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Matching. I activated complex E chemical equilibrium A reaction rate H entropy J Le Châtelier’s principle D activation energy C spontaneous.

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Presentation on theme: "Matching. I activated complex E chemical equilibrium A reaction rate H entropy J Le Châtelier’s principle D activation energy C spontaneous."— Presentation transcript:

1 Chemistry Chapter 19 – Reaction Rates and Equilibrium Study Guide - Answers

2 Matching. I activated complex E chemical equilibrium A reaction rate H entropy J Le Châtelier’s principle D activation energy C spontaneous F inhibitor G elementary reaction B free energy

3 11. In which the following physical states does a given substance have the highest entropy?
Solid Gas Liquid All of the above

4 12. A reaction that requires free energy
Must be endothermic Is nonspontaneous. Must correspond to a decrease in entropy. Is spontaneous.

5 13. The two factors that determine whether a reaction is spontaneous or nonspontaneous are:
Entropy and disorder. Entropy and enthalpy change. Electron configuration and energy change. Energy and the heat of reaction.

6 14. In which of these systems is the entropy decreasing?
Air escaping from a tire Snow melting Salt dissolving in water A gas condensing to a liquid Decreasing Entropy

7 15. All spontaneous processes:
Are exothermic Are endothermic Involve an increase in entropy Release free energy

8 16. If a catalyst is used in a reaction:
The energy of activation increases. The reaction rate does not change. The reaction rate increases. The equilibrium shifts.

9 17. Which of the following affects the rate of a chemical reaction?
The presences of a catalyst The temperature The concentration of reactants All of the above.

10 18. What is the expression for Keq for this reaction
18. What is the expression for Keq for this reaction? H2O(g) ⇋ 2H2(g) + O2(g) Keq = [2H2O] [H2]x[O2] Keq = [H2]2 x [O2] [H2O]2 Keq = [2H2] x [O2] [2H2O] Keq = [H2O]2 [H2]2 x [O2]

11 19. In an equilibrium reaction with a Keq of 1 x 108
Reactants are favored. Reaction is nonspontaneous. Products are favored. Reaction is exothermic.

12 20. What is the effect of adding more CO2 to the following equilibrium reaction? CO2 + H2O ⇋ H2CO3
More H2CO3 is produced. More H2O is produced. The equilibrium is pushed in the direction of reactants. No change.

13 21. Doing which of the following generally increases the entropy of a substance?
Freezing it Dissolving it in water Condensing it All of the above

14 22. The Keq of a reaction is 4 x 10-7. At equilibrium:
The reactants are favored. The products are favored. The reactants and products are present in equal amounts. The rate of the reverse reaction is much greater than the rate of the forward reaction.

15 23. Two opposing reactions (A + B ⇋ C + D) occurring simultaneously at the same rate is an example of: Reversibility. Chemical equilibrium. Neither a nor b Both a and b

16 Keq = 9.6 Keq = [NO]2 x [Cl2] = (1.2)2(0.60) = 9.6 [NOCl]2 (0.30)2
24. Calculate the value of Keq for this reaction at equilibrium NOCl(g) ⇋ 2NO(g) + Cl2(g) An analysis of the equilibrium mixture in a 1-L flask gives the following results: NOCl, 0.30 mol; 1.2 mol; Cl2, 0.60 mol. Keq = [NO]2 x [Cl2] = (1.2)2(0.60) = 9.6 [NOCl] (0.30)2 Keq = 9.6

17 [CH3OH] = 1.58 mol/L 2.2 x 102 L2/mol2 = [CH3OH] . [CO] x [H2]2
25. Carbon monoxide and hydrogen are combined in the commercial preparation of methyl alcohol CO(g) + 2H2(g) ⇋ CH3OH(g) At a certain set of conditions, the equilibrium mixture contains mol/L of CO, 0.60 mol/L of H2, and the equilibrium constant is 2.2 x 102 L2/mol2. What is the concentration of CH3OH in the equilibrium mixture? 2.2 x 102 L2/mol2 = [CH3OH] . [CO] x [H2]2 2.2 x 102 L2/mol2 = [CH3OH] (0.020 mol/L)(0.60 mol/L)2 [CH3OH] = 1.58 mol/L

18 This reaction is not spontaneous, because G0 is positive!
26. The following reaction has an H0 of 53 kJ/mol and a S0 of kJ/(K x mol) at 25 °C A + B → 2C Is this reaction spontaneous? (Calculate ΔG.) 273 °C + 25° = 298 K ΔG0 = ΔH0 – TΔS0 ΔG0 = (53 kJ/mol) – (298 K)(0.070 kJ/mol·K) ΔG0 = (53 kJ/mol) – (21 kJ/mol) ΔG0 = 32 kJ/mol This reaction is not spontaneous, because G0 is positive!

19 ΔS0 = S0(products) – S0(reactants)
27. What is the standard change in entropy for the following reaction when all reactants are in specified states? H2O(l) + O2(g) → 2H2O2(l) Standard entropies S0 is J/(K x mol): H2O(l) = 69.94, H2O2(l) = 92.0, O2(g) = 205.0 ΔS0 = S0(products) – S0(reactants) ΔS0 = (2 mol)(92.0 J/mol·K) – [(2 mol)(69.94 J/mol·K) + (1 mol)(205.0 J/mol·K)] 1 mole mole mole ΔS0 = -161 J/(mol·K)

20 Favors Products Favors Reactants
28. Predict the changes in the equilibrium position for this reaction when the following changes are made: A­(g) + B(g) ⇋ 4C(g) + heat Decrease the concentration of C; Increase the concentration of A; Add more heat; Increase the pressure. In each case, state whether the change causes a shift that favors the formation of reactants or of products. Favors Products Favors Reactants

21 Initial concentration of A (mol/L) Initial Concentration of B (mol/L)
29. The rate law for the following reaction is: Rate = k[A]a x [B]b aA + bB → cC + dD From the data in the following chart, find the kinetic order of the reaction with respect to A and B, as well as the overall order. Initial concentration of A (mol/L) Initial Concentration of B (mol/L) Initial Rate [mol/(L x s)] 0.50 0.05 2 x 10-3 0.10 4 x 10-3 0.20 8 x 10-3 0.01 1 x 10-3 64 x 10-3 Doubling A doubles the rate – first order in A. Doubling B increases the rate 8 times (23 = 8) – third order in B. First order + third order = fourth order overall.


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