Chem 131, Study guide pt.2
Chem 131, Study guide pt.2 Chem 131
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This 13 page Study Guide was uploaded by Daria Hebron on Monday May 9, 2016. The Study Guide belongs to Chem 131 at Towson University taught by Sonali Rosa in Spring 2016. Since its upload, it has received 40 views. For similar materials see Chemistry in Biology at Towson University.
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Date Created: 05/09/16
Study Guide –Chemistry Part 2 Chapter 610, 21 Chapter 6: Energy in Chemical Reactions Definitions and Terms: Energy: capacity to do work Radiant energy: energy comes from the sun Thermal energy: energy associated with random motion Chemical energy: is stored within the structural units of chemical substances Work : directed energy change from a process Potential energy : energy available by virtue of an objects’ position Law of Conversation: the total quantity of energy in the universe is assumed constant Energy Changes in Chemical Reactions: Exothermic process is a process that gives off heat Endothermic process is a process that absorbs heat Heat is both absorbed and release from the system Equations: ∆U sys ∆ U surr ∆U= U(products)U(reactants) S (s) + O2 (g) →SO (2 The product is 1 mol SO 2 energy content of (1 mol S+ O ),2the reactants ∆U= q+w W=fd forcexdistance W= P∆ V P x V= F/d x d =Fd=w Examples: Find the work done by the gas if it expands against a constant pressure of 2.9 atm. The gas expands in volume from the 5.0L to 8.0L at a constant pressure temperature. Step 1. Figure out what equation to use. You need this w= P∆ V Step 2. W= P∆V p= (1.2atm) → atm is the units for pressure ∆V = the final volume – the initial volume W= (1.2atm)(8.0L5.0L)= 3.6atm. Step3. Convert atm to joules by: . . 3.6L atm x 101.3J/ 1L atm = 3.64x10 2 Example: Calculate the energy change of the gas if the work done in a cylinder is 485 J. The heat transfer of is 145 J from the gas to the surroundings. ∆U=q+w →145J+ 485J *the negative sign indicates the heat is remove from the system to the surroundings. The answer is the energy of the gas increased by 340J Example: 3 o A sample of gas occupies 2.12 x 10 mL at 32 C and 840 mmHg. What is the volume will it occupy at the same temperature and 320 mmHg? The temp is unchanged * P 1 =1 V 2 2 3 P 1=0 mmHg V =2.12x11 P =320mmHg 2 =? 2 3 3 V 2 P 1 x 1 2→ (840mmHg x 2.12 x 10 mL )/320mmHg= 5.56 x 10 Example problem of thermochemical equation: Calculate the heat evolved when 156 g of white phosphorus burns in air according to the equation: P 4s) + 5O (g2 →P O (s4 10 ∆H= 3013kJ/mol Step 1: Find the molar mass of the P . Which is 60g 4 Step 2: Convert to moles 266g x 1mol /60.0g= 4.43mol P 4 Step 3: Convert mole to kJ/mol 4.43mol of P x43013kJ/mol = 0.00147kJ Calorimetry : C=ms mass x specific heat q=ms∆t . o o (32.0g)(5.163J/g C)= 165.2J/ C Chapter 7 Definitions and Terms: Heisenburg uncertainty principle: states it is impossible to know simultaneously both the momentum (mass times velocity) and the position of a particle with certainty. Orbitals Angular Momentum Quantum Number s 0 p 1 d 2 f 3 Each orbital can have a maximum of 2 electrons. If L = 0 then m= 0. If n=1 then [(2x1)+1] → use this equation to find the magnetic quantum l number. Atomic orbitals: s orbital = spherical shape p orbital = dumb bell shape d orbital = clover leaf shape f orbital = double clover List the values of n, l, and m for orbitals in the 3d subshell. n=3 l=2 m l=2, 1,0,1,2 List the total number of orbitals associated with n=2. Step 1. Figure out the l values for the n value. 2s (n=2, l =0, m=l) 2p (n=2, l=1, m = l,0, 1) Step 2. Determine the ml values. The total number of orbitals equal to the sum of all the m l value. =4 →n =4 Electron Configuration 2 2 6 2 1 Examples : Aluminum 13 1s , 2s , 2p , 3s , 3p Electrons n l m l m s 1 1 0 0 +1/2 2 1 0 0 1/2 3 2 0 0 +1/2 4 2 0 0 1/2 5 2 1 1 +1/2 6 2 1 0 1/2 7 2 1 1 +1/2 8 2 1 1 1/2 9 2 1 0 +1/2 10 2 1 1 1/2 11 3 0 0 +1/2 12 3 0 0 1/2 13 3 1 1 +1/2 Pauli Exclusion Principle: 7N= 1s , 2s , 2p ↑ ↑ ↑ ↑ ↑ ↓ ↓ 1s 2s 2p Diamagnetic: do not contain net unpaired spins and repelled by a magnet Paramagnetic : do contain a net unpaired spins and are attracted by a magnet Wavelength wave: is a vibrating disturbance by which energy is transmitted wavelength: distance between identical points on successive waves frequency (nu or v): number of waves that pass through a particular point in one second amplitude : the vertical distance from the midline of a wave to the peak light: electronmagnetic radiation, has electrical component and a perpendicular magnetic Equations: 8 c=λ v c=3.00 x 10 m/s λ=wavelength v= frequency c= speed Problem: What is the frequency of the radiation if the wavelength of the light is centered at 587nm? v= c/λ Make sure to convert nanometers to meters if asked. 587nm x 1x 10 m= 5.87x10 m 7 8 7 14 v=3.00x10 /5.87x10 m= 5.11x10 Hz Equation: E(energy)= h(planck’s constant6.63x10 ) v(frequency) or E=h (c/λ) Problem: Calculate the energy in joules of a photon with a wavelength of 5.20x10 nm. 5 1 step: convert the wavelength from nanometers to meters. 5 9 4 5.20x10 nm x 1x10 m = 5.20x10 m (6.63x10^34)(3.00x10^8)/5.20x10^4= 3.82x10^22 Equation KE=hvW(work) Problem: The work function of Mg is 6.25 x 10^15.(a) Calculate the minimum frequency. (b) Calculate the kinetic energy if the frequency is 3.00x10^16. 1. Set KE=0. Now equation is w=hv w(6.25x10^15)=h(6.63x10^34)( v=?) v=(6.25x10^15)/ (6.63x10^34)= v =9.42x10^18 2. KE=h(6.63x10^34)v(3.00x10^16)w(5.86x10^19)= 1.93x10^17 Equation: 2 18 2 2 E= ─R (1Hn ) R = 2H18x10 OR ∆E=hv=R (1/nH 1/n i) f Problem: What is the wavelength in nanometers of a photon emitted during a transition from n=8 state to i nf4 st 18 2 2 19 1 step: ∆E= R 2.1H 0 1/8 1/4 )=1.02x10 ∆E= 1.02x10 19 2 step: ch/∆E ( 3.00x10 )(6.63x10 )/1.02x10 =1.95x10 19 6 Equation: λ=h/mu m=mass u=velocity (m/s) Problem: Calculate the wavelength of the “particle” of 5.50x10^2 tennis ball traveling at 55m/s. λ= 6.63x10^34 /(5.50x10^2 x 55m/s)= 2.19x10^34 Chapter 8 Definitions: Representative elements: main elements Groups 1A7A Valence electrons: outer electrons of an atom, which are involve in chemical bonding Paramagnetic : two unpaired electrons Diamagnetic : no two unpaired electrons 1 + Na:[Ne] 3s Na :[Ne] Ca:[Ar]4s Ca :[Ar] 2 1 3+ Al:[Ne]3s 3p Al :[Ne] 1 2 H:1s H:1s or [He] F:1s 2s 2p 5 F:1s 2s 2p or [Ne]→ all of them is isoelectronic, have same number of electrons 2 2 4 2 2 2 6 O:1s 2s 2p O :1s 2s 2p or [Ne] N:1s 2s 2p 3 N : 1s 2s 2p or [Ne] Atomic Radius: is onehalf the distance between the two nuclei in two adjacent metal atoms Decreases from left to right Increases down the periodic table Ionic radius: the radius of a cation or anion. If the atom forms an anion, its size (radius) increases. Problem: Which one is larger? O or F O 2 Sr or Ca 2+ Sr2+ 3+ 2+ 2+ Co or Co Co Li or K + + 3+ Au or Au N or P 3 Ionization Energy: minimum energy required to remove an electron from a gaseous atom in its ground state. Increases left to right Decreases top to bottom Electron affinity (EA): the negative of the energy change that occurs when an electron is accepted by an atom in gaseous state to form an anion Electronegativity : ability of an atom to attract toward itself the electrons in a chemical bond Increases left to right Decreases top to bottom Chapter 9 Definitions: Ionic bond: the electrostatic force that holds ions together in an ionic compound Covalent bond: two electrons are shared by two atoms Bond length: the distance between the nuclei of two covalently bonded atoms in a molecule Polar covalent bond: when electrons spend more time in the vicinity of one atom than the other Problem: Identify the following bonds as, ionic, polar convalent, or convalent: HCl= polar covalent KFionic Writing Lewis Structure: Rules: 1. Identify the least electrons and make in central atom 2. Determine total valence electrons 3. Connect all other elements directly the central atom 4. Complete octets on outer elements first 5. Count how many electrons you need 6. If you required add lone pairs to the central atoms make double or triple bonds 7. Count electrons again + 8. For cations, subtract the number of positive charge from the total (NH ) and 4or anions , add the number of negative charges to the total (CO ) 32 Formal Charge To identify the formal charge, count the amount of electrons around each element. For example, + NH ,3the lewis structure will be Nitrogen as the central atoms. The hydrogen atoms will be single bonded to nitrogen. The nitrogen will have one lone pair as well. The hydrogen are supposed to have 1 electron assign to it and the valence electron is one too, so the overall charge of all of the hydrogens is zero. For the central atom, there are 5 valence electrons that supposed to be assign to it. If you count the electrons on the lewis structure five electrons are assigned to nitrogen. 55=0. The formal charge of nitrogen is zero. *The cations formal charge must equal positive charge and the anions sum of formal charge must be equal to a negative charge. Resonance Structure: one of two or more lewis structures for a single molecule that cannot be represented accurately by only one lewis ( e.g O ) a3other version of the definition = equivalent structures formed form delocalization of electrons are resonance structures. Exceptions to the octet rule: Hydrogen can never be the central atom Carbon, oxygen, and nitrogen can form multiple bonds Bond Enthalpy : enthalpy change required to break a particular bond in 1 mole of gaseous molecules o ∆H =∑BE(reactants) ∑BE(products) Chapter 10 Molecular geometry : 3d arrangement of atoms in a molecule Valence shell : outermost electronoccupied shell of an atom, its holds the electrons that are usually involved in bonding. VSEPR ( valenceshell eletronpair repulsion )Model: geometric arrangements of electron pairs around the central atom Double, triple bonds are accounted for single bonds Class of # of #of #of lone Arrangemen Geometry examples molecule electron bonding pairs t of electron pairs pairs pairs AB 2 3 2 1 Trigonal bent SO 2 planar AB 3 4 3 1 tetrahedral Trigonal NH 3 pyramidal AB 2 2 4 2 2 tetrahedral bent H 2 AB 4 5 4 1 Trigonal Seesaw or SF 4 bipyramidal distorted tetrahedon AB 3 2 5 3 2 Trigonal tshaped CIF 3 bipyramidal AB E 5 2 3 Trigonal linear I 2 3 3 bipyramidal AB 5 6 5 1 octahedral Square BrF 5 pyramidal AB 4 2 6 4 2 octahedral Square XeF 4 planar Examples : AsH :3Lewis Structure model the As is the central atom. The 3 hydrogen atoms are single bonded to the central atom. As have one lone pair. All together, they are 4 electron pairs so the electron geometry is tetrahedral. The molecular geometry is trigonal pyramidal because the central atom has a lone pair. SF 6: The molecular geometry is octahedral because it has 6 electrons assign to the central atom. The central atom is Sulfur and the 6 fluoride atoms are single bonded around the central atom. Each fluoride atom completes the octet rule by adding 6 additional electrons. XeF T2: central atom is Xe. The (2) fluoride atoms are single bonded to the cental atom. The geometry is linear because the lewis structure illustrates all three elements all the same number of electrons. Dipole Moments : HF= you can’t see it but on the fluoride atom, there are more electrons on surrounding it, meaning its more electronegativity. This relationship is polarcovalent. BrCl=is polar so they have a dipole moment BF = is nonpolar molecular and has no dipole moment 3 Polar molecules : diatomic molecules containing atoms of different elements like HCl, CO, and NO have dipole moments. Nonpolar molecules: diatomic molecules containing atoms of the same elements like H O and 2, 2, F 2have dipole moments Hybridizations of Atomic Orbitals : Hybrid orbitals: atomic orbitals obtained when two or more nonequivalent orbitals of the same atom combine in preparation for covalent bond formation. Hybridization: mixing of atom orbitals in an atom, central atom mostly Examples : 3 NH :3hybridization is sp . How? Because electron configuration of only the central atom, nitrogen, is 1s ,2s ,2p . The sp orbitals form the bonds with H atoms and its complete filled ↑ ↑ ↑ ↑ ↑ ↓ ↓ 2 2 3 1s 2s 2p BeCl 2 The electron configuration is 1s , 2s . Since the 2p orbital is empty one electron (↑) will go to the p orbital. So the hybridization is sp ↑↓ ↑ 2 2 1 1s 2s 2p ↓ ↑ ↑ ↑ ↓ 2 1 1 1s 2s 2p =sp Pure atomic Hybridization of Number of Shape of hybrid examples orbitals of the the central atom hybrid orbitals orbitals central atom s,p sp 2 linear BeCl 2 2 s, p, p Sp 3 Trigonal planar BF 3 S,p, p, p Sp3 4 Tetrahedral CH 4+ S, p, p ,p, d Sp d 5 Trigonal PCl5 bipyramidal S,p, p, p, d, d Sp d2 6 octahedral SF6 Sigma bonds(σ bonds): covalent bonds formed by orbitals overlapping endtoend, with electron density concentrated between the nuclei of the 3bonding atoms Pi bond (π bond): covalent bond formed by sideways overlapping orbitals with electron density concentrated above and below the plane of the nuclei of the bonding atoms. Chapter 21: Proton: 1H electron: 1 0 positron: +1ᵦ a particle: 2e or a42 Examples : 56 4 52 25Mn → a 2x = V 23 0 14 14 X → 1 + 7 = C6
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