Chemistry 105 Chemistry 105
Popular in Chem 105: Principles of Chemistry
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This 12 page Study Guide was uploaded by Emma Silverman on Sunday March 8, 2015. The Study Guide belongs to Chemistry 105 at Washington State University taught by Finnegan in Spring2015. Since its upload, it has received 362 views. For similar materials see Chem 105: Principles of Chemistry in Chemistry at Washington State University.
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Date Created: 03/08/15
CHEM 105 Tobics for the second exam Spring 2015 g Laws all are variations on the ideal gas law PVnRT Standard temperature and pressure STP T0 degrees celsius 273K and P1atm gas law problems PVnRT or PVnTR where P is pressure V is volume n is of moles R is a constant and T is temperature in K R623637 LtorrmoK R00820575 LatmmoK Example problems 1 A sample of argon occupies 7532 mL at 367 C and 6338 torr How many moles of argon are present a PVRTn b 63387532623637309850247 mol 2 6258g of oxygen gas is sealed in a rigid container with a volume of 1765L The temperature is 35000 What is the pressure of the gas in atm a nRTVP b 6258321955625 mol of 02 gt n c 19556250820575352731517652802 atm Gas density Density MPRT where M is molar mass P is pressure R is a constant and T is temperature Example Problems 1 What is the density of sulfur dioxide gas at 555C and 1602 kPa Answer in units of gL A Gas constant of kPa 83144621 B Molar mass of sulfur dioxide 64066 C Density MPRT 64071581051082057532865 3756 gL 2 What is the density of neon gas at STP Express your answer in gL DO NOT include the units in your answer A 27315K B 1atm c DMPRT D 201810820575273159003gL Molar mass of an unknown gas Example problem 1 The mass of an evacuated 255mL flask is 143187 g The mass of the flask filled with 265 torr of an unknown gas at 25 degrees Celsius is 143289 g a start with these 2 equations PV nRT n mass mw b PV mass molar weight RT c molar weight mass x RT PV d molar weight 143289g 143187g x 008206 LatmmoleK x 298K 265 torr x 1atm760torr x 0255L e molar weight 280 gmole Gas stoichiometry reactions with gas phase reactants andor products 1 2KC03s gt 2KC s 302g If 9264 g of potassium chlorate are heated and the reaction goes to completion what volume of oxygen gas at 220C and 1055 atm is produced 391 gmol 3545 gmol 316 gmol 12255gmol 92641225532 mols of 02 n nRTPV 1 1339045290082057527315221 055 2603 L 2 4NH3g 502 g gt 4NOg 6HZO If 408 L of ammonia gas and 550 L of oxygen gas both at 250 C and 1100 atm react according to this equation what volume in L of NC at 250 C and 1100 atm will be produced Find limiting reactant gt PVRTn 408 L NH3 4L NO 4L NH3 408 L NO lt Limiting reactant Answer 550 L 02 4L NO5L 02 440 L NO 3 4NH3g 502 g gt 4NOg 6HZO If 408 L of ammonia gas and 550 L of oxygen gas both at 250 C and 1100 atm react according to this equation what is the total volume in L or all gasses at 250 C and 1100 atm at the end of the reaction Not going to be any ammonia we ll end up with 408 L of NO and also 02 408 L NH3 5L O24L NH3 510L 02 used 550L 510 L 40 L 02 remains 40 02 408 L N0 448 L Kinetic Molecular Theory explain the gas laws in terms of the motion of the gas molecules Kinetic Molecular Theory 58 Know the five assumptions of KMT Know the KMT defines pressure and temperature Know how KMT explains the gas laws KMT tells us that gas particles behave independently of each other They occasionally collide and exchange energy but that is it This is useful as it means that we can deal with each gas in a mixture separately We assume gases behave the way they do because there are no forces operating between the molecules billiard balls we say there are no forces fthere were no forces operating between them there would be no liquids or solids We can ignore the forces for gases because they have so much kinetic energy that we can ignore the forces in between them End of chapter 5 talks about the fact that there are forces and molecules that DO occupy volume Low temperatures actually have an effect of that Same with high pressures Every gas in a mixture of gases behaves independently as if the other gases weren t even there That means we don t have to consider the other gases when looking at one gas the relationship between energy and molecular speed The atoms in a sample of two different elements at the same temperature do not have the same average velocity If the element is lighter they move faster the relationship between energy and temperature The atoms in a sample of two different elements at the same temperature have the same average kinetic energy effusion and diffusion Graham s Law The ratio of effusion rates of two different gases rateArateBsqrtMbMa where Mb is the molar mass of gas B and MAis the molar mass of gas A rates velocities vs times Examples of problems 1 What is the relative rate of effusion for hydrogen gas compared to nitrogen gas a rH2rN2 sqrt MmMHZ sqrt 28022016 3728 b Bigger molar mass goes on the top of the fraction 2 If 025 mol of hydrogen effuse from a container in 4632 s how much time would be required for 025 mol of nitrogen gas to effuse from the same container under identical conditions a tN2tH2 Sqrt MNZIMHZ b you need a bigger number because it s going to take N2 a longer amount of time than H2 c 3728 4632 1727 seconds Partial pressure vapor pressure collecting gas over water Example problem 1 All substances have a vapor pressure partial pressure above the solidliquid CaCO3s gt CaOsCOZg CaCO3 is heated and the CO2 is collected over water 3500mL of gas at 2500C and 6900 torr are collected What mass of CO2 is produced PH202378 torr from table 54 PCO2690O torr 2378 torr 66622 torr nPVRT 66622 torr3500mL1000LmL623637torrmoK29815K 055199 Energy units of energy We usually measure energy in Joules or kiloJoules the conservation of energy The system and the system s surroundings Energy is transferred between the system and its surroundings If it loses energy the system gains that amount of energy and vise versa heat and work Heat is called q and work is called w AEE E w products reactants q gheat positive if the system gains thermal energy negative if the system loses thermal energy wwork positive if work is done on the system negative if they system does work E positive if energy flows into the system negative if energy flows out of the system PV work Pressure volume work lgt WPAV Example problem book 1 To inflate a balloon you must do pressurevolume work on the surroundings If you inflate a balloon from a volume of 0100L to 185L against an external pressure of 100 atm how much work is done in joules a AVV2V1 175L wPAV 100amt175L 175Latm1013J1Latm177J heat capacity When a system absorbs heat q its temperature changes by AT Specific heat Specific heat capacity is measured in Jg C This is a specific number that is usually given 995 Enthalpy definition book The value of AH for a chemical reaction is the amount of heat absorbed or evolved in the reaction under conditions of constant pressure endothermic and exothermic book An endothermic reaction has a positive AH and absorbs heat from the surroundings An endothermic reaction feels cold to the touch An exothermic reaction has a negative AH and gives off heat to the surroundings An exothermic reaction feels warm to the touch Example problems 1 BZH6g 6Cl2g gt ZBCI3g 6HClg a AH 7554kJ b How much heat is produced when 1250 g of chlorine react with excess diborane c 1250g Cl2 1molCl27090g7554kJ6molCl2 2220 kJ Ignored negative sign because they wanted to know about energy 2 2 CZH6 g 7 02 g gt 4 CO2 g 6 H20 g AH285558kJ a What mass of ethane must be burned to heat 5500L of water from 2250 C to 5500 C Assume no loss of energy in the transfer b The specific heat capacity of water is 4184 Jg C The density of water is 1000 gmL c 4184550005522574789kJ d 74789kJ2mol285558kJ300681mol 15749905g Thermochemical equations Enthalpies of of reaction Aern AHmnZAHfproducts ZAHfreactants Example problem 1 Use the standard enthalpies of formation to determine the value of Aern for a C3H60 I 4 02 g gt 3 002 g 3 H20 g b Products Reactants c 33935 3 2418 2484 40 16575kJ Calorimetry bomb calorimeters A bomb calorimeter measures changes in internal energy for combustion reactions qcalCcal qcalqrxn Example problem book 1 When 1010g of sucrose undergoes combustion in a bomb calorimeter the temperature rises from 2492 C to 2833 C Find AErxn for the combustion of sucrose in kJmol sucrose The heat capacity of the bomb calorimeter determined in a separate experiment is 490kJ C a ATTfTi 28332492341 b qcalCcalAT c qcal490kJ C341 C167kJ d qrxnqcal167kJ e AErxnqrxnlmolC12H22011566x103lemolC12H22011 coffee cup calorimeters A coffee cup calorimeter measures enthalpy changes for chemical reactions in solution Example problem book 1 Magnesium metal reacts with hydrochloric acid according to the balanced equation Mgs 2Hlaq gt MgCl2aq H2g In an experiment to determine the enthalpy change for this reaction 0158g of Mg metal is combined with enough HCI to make 1000mL of slution in a coffeecup calorimeter The HCI is sufficiently concentrated so that the Mg completely reacts The temperature of the solution rises from 256 C to 328 C as a result of the reaction Find Aern for the reaction as written Use 100gmL as the density of the solution and C 418Jg C as the specific heat capacity of ssolution the solution a 08 solu on 418Jg C msoln1000ml soln100g1mLsoln 100x102g b ATTfTi 328 25672 C c qsolnmsolnCssolnAT 100x102g418Jg C72 C3Ox1O3J d qrxnqsoln3OX1O3J e AernqrxnlmolMg 3Ox1O3JO158gMg1mol2431gMg46x1O5JmolMg f 1 mol gt Aern46x1O3J Hess s Law Standard molar enthalpies of formation The change in enthalpy for a stepwise process is the sum of the enthalpy changes of the steps The waveparticle duality of light kvc Chapter 7 To understand the quantummechanical model of an atom we have ot know about electromagnetic radiation Light behaves as a wave It has frequency v and a wavelength k that are related by its speed pp 296298 c299792458x108ms Ehv where h 6626x103934Js Atomic spectra line spectra The Bohr model of the atom absorption vs emission relative energies of the levels Atomic line spectra spectrum 73 Atoms and monatomic ions can absorb and emit light This energy is emitted or absorbed by the electrons in the atomion Only specific energies frequencies of photons can be emitted or absorbed because the electrons are restricted to certain energy levels The energy of an electron in an atom is quantized The energy levels in hydrogen could be fit to an equation Ryberg equation on page 307 Later it was found that these equations could be extended to predict the spectra of any oneelectron ion He U 07 Na etc But multielectron systems produce much more complicate spectra Niels Bohr realized that the spectrum of hydrogen could be explained if the electrons were restricted to specific orbits specific distances from the nucleus The lines seen corresponded to the energy differences between these allowed orbit distances This is illustrated in Figure 712 but it is misleading the energy levels get closer together as n increases E 2178x10quot8J1n2 n is a positive integer What Bohr could not explain was why the electrons would be restricted to these specific orbits The emission spectra consists of bright lines as specific wavelengths with darkness in between The result of light emitted is a bright series of lines When an atom absorbs energy it often reemits that energy as light 306 Example problem Which of the following electron transitions correspond to the emission of light Learning Catalytics Rank these electron transitions in order of increasing energy n4 gt n5 n1 gt n3 n1 gt n2 n2 gt n3 n4 gt n5 lt n2 gt n3 lt n1 gt n2 lt n1 gt n3 Waveparticle duality of matter de Broglie s hypothesis khmv how electrons as waves explain energy quantization in an atom Louis de Broglie if light could be both a wave and a particle then particles such as electrons might also be waves where m is the mass of the particle in kg and v is the speed of the particle in ms The quantum mechanical atom wave functions electron clouds and orbitals quantum numbers n l ml ms orbital shapes 5 p d f shells and subshells orbital energy levels in multielectron atoms shielding and effective nuclear charge orbital penetration zsin1Tr18OkGn where x is the amplitude and G is the angle in degrees There is more than one wave but each one has the same equation with a different value of n n is a quantum number n123 Note that the number of nodes z0 increases as n increases The Schrodinger equation defines the condition that an electron wave must meet in order to be stable inside an atom A solution to the Schrodinger equation is called a wave function Lp a mathematical description of the electron wave Each wave function represents an atomic orbital an allowed energy level of the electron Mathematically all wave functions are quite similar They differ only in a set of coefficients for the parts of the equation These coefficients are called quantum numbers A set of three quantum numbers defines a specific atomic orbital n associated with the shell or level related to the size of the orbital n is always positive I l determines the subshell sublevel l is related to the shape of the orbital I may be any integer from O to n1 sorbitals have lo porbitals have l1 dorbitals have l2 f orbitals have l3 ml specifies the orientation of the orbital ml is an integer between I and l ms the spin quantum number either 12 or 12 Electron configurations ground state electron configurations of atoms 3 block elements are in groups 1A and 18 as well as helium p block elements are in groups 3A8A besides helium d block elements are the transition metals and the coefficient is one less than the period fblock elements are the lanthanides and actinides the coefficient is two less than the period full doing a full ground state electron configuration means you start with 1s Hydrogen would be 131 because it is the first 13 element Each 3 has 2 exponents p has 6 d has 10 and f has 14 If you want to known the electron configuration of Cl you would do 1322322p63323p5 This is because Chlorine is the 5th element in period 3 and in the p block elements A good way to check if you did it correctly is to count the exponents The number should equal the atomic number of the element noble gas notation The noble gas notation is a shortcut You put the noble gas just before it on the periodic table and only write the electron configuration of everything following For Chlorine it would be Ne3s23p5 orbital box diagrams This is the same idea but instead of exponents there are boxes Each box gets an up arrow and a down arrow This chart should help B C N I 152 252 irregularities in ground state electron configurations of transition metals Anomalous electron configurations Full sublevels subshells s p d f are more stable lower energy than partially filled sublevels Halffilled sublevels are more stable lower energy than any other partially filled configuration This explains some anomalies in the ground state electron configurations CrAr4s 3d5 rather than Ar4s23d4 CuAr4s 3d1O rather than Ar4s23d9 You should be aware that such anomalies exist You DO NOT need to memorize them No exam questions on them excited state electron configurations Any other configuration of the electrons would be an excited state 3Which of the following orbital diagrams represents an excited state of the hydrogen atom 1 15 15 15 25 lt this one paramagnetic vs diamagnetic If an atom or ion has unpaired electrons it is paramagnetic If all the electrons are paired the atom or ion is diamagnetic Example problems 1 Classify the ground state of sulfur as paramagnetic or diamagnetic a Paramagnetic 2 Classify the ground state of zinc as paramagnetic or diamagnetic a Diamagnetic Valence electrons vs core electrons valence shell Valence Electron electrons that participate in bonding and ion formation These are the electrons in the outermost shell and any electrons in incomplete d or f subshells Core Electron the electrons that occupy orbitals that were full in the previous noble gas atom plus any filled d and fsubshells
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