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Chem 113 Exam 1 Study Guide

by: Caroline Hurlbut

Chem 113 Exam 1 Study Guide Chem 113

Marketplace > Colorado State University > Chemistry > Chem 113 > Chem 113 Exam 1 Study Guide
Caroline Hurlbut
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This study guide covers all lecture material leading up to the exam, including the last two lectures before the exam.
General Chemistry II
Ingrid Marie Laughman
Study Guide
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This 5 page Study Guide was uploaded by Caroline Hurlbut on Sunday February 7, 2016. The Study Guide belongs to Chem 113 at Colorado State University taught by Ingrid Marie Laughman in Spring 2016. Since its upload, it has received 62 views. For similar materials see General Chemistry II in Chemistry at Colorado State University.


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Date Created: 02/07/16
Exam 1 Study Guide 1. What are the first three laws of thermodynamics? 2. Define chemical spontaneity. 3. What is the relationship between entropy and accessible microstates? 4. What are some ways that entropy can increase? 5. What are the trends for spontaneity, Gibbs free energy, and entropy of the universe? 6. Given the reaction Cu2S(s)—>2Cu(s) + S(s), calculate the entropy of the reaction. ∆S S(s)=321 J/K/mol ∆S Cu(s)=33.2 J/K/mol ∆S Cu2S(s)=120.9 J/K/mol 7. Explain why termolecular reactions are not possible. 8. What is an elementary reaction? 9. Given the reaction 2H2(g) + O2(g)—>2H2O(g), define the system and surroundings. 10. Given the reaction H2O(l)—>H2O(g), would you expect entropy to increase, decrease, or remain the same? Explain. 11. Given the following energy profile, answer these questions: a.) Is this reaction endothermic or exothermic? b.) Is this an elementary reaction? Why or why not? c.) What is the rate-determining step of this reaction? (Answer ts1 or ts2.) d.) Identify a possible activated complex for this reaction. 12. What is the purpose of a driving reaction? 13. Given the reaction NH4NO3(s)—>NH4+(aq) + NO3-(aq) at 25˚C, calculate the free energy of the reaction and determine whether or not the reaction is spontaneous. ∆Hf ∆Sf NH4NO3(s) -365.56 kJ/mol 151.08 J/K/mol NH4+(aq) -132.51 kJ/mol 113.4 J/K/mol NO3-(aq) -205.0 kJ/mol 146.4 J/K/mol 14. If a reaction is spontaneous only at high temperatures, what does this tell you about the signs of ∆H and ∆S? 15. The reaction NO2(g) + CO(g)—>NO(g) + CO2(g) has two steps. The first step is NO2(g) + NO2(g)—>NO3(g) + NO(g). Suggest a plausible second step for this reaction. 16. Given the coupled reaction and the values for ∆S, calculate the net change in enthalpy for the formation of one mole of lead(II) sulfate. H2SO4(l)—>SO3(g) +H2O(l) ∆H=113. kJ Pb(s) + PbO2(s) + 2SO3(g)—>2PbSO4(s) ∆H=-775. kJ 17. Given the standard entropies of each compound, calculate the standard molar entropy of the following reaction: C3H8(g) + 5O2(g)—>3CO2(g) + 4H2O(l) ∆S C3H8(g)=269.9 J/K/mol ∆S O2(g)=205. J/K/mol ∆S CO2(g)=214. J/K/mol ∆S H2O(l)=69.91 J/K/mol 18. Calculate the temperature at which hydrazine, N2H4, becomes spontaneous (boiling point) using the given ∆H and ∆S values: ∆H N2H4(l)=50.9 kJ/mol ∆H N2H4(g)=95.4 kJ/mol ∆S N2H4(l)=121.2 J/K/mol ∆S N2H4(g)=238.4 J/K/mol 19. Given that the reaction 2NO(g) + O2(g)—>2NO2(g) happens over a certain period of time with certain concentrations, set up how the rates of reaction for NO and NO2 would be calculated. 20. Given the Lewis structures of the following two molecules, predict which one has the higher entropy. B A Answers 1. 1st law: Energy is conserved and cannot be created or destroyed. 2nd law: The entropy of the universe (or an isolated system) always increases during spontaneous processes. 3rd law: As temperature drops toward absolute zero (0 K), entropy decreases to 0 and the number of microstates decreases to 1. (Note: an entropy of zero can only be achieved by a perfect crystal at 0 K.) 2. Chemical spontaneity is defined as a reaction will proceed once started as long as there are reactants. The free energy of the products must also be less than the free energy of the reactants (∆G<0). 3. Microstates are the number of ways to arrange the available energy in a system, so if there is more energy, there are more ways to arrange it. Therefore, as entropy increases, the number of accessible microstates also increases. 4. A number of factors including temperature, volume, the number of particles, and the size of the particles (as each factor increases) can all be involved in increasing entropy. 5. The spontaneity of a reaction can be determined by the sign of the Gibbs free energy. The trends for this are: If ∆Gsys<0, then ∆Suniv>0 and the reaction is spontaneous If ∆Gsys>0, then ∆Suniv<0 and the reaction is non spontaneous (no reaction) If ∆Gsys=0, then ∆Suniv=0 and the reaction has reached chemical equilibrium, meaning that the forward and reverse reactions occur at the same rate (not that they are both spontaneous). 6. Use the formula ∆Srxn=(#moles x ∆S) products - (#moles x ∆S) reactants. ∆Srxn = [(120.9 x 1)] - [(33.2 x 2) + (32.1 x 1)] ∆Srxn = 120.9 - 387.4 ∆Srxn = 22.4 J/K 7. In order to react, molecules must have the right collision, energy, and geometry. For three molecules to have all of these things and react is so unlikely that it is impossible for a reaction with more than two reactants to occur. 8. An elementary reaction is reaction in which one or more reactants react directly to form products in a single reaction step and with a single transition state. One or more elementary reactions can make up an overall reaction. 9. The system is the actual reaction, in this case 2H2(g) + O2(g)—>2H2O(g). The surroundings are everything else not part of the reaction. For example, since this reaction is happening in the gas phase, the walls of the container would be part of the surroundings. 10. Since water is physically changing from a liquid to a gas, entropy would be expected to increase, because molecules in the gas phase have more accessible microstates and therefore more entropy. 11. a.) The reaction is exothermic. b.) This is not an elementary reaction because there are two steps involved. c.) The rate-determining step is ts1 because it has the higher activation energy. d.) Ts1 is an activated complex in this reaction, and so is ts2. 12. The purpose of a driving reaction is to provide enough ∆G to make a separate reaction run. 13. Use the formula ∆Srxn=(#moles x ∆S) products - (#moles x ∆S) reactants. First, calculate ∆H: ∆H = [(-132.51 x 1) + (-205.0 x 1)] - [(-365.56 x 1)] ∆H = (-337.51) - (-365.56) ∆H = 28.05 kJ/mol Next, calculate ∆S: ∆S = [(113.4 x 1) + (146.4 x 1)] - [(151.08 x 1)] ∆S = (259.8) - (151.08) ∆S = 108.72 J/K/mol Finally, use the Gibbs free energy formula, ∆G=∆H - T∆S, to calculate ∆G: ∆G = 28.05 - 298.15(108.72) *Convert Celsius to Kelvin ∆G = 28.05 - 298.15(0.10872) *Convert J to kJ ∆G = 28.05 - 32.41 ∆G = -4.36 kJ 14. If a reaction is spontaneous at high temperatures only, we can use the Gibbs free energy equation, ∆G=∆H - T∆S, to determine the signs of ∆H and ∆S. If the reaction is spontaneous at high temperatures, ∆G is negative and T is positive. This means both ∆H and ∆S must be positive in order to have a negative ∆G at a high temperature. 15. When given the net reaction, you want the second step combined with the given first step to cancel to/add up to the net reaction. A plausible second step for the reaction given would be NO3(g) + CO(g)—>NO2(g) + CO2(g). 16. Both reactions have SO3 in common, except step two has 2 moles of SO3 on the product side and step one has one mole of SO3 on the reactant side. 2x[H2SO4(l)—>SO3(g) +H2O(l)] [∆H=113. kJ]2x *Cancel SO3 Pb(s) + PbO2(s) + 2SO3(g)—>2PbSO4(s) ∆H=-775. kJ ∆H = 226 kJ - 775 kJ ∆H = -549 kJ 17. Use the formula ∆Srxn=(#moles x ∆S) products - (#moles x ∆S) reactants. ∆Srxn = [(214 x 3) + (69.91 x 4)] - [(269.9 x 1) + (205 x 5)] ∆Srxn = (921.64) - (1294.9) ∆Srxn = -373. J/K 18. Use the formulas ∆Srxn=(#moles x ∆S) products - (#moles x ∆S) reactants and ∆G=∆H - T∆S. First, write an equilibrium equation for hydrazine: N2H4(l)—>N2H4(g) (Note: write the equation going from liquid to gas, since entropy is always increasing.) Next, calculate ∆H: ∆H = (95.4 kJ/mol x 1) - (50.9 kJ/mol x 1) ∆H = 44.5 kJ/mol Now calculate ∆S: ∆S = (238.4 J/K/mol x 1) - (121.2 J/K/mol x 1) ∆S = 117.2 J/K/mol Finally, calculate T by rearranging ∆G=∆H - T∆S: T = ∆H - ∆G ∆S T = (44.5 kJ/mol) - (0) *Set ∆G=0 (117.2 J/K/mol) T = (445000 J/mol) *Convert kJ to J (117.2 J/K/mol) T = 380. K T = 107.˚C *Convert Kelvin to Celsius 19. Rate of reaction for NO: -(1/2)(∆NO/∆t) *Multiply by 1/coefficient (Note: the rate of reaction for NO is negative because NO is a reactant, and the concentration of reactants decreases over time.) Rate of reaction for NO2: (1/2)(∆NO2/∆t) *Multiply by 1/coefficient (Note: the rate of reaction for NO2 is positive because NO2 is a product, and the concentration of products increases over time.) 20. Molecule A has the higher entropy because it is the more structurally complex and “spread-out” molecule.


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