred ie the gas the system did work on the piston the smmundmgs Consider the compression ofan ideal gas from pm to pzvz cm us Whom Compression gt Aplot i i i on n i i ii r PVCurve 22 p A 20 Staten E i 2 in Area under a A W Cuwe Slate m Voiume L V since V2 lt V the area under the PV curve lt 0 and w area under the PV curve gt 0 The piston the Smmlmdmgs did work on the ideal gas the sysmm r A 39 39 39 39 tothe nal stateState 2 The magmde ufwzspath dependent Example Consider the expansion ofan ideal gas by two di erent pathways in which the initial state pm and the nal state pzvz are the sarne in both pat ways 1Expansion intoayacuurn Pm0 Vi Vi w r Pde r OdV 0 Vi 39Vi cm us madam 2 Expansion against a constant ada nal pressure P constant V1 V1 W Pde Pm W Psz Vt 39V 39V w r PWAV Note Under the Sandman nfcnnstampessm w 7 PAme an Ideal gas c First Law ofThermodynamics Consider acycllc process for an ideal gas in a cylinder with a ictionless piston V22EIL Cyclic Process 1 Pressu39e 21m F1 10 7 n W PIN 39V E 7 cm 5m 5 a 2 u A t t t t 0 a m 15 29 25 v vatnmeu CHEHS Thamochemish39y The net work w done in one cyclic is the sum of the work associated each step net w wa wb wC wd P1V2 V1 P2V1 V2 net Payment for the net work wnet is in the form of heat qnet In Step a heat must be added in order to increase the volume of the gas at constant P and n To lower the pressure of the gas under the condition of constant V and n in Step b heat must be removed In Step c the gas must lose heat and in Step at the gas must gain heat qnet qaqbqcqd For a cyclic process wnet qnet qnet wnet 0 In those cases where q gt w the excess heat is absorbed by the system and the internal energy of the system will increase When q lt w some of the internal energy of the system will used to do the work and there will be a reduction in the internal energy of the system E is the symbol for internal energy Example of Internal Energy When an ideal monatomic gas is the thermodynamic system then E internal energy is the kinetic energy of translation ET 3RT2 The change in the internal energy of the system is AE q w where q is the heat absorbed by the system and w is the work done on the system by the surroundings The First Law of Thermodynamics states i AE q w ii E is a state function A state function is any lnction in which the change in going from the initial state state 1 to the nal state state 2 depends only on the conditions of the initial and nal states but not on the path taken between the two states State functions are path independent Example Consider the expansion of one mole of an ideal monatomic gas from the same initial state State 1 to the same nal state State 2 by two diiTerent paths As can be seen from the PV curves below the amount of work done in the process associated with Path 1 is diiTerent from the work associated with Path 2 However since the initial and nal states are the same in both paths AE for both paths must have the same value Path 1 AE qw 6481J 6481J 0 Path 2 AE q w 2906J 2960J 0 Em us Wham Expansion of One Mole of an Ideal Gas from State 1 to State 2 by Two Different Paths sune m Pam n am an Pyesswe arm w 2906 J Stats 2 2 3 Volume Lt DEntha1py H qv is the heat absorbed by the system in a constant volume process q 39 39 39 39 quot thesystemm 1 A Constant Volume Process M qv w at Gunman vulume w 2 0 and AE qr 0 E qv Thus q is a state function 2 A Constant Pressure Process M q w at constant pmssurz w PAV and ME q PAV Rearranging Thamochemistry mm qp AEPAV E2 E1 PV2 V1 qp E2PV2 E1PV1 Hz Hl AH qp AH Since the magnitude of qp depends only on the conditions of the initial and nal states qp is a state lnction H is called enthalpy and is de ned as H E PV Enthalpy H is a state function At Constant Pressure AH qI 3 When a Chemical Reaction is the Thermodynamic System the Relationship between AE and AH a Reactants and or Products are Gases Consider the reaction at constant T and P N2 g 3 H2 g a 2 NH3 g AH 9338 kJ P1 1 atm P2 1 atm 5 l V1 at qp 9338 kJ I 0 V at 25 C Initial and Final States m 2 have the same T amp P 25 C State 1 state 2 qp AH AEPAV 1 Since T and P are the same in the initial and nal states then a change in volume AV will result from a change in the total moles of the gases An Assuming ideal gas behavior State 1 P1 V1 nNZ nHZRT and State 2 P2 V2 nNHgRT PAV PV2 V1 nNH3 nNZ nHZRT AnRT 2 Substituting eq 2 into eq 1 AH AEAnRT CHE 115 Thamochemistry where An is the TOTAL moles of GASEOUS products minus the TOTAL moles of GASEOUS reactants b Reactants andor Products are Liquids or Solids For solids and liquids AV 2 0 and PAV w 0 andthus AHAE0 AE Note Physical states g gas l liquid s solid aq aqueous solution Example Calculate AH at 25 C for the reaction below C6H6 1 152 02 g 6002 g 3H20 1 AE 3264 k AHiAEiAnRT 15 3 An 7 ncoz no2 7 6 7 3 male 3 V J 7 Id 7 AH 3264M 1 mole 8314 298K 7 3268M 2 WE K 1000 E Heats of Reaction Heat of reaction under the condition of constant volume qV AE Heat of reaction under the condition of constant pressure qp AH 1 Heat of Combustion The heat change associated with the burning of a compound Example 204H109 13 029 gt 80029 10 H200 Hocomb5314KJ Thermochemical Equation Heat Change for Reaction as Written The heat of combustion is measured in a bomb calorimeter CHE 115 Thamochemish39y Bomb Calorimeter A weighed sample of a compound is placed in the bomb The bomb is charged with 25 atm of O2 and ignited The heat given up by the reaction qrmlion lt 0 is absorbed by the calorimeter qcal gt O qreaction goal The change in temperature of the calorimeter is measured and the heat absorbed qcal Ccal AT by the calorimeter is calculated Since the volume of the bomb in which the reaction takes place is constant qreaction qv AE Coalor Cwafer Coal qcal qreaction AE Coal where Ccalor is the heat capacity of the calorimeter components bomb bucket thermometer etc and Cver is the heat capacity of the water in the bucket The heat of reaction under the condition of constant pressure is calculated with the the following equation AH AE AnRT 2 Heat of Formation The heat change associated with the formation of a compound from its elements in their most stable form Example N2 g 3 H2 g 2 NH3 g AHquotf 9338 k The numeric value for the heat of a reaction is dependent on the temperature and pressure at which the reaction is carried out If we are to compare heats of reaction then these heats must be measured under the same conditions For this reason a standard set of 10 CHEHS 39I hamochemistry conditions has been chosen for reporting heats of reaction These conditions are that the reactants and products are in their standard states Standard State of gas liquid solid Physical State pure gas pure liquid pure crystalline solid Pressure partial pressure 1 atm 1 atm of 1 atm Temperature 25 C 25 C 25 C The heat change associated with the formation of a compound in its standard state from its elements most stable form in their standard states is called the standard enthalpy of formation AH f denotes standard state 1 atm and 25 C A Hof denotes formation By de nition the AH for an element is its most stableform is zero A table of heats of formation for compounds can be found in the text Example Table of Standard Enthalpies of Formation Compound AH f kJmole H20 1 28584 NH3 g 4669 CO2 g 39351 H20 g 2418 H2 g 00 H g 21794 The thermochemical equations for these AH f are known by definition Example Write the thermochemical equation for AH f CO2 g 39351 ldmole Since 3935l kl is the heat change for the formation of one mole of CO2 from the most stable forms of elemental carbon and hydrogen then Cs 02g t coZ g AH f 39351kJ 3 Hess s Law The heat change at constant temperature and pressure for a particular chemical reaction is constant and independent of the path taken from reactants to products Consider the formation of CO2 by two diiTerent pathways CHEHS Thamochemistry CO 9 12 029 12 029 02 9 c s gt 002 g The heat change is the same for both pathways to C02 These thermochemical C s 12 02 9 CO 9 AHof 41052 N The enthalpies add equations a to give to gtIge the ent a py I or e summary the 5mm squat co 9 12 02 g 002 g AHocomb 28299 kJ equation Summary equation gt C s 02 g gt C02 9 AHof 69351 kJ 4 Calculation of enthalpies that cannot be measured from enthalpies that have been measured Procedure Step 1 Write down the thermochemcial equation for which you wish to calculate the AH or other thermodynamic function as the summary equation Step 2 Arrange given thermochemical equations equations with known values for AH or other thermodynamic function so that the sum of the equations is the summary equation Step 3 Write down an appropriate value for the AH or other thermodynamic function associated with each thermochemical equation in Step 2 If the thermochemical equation is the reverse of the given equation change the sign of the AH Step 4 Add up the values for the AHs or other thermodynamic functions in Step 3 Example Calculate the enthalpy of formation AHquotf for benzene C6H6 at 25 C given 06H 1 152 02 g 6002 g 3H20 1 AH 32677kJ 0comb and AHquotf CO2 g 39351 ldmole AHquotf H20 1 28584 kJmole Note The thermochemical equations associated with the enthalpies of formation for CO2 and H20 are known by de nition Cs 02 g coZ g AHquotf 39351kJ HZ g 12 02 g H20 1 AHquotf 28584 kJ CHEHS Thamochemsh39y 3 H2 9 3 2 O2 9 3 H20 I AH 3 mole 28584 kJmole Step 6 35 602 9 6 0029 AH 6 mole 39351 kJmole Stem 6 C02 9 3 H20 0 06 D 152 02 g AH 32677 kJ Step1 gt 6 C s 3 H2 9 gt C6H6 I AHO 7 AH CSHS 1 4912 kJmole Altemate Method for Calculating AH AH reaction VproductAH f prodz wt 7 Z VreacmmAH f VGQCIQI ZI where v called the mole number is the stoichiometric coef cient in the thermochemical equation Example Calculate the enthalpy of formation AHquotf for benzene C6H6 at 25 C given C6H61 152 02 g 6 CO2 g 3 H201 AHquotwmb 32677kJ and AHquotf CO2 g 39351 ldmole AHquotf H20 1 28584 kJmole AHlt mmb JCOZAI Ff C02 l VHZOAHf H20 Vc H AITf CsHs VOZAHf 02 32677k 7 6male 39351 M 1 3male 28584 lmaleAH fC6H6 1 1 25maleoo male mole male AH fC6H6 1 7 4912 H male F Molar Heat Capacity 1 CV Heat Capacity for a Constant Volume Process Consider increasing the temperature of one mole of an ideal monoatomic gas by AT degrees in a constant volume process E2 32 RT AT E1 32 RT The amount of heat required AE E E 32 RAT q to raise the temperature of 2 1 V the gas by AT degrees in a constant volume process CHE 115 Thamochemish39y V E 3 AT AT CV for an ideal monatomic gas is C 3R 1247 V 2 mole K Values of CV for some other gases Compound H2 Cl2 HCl CV JmoleK 2033 2510 2100 Rearranging eq 3 AE CV AT 4 Eq 4 is valid for one mole of a real gas in a constant volume process and for one mole of an ideal gas under all conditions 2 Cp Heat Capacity for a Constant Pressure Process For a constant pressure process the change in volume of one mole of an ideal gas is the result of a change in temperature PAV PV2 V1 RT2 T1 RAT 5 The heat absorbed by one mole of an ideal gas in a constant pressure process qp AH AEPAV 6 and Rearranging eq 7 AH CP AT 8 Eq 8 is valid for one mole of a real gas in a constant pressure process and for one mole of an ideal gas under all conditions Substituting eqs 4 5 and 8 into eq 6 CF AT CV AT RAT and dividing by AT CP CVR 9 CHE 115 Swichiometry I Stoichiometry A Laws of Chemical Combination A law is a statement that describes the general behavior of nature Law of Conservation of Mass The mass of the products is equal to the mass of the reactants Law of De nite Pr0p01ti0n A given compound however it is obtained contains its component elements in a xed proportion by weight Example For any given weight wt of water H20 wt ofH in the given wt oszO 0 126 wt ofO in the given wt oszO T I Law of Multiple Proportions When two elements combine to form more than one compound the different weights of one element that combined with a xed weight of the other element are in a ratio of a small whole number Example 16 g of oxygen combine with 2 g of hydrogen to form water H20 and 32 g of oxygen combine with 2 g of hydrogen to form hydrogen peroxide H202 For the same xed weight 2 g of hydrogen wt ofO in H202 7 32g 7 2 wt ofO in H20 16g B Atomic Theory John Dalton 1803 A theory is an explanation of the observed behavior law in terms of a model Postulates underlying hypotheses or assumptions 0f the Atomic Theory 1 A chemical element is composed of very small discrete particles called atoms that remain unchanged during a chemical reaction 2 All atoms of the same element are alike in all respects thought to have the same mass and atoms of different elements have different masses Today we known that atoms of the same element may have different masses but these atoms will have the same number of protons Thus atoms of different elements will differ in the number of their protons 3 Chemical compounds consist of molecules which are a stable grouping of an integral number of atoms of each component element Example A water molecules contains two hydrogen atoms and one oxygen atom The chemical formula H20 for water indicates this composition ie subscript l CHE 115 Swichiometry states the number of atoms of that element in the molecule Explanation of the Laws of Chemical Combination Law of Conservation of Mass Since atoms are conserved in a chemical reaction Postulate l the number of atoms of each element in the products will equal the number of atoms of each element in the reactants H2 1202 a H20 Law of De nite Proportion Consider a compound that contains hydrogen H and oxygen 0 A molecule of the compound contains x hydrogen atoms and y oxygen atoms ie HXOy and x and y are whole numbers Postulate 3 Each hydrogen atom has a mass h and each oxygen atom has a mass 0 Postulate 2 If a given weight of the compound HXOy contains N molecules then wt ofH in the given wt of HXO Nxh X y r constant wt ofO in the given wt of HXOy Nyo yo Law of Multiple Proportions Consider the compounds HXOy and HWOZ x y w and z are whole numbers Postulate 3 Each hydrogen atom has a mass h and each oxygen atom has a mass 0 Postulate 2 If a xed weight of hydrogen contains N hydrogen atoms then the number of oxygen atoms needed to form HXOy is given by the equation Number of HXOy molecules that can be formed from N hydrogen atoms g x Number of oxygen atoms in 3 molecules ofoOy w H X x x and the number of oxygen atoms needed to form HWOZ is Number of oxygen atoms in E molecules ofHWOz z W w w The weight of oxygen needed to combine with the given weight of hydrogen to form HXOy is equal to the mass of one oxygen atom times the number of oxygen atoms needed ie oyNx and the weight of oxygen needed to combine with the same given weight of hydrogen to form HWOZ is ozNw For the same xed weight of hydrogen the ratio of the weight of oxygen needed to form HXOy to the weight of oxygen needed to form HWOZ 1s CHEHS Swichiometry wt ofO m HxOy 0 yw wt ofO in HWOz 0 E xz C Law of Combining Volumes Gaseous volumes measured at the same temperature temp and pressure 1 volume of hydrogen 1 volume of chlorine 2 volumes of hydrogen chloride 2 volumes of hydrogen 1 volume of oxygen 2 volumes of water Law of Combining Volumes When gases react the volumes produced or consumed measured at the same temperature and pressure are in the ratio of small whole numbers Explanation proposed by Avogadro Avogadro s Hypothesis Equal volumes of gases measured at the same temperature and pressure contain the same number of molecules Implication At the same temperature and pressure 1 volume of hydrogen 1 volume of chlorine 2 volumes of hydrogen chloride Let N the number of molecules in one volume at the temperature and pressure N molecules of hydrogen N molecules of chlorine 2N molecules of hydrogen chloride Divide by N 1 molecule of hydrogen 1 molecule of chlorine 2 molecules of hydrogen chloride The chemical formula for hydrogen chloride has been experimentally determined to be HCl Inserting the chemical formulas for hydrogen chlorine and hydrogen chloride H c1y 2 HQ The Law of Conservation of Mass demands that x and y are 2 H c1 2 HQ Some elements that exist as diatomic molecules H N 0 01 Br D The Mole Concept CHE 115 Swichiometry According to the Dalton s Atomic Theory atoms of one element or molecules of one compound can be distinguish from atoms of other elements or molecules of other compounds by their absolute masses Since the absolute mass of an atom or molecule could not be measured in the early part of the nineteenth century a set of relative masses were established to characterize elements and compounds The relative mass of element A RM A is de ned with respect to the relative mass of a reference element RMref RMA Absolute mass of an atom of element A RMref Absolute mass of an atom of reference element Example When 0600 g of carbon are burned in pure oxygen 220 g of carbon dioxide are formed The chemical formula for carbon dioxide is C02 RM0 7 g Absolute mass of an atom of oxygen 1 RMC c Absolute mass of an atom of carbon c 0600g N where N is the number of carbon atoms in the 220g of CO2 220g 7 0600g 7 160g 0 2N 2N If the value of RMc is arbitrarily set at 1200 then RMo g IZN 0800g RMC c 06 g 0600 g N RMO RMC 039800g HMW 1600 0600g 0600g In a similar manner the relative masses of other elements and compounds were determined Substance Relative Mass H l 00 B 10 81 H2 200 CO2 44 00 H20 18 00 Note The relative mass of each substance is characteristic of the substance in the same way that the absolute mass is characteristic of the substance Realranging eq 1 CHE 115 Swichiometry o RMC Total number of atoms in the relative mass N A Relative mass of an element or compound constant N Absolute mass of an atom of the element or molecule of the compound A N A is de ned as the number of atoms of carbon in 120000 g of carbon12 isotope and that number called Avogadro s Number is 60221023 The quantity of material that contains Avogadro s Number 60221023 of particles is called a mole dozen 12 particles mole 6022391023 particles The weight of one mole of atoms is equal to the atomic weight AW or molar mass of the element Units gmole The AWs of the elements are listed in the periodic table These AWs are the relative masses of the elements The weight of one mole of molecules is equal to the molecular weight MW or molar mass of the compound Units gmole The molecular weight is equal to the sum of the atomic weights of all the atoms in the chemical formula for the compound Example Calculate the MW of H20 MWHZO ZAWH AWO 20008 g 15999 g 18015 g mole mole mole The number of moles in a speci c weight wt of a substance is given by the equation L wt of substance g molar mass of substance gmole moles of a Example Find the number of moles in 1802 g of H20 wto HO mole H20 01000moleH20 E Chemical Formulas CHE 115 Swichiometry l Covalent Molecular Compounds The valence is de ned as the combining capacity of the atom Compound Atom Valence HCl HCl H 1 Cl 1 Cl BCl3 ClBCl B 3 H20 OH O 2 H Group in Periodic Table Common Valence 2 l3 14 15 l6 l7 t NUJAUJNt n When writing the chemical formula for a covalent compound the valence of each atom must be satis ed Example Write the chemical formula for hydrogen sul de Element Group Valence H 2 Ionic Compounds Ionic compounds do NOT exist as molecules but rather as a collection of ions Positive ions are called cations and negative ions are called anions The chemical formula indicates the simplest ratio of the ions in the ionic compound Example Sodium chloride NaCl Na Cl 7 has one sodium cation for every chloride anion The molar mass of an ionic compound is called the formula weight FW units gmole See handout of Common Ions CHE 115 Swichiometry When writing the chemical formula for an ionic compound the total positive charge of the cations must equal in magnitude the total negative charge of the anions Example Write the chemical formula for lead hydroxide See handout of Common Ions Ions Charge Pb 2 OH39 l Pb 2 OH39 PbOHZ F Oxidation Number The oxidation number oxid no is the charge an atom appears to have in a compound Rules for Determining Oxidation Numbers 1 In free elements the oxidation number of each atom is zero Example In C12 the oxidation number of each Cl atom 0 2 In simple ions the oxidation number is equal to charge on the ion Example For Ca2 the oxidation number of Ca 2 3 In most compounds that contain oxygen the oxidation number of each oxygen atom is 2 Example In H20 oxidation number of O 2 Exception In peroxides OO the oxidation number of each oxygen atom is 1 Example In HOOH oxidation number of each oxygen atom l 4 In most compounds that contain hydrogen the oxidation number of each hydrogen is 1 Example In H20 oxidation of each hydrogen atom 1 Exception In metal hydrides MH the oxidation number of each hydrogen atom is Example In NaH the oxidation number of H l 5 All oxidation numbers must be consistent with the conservation of charge In neutral molecules the sum of the oxidation numbers of all the atoms must equal zero In a complex ion an ion with two or more atoms the sum of the oxidation numbers of all the atoms must equal the charge on the ion Example H20 CHE 115 Swichiometry 20xid no afH axid no of O 2l 2 0 Example Determine the oxidation number of the nitrogen atom in the nitrate ion NO3 39 axid no afN 30xid no of O axid no afN 32 l axid no afN 5 G Chemical Equations 2 H2 02 a 2 H20 The species on the le side of the arrow are called the reactants and the species on the right side are called products The numbers to the left of the chemical formulas are called stoichz39ometrz39c coef cients For a chemical equation to be valid it must satisfy the i Law of Conservation of Mass ii Law of Conservation of Charge iii and experimental fact When criteria i and ii are satisfied the chemical equation is said to be balanced Chemical equations for acidbase precipitation and complexformation reactions may be balanced by inspection Example Write a balanced chemical equation for the precipitation of lead sulfate from an aqueous solution of lead nitrate and sodium sulfate lead nitrate Pb 2 NOg39 PbNOZ lead sulfate Pb 504239 PbSO4 sodium sulfate 2Na 504239 NaZSO4 sodium nitrateNaNO339 NaNO3 PbNOZ aq NaZSO4 aq v PbSO4 s NaNO3 aq Where g vapor state 1 liquid state s solid state and aq aqueous solution PbNOZaq NaZSO4aq a PbSO4s 2NaNOaq Ionic compounds in aqueous solutions exist as dissociated ions ie PbNO32 s excess water a Pb aq 2 N0339 aq HO PbNoa2 s 2 mm 2 N03 aq CHE 115 swichiometry Thus the above chemical equation can be written as Pb2aq 2NO339aq 2Naaq 804239 aq a PbSO4s 2N0339aq 2Naaq The net ionic equation includes only those ions involved in the reaction for the precipitation reaction is Pb2aq s042 aq a PbSO4s H OxidationReduction Reactions Oxidation refers to a chemical change in which there is an increase in an oxidation number ie electrons are lost Reduction refers to a chemical change in which there is a decrease in an oxidation number ie electrons are gained An example of an oxidationreduction reaction Oxidation oxid no am 39gt 393 1 0 42 12 per CZHS 02 gt C02 H20 Equation not ba d lance Reduction To balance a chemical equation for a simple oxidationreduction reaction assign a stoichiometric coefficient of one to the substance with the largest number of atoms and then assign the remaining coefficients so that the equation satis es the Laws of Conservation of Mass and Charge Example Balance the above chemical equation for the combustion of C2H6 lCZH6 02 a C02 HZO 1C2H6 02 a gcozgHzo 1czH6 72oZ a 2coZ 3 H20 In some cases it will be advantageous to express the stoichiometric coefficients as integers Multiply the coefficients in the above equation by 2 2CZH6 702 a 4coZ 6 H20 CHE 115 Swlchlometry I Stoichiometric Calculations Example How many grams of N2 and H2 are needed to prepare 340 g of NH3 Unknown Wt of H2 and Wt of NZ needed to prepare 340 g ofN39H3 Knowns Wt ofN39H3 340 g H2 and NZ react to form NH3 Concepts stoichiometry moles MW balanced chemical equations Step 1 Write a balanced chemical equation N2 HZ NH3 An oxidationreduction reaction HZ 1NH3 NZ HZ 1NH3 N2 3 HZ 2 NH3 Step 2 Find moles of the given substances knowns wIafNH3 340g MW 170 g NH 3 mole mole NH3 200 mole NH3 Step 3 Find moles of the sought substances with the aid of an appropriate mole ratio and the answers in Step 2 The mole ratio is determined from the stoichiometric coef cients or subscripts in a chemical formula Units moles N2 on left side of equal sign are the same as units on the right Shh Stalchlometnc ooef clerrt f mole N 1 mole N 2 2 mole N2 200 mole NH3 mole NHs 200 mole NH3 2 mole NH3 100 mole N2 4 I Y Stuchlometnc coef clent M 0 e Ratio 3 mole H2 mole H2 200 mole NH3 2 mole NH3 300 mole H2 Step 4 Express answer in desired units 10 CHE 115 swichiometry wt N2 male N2MWNZ 100 mole N2 280 g 280g N2 mole wt H2 male H2MWHZ 300 mole Hz 20 g 600g H2 mole Example Limiting Reactant Problem Find the maximum weight of nitric oxide NO that can be prepared from 680 g ofNH3 and 128 g of 02 Unknown maximum Wt of NO that can be prepared Knowns Wt ofN39H3 680 g Wt of OZ 128 g and NH3 and O2 react to form NO and H20 Concepts mole ratio moles MW balanced chemical equations Step 1 N39H3 02 a NO HZO 1NH3OZ a 1NO 32HZO 1NH3 540Z a 1N0 32HZO 4NH3 502 a 4N0 6HZO Step 2 wt NH male Ast 3 400m0le Ast MWNH 170 male wt 0 male 02 2 400m0le O2 MWO 320 male Step 3 One of the reactants is limiting ie it is the reactant that is completely consumed and thus the moles of the limiting reactant determine the maximum number of moles of product that can be produced To determine which reactant is the limiting reactant consider all cases Case 1 Assume that NH3 is the limiting reactant ie the 400 moles ofN39H3 are consumed 4 male NO 4 male NH3 male N0 400m0le NHS 400m0le NO Case H Assume that 02 is the limiting reactant ie the 400 moles of 02 are consumed 4 male NO 5 male 02 male N0 400m0le OZ 320m0le NO The case with the smaller or smallest number of moles of product is the correct case In this example Case II is the correct case Since NH3 is not the limiting reactant how many moles of NH3 will remain after the 02 has been 11 CHE 115 swichiometry consum e d Step 4 thO mole NOMWNO 320mole NO300 g 960g NO mole The empirical formula is the simplest integral ratio in which the atoms combine molecular formula empirical formula where i is an integer Example Molecular Formula Empirical Formula i CZH6 H3 2 H6B3N3 HZBN 3 C6H6 CH 6 where MW is the molecular weight of the compound and FW is the formula weight for the empirical formula Procedure for Determining Empirical Formula Step 1 Final moles of substances formed from given sample of the compound Step 2 Final moles of each element in the given sample of the compound Step 3 Obtain ratios of the moles of each element to the element with the smallest number of moles If these ratios are not roughly whole numbers multiply them by an integer such that the products are whole numbers Example The composition of a compound that contains phosphorus and sulfur is found to be 436 P and 564 S by weight What is the empirical formula of the compound Unknown empirical formula of the compoun d Knowns compound is 436 phosphorus and 564 sulfur by weight Assume a 100 g sample of the compound Concepts stoichiometry moles molar mass empirical formula 12 CHEHS swichiometry Step 2 Assume 100 g of the compound 436 P wIP 100 mm le 436 P g p 100 g 564 S wIS 100 mm le 564 S g p 100 g male P 141m0le P and male S 176m0le S 310 g 320 g molz W712 Step 3 The element with the smaller number of moles is phosphorus 141m0le P 100 100x44 141m0leP M125 125x45 141m0leP PLOOSIJS P4S5 Example A 212 g sample of a compound that contained carbon hydrogen and oxygen was burned in excess oxygen The H20 that was produced weighed 1080 g and the CO2 weighed 2640 g The MW of the compound was determined to be 212 gmole What is the molecular formula of the compound Unknown molecular formula CKHYOZ ie what are the values for x y and z Knowns 212 g sample of CXHYOZ 1080 g of H20 and 2640 g of CO2 produced from 212 g sample of CXHYOZ MW of CKHYOZ is 212 gmole Concepts mole ratio molar mass moles empirical formula molecular formula CxHyOz 02 x CO2 H20 NOT BALANCED Step 1 2640g CO2 male CO2 0600m0le C02 4404 molz 1080g H20 male H20 0600m0le H20 180 molz Step 2 All of the carbon in the 212 g sample of CKHyOZ ends up in the 2640 g of C02 males of C in 2640g of CO2 males afC in 212g 0f CxHyOZ CHE 115 swichiometry males afC in 2640g of C02 0600m0le C0211 male C l 0600m0le C male CO2 moles of C in 212g 0f CXHyOZ 0600mole C All of the hydrogen in the 212 g sample of CXHYOZ ends up in the 1080 g of H20 males afH in 1080g afHZO males afH in 212g 0f CXHyOZ M 120m0leH males afH in 1080g afHZO 0600m0le H20 1 male H20 moles ofH in 212g 0f CXHyOZ 120mole H To find the moles of oxygen in the 212 g sample of CXHYOZ utilize the concept that the total equals the sum of the parts wt of sample of CXHyOZ wt C wt H wt 0 in the sample 212g afoHyOZ 0600m0le Cl201 g 120m0le H1008 g wIO mole mole wt 0 128g 0 moles ofolquot mg ofCXHyOZ M 0800moleO 1600 ml Step3 W 100 10003 3 0600m0le C M 200 20003 6 0600m0le C W 133 13303 4 0600m0le C CLOOHLOO 0133 CSHS 04 Empirical Formula malecularfarmula empi calfarmulal cm 115 Solid andLiquid States X Solid and Liquid States A Types of Solids l Crystalline Solid Crystals result from an orderly arrangement of atoms ions or molecules Crystals have planar surfaces that intersect at speci c angles Example A crystal of CsCl is cubic The unit cell smallest repeating unit in the crystal of CsCl is illustrated below Unit Cell Cubic The physical properties of crystals are anisotropic not the same in all directions Crystalline solids have sharp melting points For example the melting point of CsCl is 1290 C Several Types of Unit Cells The lengths of the edges of the unit cell are a b and c The angle between the b and c edges is called a the angle between the a and c edges is called B and the angle between the a and b edges is called 7 x o I b b I a Cubic Body Centered Cubic b abc Angle q s y 90 Tetragonal abfe Angloq Bv90 CHE 115 Solld and quuld Slates J b I Orthorhombic Monoclinic Triclinic I btc a b c ahbaec Angleu3vm Angleqy90 90 Angloq valso 2 Amorphous Solids Amorphous solids result from the unorderly arrangement of atoms ions or molecules Amorphous solids do not have characteristic shapes Example Window glass The physical properties of amorphous solids are isotropic same in all directions Amorphous solids melt over a broad temperature range For example Pyrex glass so ens at 800 C but does not completely liquefy until 1250 C B Phase Change 1 Sublimation Solid Vapor Molecules with KET kinetic energy of translation greater than the energy of the van der Waals forces in the solid will escape from the surface of the solid and enter the vapor phase When molecules that are in the vapor phase and have a KET less than the energy of the van der Waals forces in the solid collide with the surface of the solid the molecules will stick to the surface Vapor Phase gt39 O molecule I x Solid Phase CHE 115 Solid and Liquid Slates When the number of molecules leaving the solid and entering the vapor phase equals the number of molecules leaving the vapor and entering the solid phase then the system is in equilibrium solial vapor anal Gvapor Gsolid anal AGsub Gvapor Gsolid 0 The pressure exerted by the vapor that is in equilibrium with the solid is called the vapor pressure of the solid The magnitude of the vapor pressure depends on the magnitude of the van der Waals forces in the solid and the temperature Over a narrow temperature range the vapor pressure is related to temperature by the equation lnP AH 39TT39 subll 1nP TT vapor R 2 1 1 vapor2 where AHsub the enthalpy of sublimation is the amount of heat required to convert one mole of a solid into a vapor PWW2 is the vapor pressure of the solid at temperature T2 and Pvmr1 is the vapor pressure of the solid at temperature T1 At equilibrium AGsub AHsub TASsub 0 AHMb TASmb Dividing byT AHmb Assub T 2 Evaporation Liquid Vapor Molecules with KET greater than the energy of the van der Waals forces in the liquid will escape from the surface of the liquid and enter the vapor phase When molecules that are in the vapor phase and have a KET less than the energy of the van der Waals forces in the liquid collide with the surface of the liquid the molecules will stick to the surface CHE 115 Solid and Liquid Slates f A Liquid Phase When the number of molecules leaving the liquid and entering the vapor phase equals the number of molecules leaving the vapor and entering the liquid phase then the system is in equilibrium liquid vapor and Gvapor Gliquid and Acvap Gvapor Gliquid The pressure exerted by the vapor that is in equilibrium with the liquid is called the vapor pressure of the liquid The magnitude of the vapor pressure depends on the magnitude of the van der Waals forces in the liquid and the temperature Over a narrow temperature range the vapor pressure is related to temperature by the equation T2 T1 nP R TZT1 nP V P vapor1 AH vapor2 where AHWP the enthalpy of vaporization is the amount of heat required to convert one mole of a liquid into a vapor At equilibrium AG AH T AS 0 up vap vap v AH TAS vap vap Dividing by T Em us samanwu suus AHV As 39 Vnp T 3 Phase Diagram A whzch two Phase Diagram for H20 5000 mum livnr um smprualu i ulhbnum a cum 4mm 7 g 3mquot quul 9 is g i 3 m g 2000 a 39 i Vapor 100 ml arqu mum e 35 a a 39 J Wu Pam 3 saw a W s Tail i 750 o 50 100 150 200 250 Temperature lt2 Salient Features of a Phase Diagram 39 L 39 hi h i phases are 39 equilibrium The rzplepumt is the temperature and pressure at which all three phases are in equilibrium Vapo 39 39 Example The Vapor pressure ofwaler at we is 355 ton see above phase diagram for H20 i i i 4 5011171ngan 39 The L external pressure i apu pie uie nra39 A I CHE 115 Solid and Liquid States Example The boiling point of water at an external atmospheric pressure of 100 atm is 100 C If the external pressure is reduced to 355 torr then water will boil at 80 C see above phase diagram for H20 39 Melting points at speci c external pressures can be read from the phase diagram The melting point is the temperature at which liquid and solid coexist ie the temperature at which the vapor pressure of the solid equals the vapor pressure of the liquid Example The melting point of water at an external atmospheric pressure of 100 atm is 0 C If the external pressure is increased to 4636 torr then water will melt at 005 C see above phase diagram for H20 39 Points off the curves in the phase diagram indicate the temperatures and pressures at which only one phase exists C Types of Crystalline Solids 1 Molecular Crystals Component unit molecules Cohesive forces of attraction van der Waals forces Propelties so low melting points volatile nonconductors Example H3C2GaNCHCH32SnCH332 mp 151 C CHEHS Solid and Liquid States 2 Ionic Crystals Component unit ions Cohesive forces of attraction electrostatic attractions between the ions ionic bond Propelties hard and brittle high melting points nonvolatile nonconductors in the solid state Example SnO2 mp 1630 C 3 Covalent Crystals Component unit atoms Cohesive forces of attraction covalent bonds Propelties very hard very high melting points nonvolatile insulator Example Diamond mp 3500 C O 1 4 Metallic Crystals Component unit atoms Cohesive forces of attraction electrostatic attractions Propelties malleable and ductile high melting points nonvolatile good conductors Band Theory An Explanation of Bonding in Metallic Crystals 7 CHE 115 Gravirnetnc Analysis VIII Gravimetric Analysis Gravimetric analysis is the quantitative isolation of a substance by precipitation and the weighing of the precipitate Procedure Example Analysis of Ca2 39 Dissolve a weighed sample in water w Weighed Sample 6360J s Nazcos s gt 632 sq Na sq 003239 39 Precipitate the sought substance Ca2 with the addition of a precipitating reagent 020423 Caz aq C2042 aq CaCzO4HZO s 39 Filter dry and weigh the precipitate 39 Calculate moles of precipitate moles of sought substance and percentage of sought substance in the sample In a success ll gravimetric analysis 39 The sought substance must be completely isolated from the remainder of the sample To achieve this goal 0 Precipitates with very low solubilities KSp is very small are selected and CaCZOAHZO s 4 Caz aq i C2042 aq KSp 7 21 10 9 0 An excess of the precipitating reagent is added 39 The weighed precipitate must be a pure substance of known chemical composition 39 And the precipitate must be easily ltered A Mechanism for Precipitation 39 Nucleation Several ions of the precipitate come together to form a microsize particle called the nucleus 39 Growth The particle grows with the addition of ions of the precipitate until the system comes to equilibrium CHE 115 Gravimemc Analysts Primary adsorbed ions are ions of the precipitate that are adsorbed on the surface of the macrosize particle As a result of the adsorption of these ions the surface of the macrosize particle is charged Ions of opposite charge are attracted to the region of the solution surrounding the particle These ions are called counter ions B Factors Affecting Particle Size As the particle size of the precipitate increases the impurities decrease and the ease of ltration increases The particle size of the precipitate is inversely proportional to the relative supersaturation Q S S Relative Supersaturation where Q is the concentration of the solute in solution at the instant the reagents are mixed S is the concentration of the solute in a saturated solution a saturated solution is a solution in which the solute in solution is in equilibrium with the solid solute and Q S called supersaturation is the concentration of the solute in excess of that found in a saturated solution When 3 is large then the rate of nucleation is much greater than the rate of growth and the particle size of the precipitate is very small CHE 115 Gravimetnc Analysis E Techniques to minimize S 39 Precipitation from a very dilute solution Q is kept small 39 Slow addition of the precipitating reagent Q is kept small In Experiment 8 the precipitating reagent is slowly generated in situ H2 2C0 aq 1 ZHZO NI13aq 1 HCOSaq 1 NH aq urea HZCZO4 aq 1 2 NH3 aq 2NH4 aq 1 C2042 aq Oxalic acid oxalate ion precipitating reagent 39 Precipitation from a hot solution Initially S is kept large When Q is very small at the end of the precipitation S is reduced by cooling the solution 39 Precipitation from an acidic solution Initially S is kept large When Q is very small at the end of the precipitation S is reduced by neutralizing the solution C Calculations Just Stoichiometry 39 Final moles of precipitate moles of precipitate W Wprecipitafe 39 Final moles of sought substance wt of precipitate 39 moles o sou ht substance f g FW mole ratio precipitate 39 Final weight of sought substance FW m wt ofsought substance wlof precipitate mole ratio 39 quotL Fmem Gravimelric Factor 39 Calculate the percentage of the sample that is the sought substance CHE 115 ProblemSolving A General ProblemSolving Strategy For Chemistry 3394 De ne the Problem De ne the unknown and use an appropriate symbol to represent it List the explicit and implicit knowns and represent them with appropriate symbols List the chemical and physical concepts represented in the problem Make a model of the problem Sketch a pictorial or graphical representation of the problem Include unknown and knowns with appropriate symbols and units Make a qualitative estimate of the value for the unknown Devise a Plan for Solving the Problem To aid you in this endeavor consider the following questions What chemical or physical principles can be used to relate the various knowns and unknown Are these relationships appropriate for this problem If there is more than one unknown or if the above relationships generate additional unknowns what additional chemical or physical principles can be used to relate the various knowns and unknowns What other information is required to evaluate a certain quantity Can the problem be reduced to a series of simpler problems Are there any simplifying assumptions that can be used to solve the problem Is there an analogous or related problem for which the solution is known Carry out the Plan Write mathematical relationships in the general form algebraically rearrange and then substitute known numeric values into the general form Make sure the units on the right side of the mathematical equation are the same as the units on the left side units of quantity for which a value is to be calculated Write the name or symbol of the quantity for which the value was calculated and express the value in the correct units and significant figures Look Back Is the answer consistent in sign and magnitude with the qualitative estimate Were the calculations in the various parts of the solution done correctly Which chemical or physical principles were used to solve the problem Which chemical or physical principles were overlooked in your initial attempts Can you devise a alternate way to solve the problem CHE 115 ProblemSolving Example A container with a volume of 105 L has a mixture of two gases CH4 methane and 02 oxygen At 65 C the partial pressures of the methane and oxygen are 0175 and 0250 atm respectively Calculate the number of grams of each gas in the mixture De ne the Problem unkowns x wt ofCH4 and y wt ofO2 knowns Pmethane 0175 atm Poxygen 0250 atm V 105 L T 65 C 273 Assume that CH4 and 02 are ideal gases concepts Ideal Gas Law PV nRT Dalton s Law of Partial Pressures PT P methane Poxygen mole wtMW OCH4 02 Partial Pressure of CH4 PCH4 0175 atm V 105 L at T 338 K Partial Pressure of 02 P02 0250 atm estimate Since Poxygen gt Pmethane at the same conditions V and T then noxygen gt merm Since noxygen gt nmeame and MWoxygen gt MWmemm then y gt X Devz39se a Plan relationships moles of CH4 nmethane xMWmethane and moles of O2 noxygen yMWoxygen PT Pmemm t Poxygen methane t noxygenRT V P nmemmRT V VMWmathRT V methane CHE 115 ProblemSolving Poxygen noxygmRT V y MWoxygenXRT V Carry out the Plan X PmemmeMWmemmVy RT X 0175 atml60 gmole105 L00821 LatmmoleK338 K X 106 g ofCH4 y PoxygenMWoxygenVy RT y 0250 atm320 gmole105 L00821 LatmmoleK338 K y 303 g of 02 Look Back Is the answer consistent in sign and magnitude with the qualitative estimate Yes Were the calculations in the various parts of the solution done correctly Yes Which chemical or physical principles were used to solve the problem Ideal Gas Law Dalton s Law of Partial Pressures mole wtMW Which chemical or physical principles were overlooked in your initial attempts Dalton s Law of Partial Pressures Can you devise a alternate way to solve the problem Yes How Can I Improve My ProblemSolving Skills 3395 Master the fundamentals of chemistry and look for relationships between the various chemical principles and concepts Not only must you be familiar with the fundamental principles and concepts but you must also be able to recall easily this information in order to solve the problem The retrieval of information is facilitated by the storage of quotchu quot of related ideas in your memory One purpose of the Look Back step in the problemsolving strategy is to help you to identify related chemical principles and concepts Take time to study the processes that are used by you and others to solve problems Take advantage of the Look Back step and review the process that you used to obtain your answer Compare your process with those of others How do the processes differ Think aloud Whimbey and Lochhead have found that vocalizing your thoughts as you work through a problem helps you to organize and clarify your ideas and reduces the likelihood of careless errors They have suggested that students work in pairs to solve problems One partner describes how he or she would solve the problem while the other partner listens The listener contributes to the process by asking questions for the purpose of clarification If you prefer to work alone then CHE 115 ProblemSolving subvocalize or write down your thoughts as you solve a problem References 1 a Genyea J J Chem Educ 1983 60 478 b Mettes C T C W Pilot A Roossink H J KramersPals H J Chem Educ 1980 57 882 c Gendell J J Chem Educ 1987 64 523 d Gilbert G L J Chem Educ 1980 57 79 2 Polya G How to Solve It 2nd ed Princeton University Press Princeton NJ 1985 3 Woods D R In Developing Critical Thinking and Problem Solving Abilities Stice J EEd New Directions for Teaching and Leaming No 30 JosseyBass Inc San Francisco 1987 p 55 4 Leibold B G Moreland J L C Ross D C Butko J A Engineering Education 1976 67 172 5 a Rubinstein M F Firstenberg I R In Developing Critical Thinking and Problem Solving Abilities Stice J EEd New Directions for Teaching and Leaming No 30 JosseyBass Inc San Francisco 1987 p 23 b Rubinstein M F Patterns ofProblem Solving PrenticeHall Inc Englewood Cliffs NJ 1975 6 Woods D R Crowe C M Hoffman T W Wright J D Engineering Education 1979 70 277 7 Whimbey A Lochhead J Problem Solving and Comprehension 5th ed Lawrence Erlbaum Associates Hillsdale NJ 1991 W R Nutt 7 29 97 CHE 115 chemical Kineth XIII Chemical Kinetics CiNo2 g NO g No2 g ClNO g AHquot 173 k AGquot 232 Id How fast does the above reaction proceed spontaneously to the right To answer this question the rate of the reaction the change in concentration of a reactant or product per unit time must be measured The study of the rates and mechanisms of chemical reactions is called chemical kinetics In chemical kinetics reactions are classi ed into two groups Heterogeneous reactions are reactions that occur at the interface boundary between two phases Example Cs 02 g quot C02 9 This reaction occurs at the surface of the carbon Homogeneous reactions are reactions that occur in a single phase Example 2 NO g 02 g 2 N02 g This reaction occurs in the vapor phase A Factors Affecting Reaction Rates at Constant Temperature 1 Concentration of the Reactants in Homogeneous Reactions a Rate Law At constant temperature the rate of a homogeneous reaction is proportional to concentrations of the reactants raised to some power Consider the general reaction aA bB cC rate olt AXBy Replace ltgtlt with and the proportionality constant k CHE 115 Chemical Kineth I l u uuwnuuuu 4 7 l Rate Law rate kMJ39lBJV 5 Speci c rate constanl There is no necessary relationship between the exponents x anal y in the experimental rate law anal the stoichiometric coefficients a anal b in the balanced chemical equation for the OVERALL reaction For the general reaction aA bB cC Change in mnlnn39ty or c x Y1 d mtao onnation are I Change In time rate of consumption of A The term rate is de ned rate 1 1 1 kAxBy b alt a alt c alt stoichiometric coe icents The rate of formation of C is related to the rate of consumption of A by the mole ratio CHE 115 Chemical Kineth quotC 3 A ckiAriBiy k iAriBiy dt 6 dt where kquot ck and 3 is the mole ratio a Example Calculate the rate of formation of H1 H2 g 12 g quot 2 H1g if the experimental rate law is d1 rate dtz kH2Iz dHI 392 dUz d1 7 1 d1 2 sz b Order of the Reaction The order of a reaction is the sum of the exponents of the concentration terms in the rate law Example Determine the order of the reaction 2 N0 g 02 g quot 2 N02 g if the experimental rate is L70 kN0202 The reaction is third order overall second order in NO and rst order in 02 c Method of Initial Reaction Rates In this method the initial rate of the reaction is measured at or near time zero where the concentrations of all the reactants are known Example Consider the following gaseous reaction cm us mumm CINO2 g NO g N02 g CINOg Plot of Molarity ofCINO2 versus Time at 27 C 6675 Negahve a me slupe a me langem m me curve 5 m nea lms 15m sma mum me 2 21m rumLegit 5ee5 4e75 0 E 3e5 391 g Negawe unne swpc e A ch031 m ulmu 5 tan emm hecuwe ala mmmemme H e a s mmmes 1 2 v m mallLsec 2e 5 r 1amp5 D j u x x x D 2 4 6 8 10 12 14 16 Time sec A quot quot39 quot 39 quot lawandrate constant for the reaction Eup Icmommm no Mu dIIN02Idtvaa uw 1 5 w 22 w 2 2510quot 5 10quot 1410quot a uw 11104 Mw dClNOZ rate T IrClNOZ NOY the data for experiment 1 CHEHS Chemical Kinequ Mel k CINO2 r2 NO 1y2 ratel kC1N02 r1 NO 1y1 1110 5K 5 5 k2610 W5310 W 2210 5 k5310 5M5310 5W 56C Take the log of both sides of the equation xlllo si x 75 x 5 log logti xlog261075M y10g5310 M 2210 5 xk 5310 M sec 5 310 5M log050 0 xlog049 0 To determine the exponent y take the ratio of the rate data for experiment 3 to the data for experiment 1 5 4394 sec k5310 5M1110 4My 2210 5 k5310 5M5310 5My sec Take the log of both sides of the equation 544107554 k 55310 5M 51110 4M log 75M logh log 75 ylog 75 2210 x5310 M x5310 M log20 0 0 ylog21 030 094 1 y 032 CHE 115 Chemical Kineth Experimental rate law Liv kClNOz NO Rearranging the rate law and substituting the data from experiment 1 into the resulting expression gives dClNOZ dt CZNOZNO 5310 5M5310 5M 2210 5 SCC L mol sec k 78103 2 Surface Area of the Interface in Heterogeneous Reactions The rate of a heterogeneous reaction is proportional to area of the interface The surface area of a solid increases as the particle size of the solid decreases In reactions such as C S 02 g quot C02 g the rate increases as the particle size of the solid the carbon in the above example decreases B Theories of Reaction Rates and Temperature Dependency of k 1 Collision Theory Consider the binuclear reaction AgBg quot Cg Dg To form the products C and D molecules A and B must come together ie collide The total number of collisions per liter per second between 1 mole of A molecules and 1 mole ofB molecules at 25 C is Total Number of Collisions E 1034 L sec 1 L sec If a collision were the only requirement for the formation of products then all gaseous reactions would proceed at the same explosive rate Since all gaseous reactions do not occur at same rate at 25 C other criteria in addition to the collision of the reacting molecules A and B must be satis ed in order to achieve product formation cm us rimirm in l l quot CL A L l 39 l l These changes require energy Consequently ifthe collision is to result in products the en t i second andthefmction H 39 39 rnolecules is equal to or greater than Ea m ml numrm oolllslnns leon nlcolllslon Iitar 51 with snamy gt or EI ul wul lull pel lllel pel u 1 c u Il39d39 020 9 ul 0054 Fraction all Molecules Wlth Energles between E and E SE a 3 l 0 l a l 120 Jlmul 12 kJmolo 7 O 5 10 15 20 Translational Energy KJmol Consider the following cases at 298 K IfE for the reaction is 240 J tllen collisions between A and B rnolecules witll KEr kinetic energy oftranslation equal to or greater than 120 Jwill result in products since the percentage ofmolecule witll Klr e 120 Jis large see the diagrarn above the number ofcollisions per liter per second that will result CHEllS Chemlcal KlneLlCS in products is large and the rate of the reaction will be very fast If Ea for the reaction is 24 1d then collisions between A and B molecules with KET equal to or greater than 12 kJ will result in products Since the percentage of molecules with KET 2 12 k is very small see diagram above at 25 C the number of collisions per liter per second that will result in products is very small and the rate of the reaction will be very slow What effect will an increase in temperature have on the rate of the reaction If Ea for the reaction is 24 k and the temperature is increased from 298 to 600 K then the percentage of molecules with KET 2 12 kJ will increase ie the percentage of molecules with KET 2 12 k is larger at 600 K than at 298 K see diagram above An increase in the percentage of molecules with KET 2 12 kJ will result in an increase in the reaction rate The following equation exhibits the dependency of the rate on Ea and T outlined above Emmy ofAc va on 39 9 rate Total number of collisions exp E I liter sec R1 I I I I r f Ideal Gas Constant Absent Temperah 3314 June The rate of the reaction increases as the temperature increases and or the energy of activation decreases Not all collisions with energy 2 E21 result in products In addition to a combined energy 2 Ea the colliding molecules must also approach in the correct spatial orientation for bond forming and breaking Example H2 12 2HI H2 I2 gt H HI I gt 2HI Chemical Kineth CHEHS me x l f J sum haulIndian oflhl eolklumwlh th pmdudx can nplthl amm nn h bun 7 In If the rate is expressed as rate kNquotANquotB where Nquot A and Nquot B are the number of molecules of A and B in 1 liter then k is Wm 94 bra uquot tr The preeXponential term is constant over a short temperature range and is represented by 39 E k A 1 f E 5mm Eq 1 is called the Arrhenius equation If we take the natural logarithm of both sides of eq 1 then the Arrhenius equation has the form 111k E42 lnA R T 1 The plot of 111k versus 1 T will give a straight line with a slope of EaR Example ClNO2 g NO g NO2 g ClNO g unk mus HEus Plot of Inlk vesus 1 for the reaction INO2 g No a r N02 9 CINO g 11 7 since Aumkmum raxmzese w E 34 D Knazmlkm mnoom um mm mm ouazs aouzs onoso uuoai DOD32 00033 now 1mm Note The rate constants at temperatures T and T2 are related by the equation 2 Actiwted Complex Theory Henry Eyring n A u u in order for products to form As an illustration the addition R A Acc Chem Rex 1930 13 58 c A ethylene to form 11dichlorocyclopropane Moss CIZC HZCCHZ HZCCCIZCHZ CHEHS Enthalpy Diagram chemical Kineth C1697ll c140 Activated Complex 23603quot 5 19011 H l H H 1365 H H 4 39 Ch H H AlF 228kJ C 39C 39 Iquot 391 Cl H 1326KCH 39 l 1672R 39 Hr 5 E n 39 t l h AH2421kJ l 39 p quot CI y i Clansi 0 C o 1510P EDA 39 Hquot39C H H 151mi H Hp The enthalpy diagram illustrates the relative enthalpies of the reactants Hr the activated complex H and the product Hp The bond lengths angles and enthalpies inthe diagram are theoretical values that were obtain from molecular orbital calculations with the aid of the PC Spartan plus package As the dichlorocarbene approaches the ethylene molecule from above and forms the activated complex the carboncarbon bond in the ethylene lengthens and the carbon carbon bonds between the dichlorocarbene and ethylene begin to develop The diiTerence in the enthalpies of the reactants and activated complex is called the enthalpy of activation Em us amnan Emalpymmmnn AH H H For reactions in solution the enthapr ofactivation is related to the actiwtion energy by E AHRT k 203 10 1Iaeo KT lExp Rf El where Ae An ms 4 I J r E FluBIde EMIWIMM K Fukui A dlrE t A reactzhg speczes Wlth the luwest uhumupzed mulecular urbztal L UM O ah the uther reactzhg Speczes Fukux K Acc Chem Res 1971 4 Boundary Surfaces of the LUMO on the Dichlorocarbene HOMO on the Ethylene and the MO Resulting from the Overlap of the LUMO and HOMO in the Activated Complex umo 2p on the Diclhlnrnurhuu Running M0 on HOMO 1 M0 an the Ethylene mm C39m39P39 CHEHS Chemical Kineth As the dichlorocarbene approaches the ethylene and forms the activated complex the empty LUMO mostly the 2py AO on the carbon on the dichlorocarbene overlaps with the lled HOMO 7 M0 on the ethylene molecule and electron density ows from the ethylene to the dichorocarbene The ow of electron density away from the ethylene weakens the ethylene TE bond and initiates the development of CC bonds between the dichlorocarbene and the ethylene C Multistep Reactions Many reactions occur in a sequence of elementary reactions or steps Example 2 Br39 aq H202 aq 2 H aq Br2 aq 2 H20 aq 1 51 1quot Br39 H202 13927 Hos H10 nale r WSW H Br39 HOBr f wgt Br2 H20 ohmenhrynawazkst mhular Oman 2H ZBr39 rhea gt39 zuzo One step in the sequence is the slower or slowest step first step in the example above This step determines the rate of the overall reaction and is called the rate determining or ratelimiting step The ratedetermining step is the elementary reaction in the sequence of steps with the largest energy of activation Ea The HOBr in the sequence above is called an intermediate An intermediate is a species that is formed in one step elementary reaction and consumed in another ie it is neither a reactant nor a product The number of species that come together to form the activated complex in an elementary reaction is called the molecularity of the reaction Both steps in the example above are trimolecular ie three species come together to form the activated complex Reactions in which two species combine to form the activated complex are said to be bimolecular When the activated complex forms from one reactant the elementary reaction is said to be unimolecular D Reaction Mechanisms A reaction mechanism is the sequence of steps that leads to the overall reaction The two steps in the example above constitute the reaction mechanism for the formation of bromine The experimental rate law contains information concerning the nature of all the steps preceding and including the ratedetermining step If the exponents in the experimental rate law are NOT l3 CHEHS Chemical Kineth equal to the stoichiometric coef cients in the balanced chemical equation for the OVERALL reaction then the reaction will occur in a sequence of steps To obtain a possible mechanism for a chemical reaction Step 1 Write a sequence of steps that add to give the overall reaction and identify the rate determining step Step 2 Derive a rate law for the proposed mechanism Write down the rate law for the rate determining step If the rate law contains concentration terms for intermediates relate these concentrations terms to the concentrations of the reactants Note Exponents in the rate law for an ELEMENTARY reaction are equal to the stoichoimetric coefficients in the balanced chemical equation for the ELEMENTARY reaction Step 3 Compare the derived and experimental rate laws If the two rate laws are identical then the mechanism is possible provided that it is consistent with our knowledge of chemistry Example For the chemical reaction 2N205 2N204 02 the experimental rate law is d N2 04 dt kaisNzOs I Is the following mechanism possible N205 NO2 N03 fast equilibrium No2 N03 k gt NO 12 02 No3 slaw N0 No3 k2gt N204 fast Step 1 Given above The second step is the ratedetermining step Step 2 dN204l dt rate of rate determining step 7 k1NOZNO3 2 The concentrations NOZ and NO are related to the concentration of the reactant N205 through the equilibrium in the first step CHE 115 AnalyucalData VI Treatment of Analytical Data Types of Error Systematic Determinate errors effect each individual measurement in the same manner and o en arise from a defect in the analytical procedure or measuring instrument Example Use of uncorrected temperatures that were measured with the thermometer for Cup Z in Experiment 5 would lead to a systematic error in the heats of solution Random Indeteminate error affects each individual measurement in a random manner and results from uncertainties associated with every physical measurement Example Calibration of the 20 mL pipet un I II 111 Volume mL 1995 1997 1999 Variation in volumes results from the uncertainty in reading the position of the meniscus relative to the mark on the pipet Accuracy refers to the nearness of the arithmetic mean of a set of measurements to the true value Precision refers to the agreement of a set of measurements among themselves The precision of a set of measurements is indicated by the deviation of the individual measurements from the arithmetic mean deviation lxi El A verage Deviation 7 Zilei El N where Z is the summation over all N values N is the number of measurements Xi is the i 11 measurement 1 l is the absolute value of the quantity inside the vertical lines and ZNx x1xzxN x arithmetic mean N N Example Determine the average deviation for the volumes delivered by a 20 mL pipet Run I II 111 Volume mL 1995 1997 1999 CHE 115 AnalyucalData ENx x W 1997mL N 3 ENix E 7 I 7 11995 19971111997 319971111999 1997imL 7 0013M arithmetic mean i average deviation a 1997 i 001 mL Relative Average Deviation RAD RAD g 1000 units ppt parts per thousand x Example Determine the relative average deviation for the volumes delivered by a 20 mL pipet See data in previous example i 1000 7 1M 1000 7 065ppt x 1997mL RADi Sample Standard Deviation Example Determine the standard deviation for the volumes delivered by a 20 mL pipet See data in previous example 2 x E2 2 2 2 7 1 7 002mL 4100 i002mL 7 002m arithmetic mean i standard deviation a 1997 i 002 mL There is a 68 probability that the next volume delivered will fall in the range of 1995 to 1999 mL 1997 i 002 mL Signi cant Figures Exact numbers are numbers that describe indivisible objects such as people cars atoms and molecules CHE 115 AnalyucalData I nexact numbers express the results of measurements of divisible entities The number of digits in an inexact number should express the accuracy of the measuring instrument or the precision of a set of repetitive measurements These digits are called signi cant gures and include all gures that are known with certainty and the first uncertain gure Example Volume delivered by 20 mL pipet rst uncertain gure K YI i 199680 002 mL a k 4 Sig Figs Standard 5 Devratron r Drop remaining uncertain gures 1997 002 mL The standard deviation of 002 mL indicates that the fourth figure in the number 199680 is uncertain All gures to the right of this figure are dropped A zero is signi cant only if it is preceded by a number other than zero Example 000100 3 sig gs 10010 5 sig gs Use scienti c notation to avoid con lsion 100 10393 10010 102 Rules Addition or Subtraction The sum or diiTerence is rounded off at the last common gure that occurs farthest to the right in all the component numbers Example 20020 02 644 267 Multiplication or Division The product or quotient is rounded off in order that it contains the same number of signi cant gures as the component number with the smallest number 3
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