AP CHEM UNIT 9 FINAL

(b)

(a) (b) (c) (d)

(c) +0.68 V

(a) -1.78 V (b) -0.68 V (c) +0.68 V (d) +1.78 V

(b) Low temperatures only

(a) All temperatures (b) Low temperatures only (c) High temperatures only (d) No temperature

(d) Cl2(g)→Cl2(l)

(a) C2H5OH(l)→C2H5OH(g) (b) NaCl(s)→NaCl(l) (c) CO2(s)→CO2(g) (d) Cl2(g)→Cl2(l)

(d) 1.0MKNO3(aq), because the mobile ions would allow charge to flow between the half-cells.

(a) Distilled water, because it would avoid contaminating the solutions in the half-cells. (b) 1.0MHCl(aq), because it would acidify the solutions and catalyze the overall cell reaction. (c) 1.0MCH3OH(aq), because it would increase the pH of the solutions and catalyze the overall cell reaction. (d) 1.0MKNO3(aq), because the mobile ions would allow charge to flow between the half-cells.

(a) E°cell=[+0.40−(−2.31)]V and ΔG°rxn=−(12×96,500×2.71)/(1,000) kJ

(a) E°cell=[+0.40−(−2.31)]V and ΔG°rxn=−(12×96,500×2.71)/(1,000) kJ (b) E°cell=[+0.40−(−2.31)]V and ΔG°rxn=−(4×96,500×2.71)/(1,000) kJ (c) E°cell=[-2.31−(+0.40)]V and ΔG°rxn=−(12×96,500×2.71)/(1,000) kJ (d) E°cell=[-2.31−(+0.40)]V and ΔG°rxn=−(3×96,500×2.71)/(1,000) kJ

(a) I=((2×0.125×96,485)/3,600)A

(a) I=((2×0.125×96,485)/3,600)A (b) I=((2×0.125×96,485)/1)A (c) I=((0.125×96,485)/3,600)A (d) I=((0.125×96,485)/1)A

(c) It must be positive, since ∆G° is negative and ∆H° is positive.

(a) It must be positive, since the reaction is thermodynamically unfavorable at 600 K. (b) It must be negative, since there are more moles of products than reactants. (c) It must be positive, since ∆G° is negative and ∆H° is positive. (d) It must be negative, since ∆G° is positive and ∆H° is positive.

(c) The reaction has an extremely large activation energy due to strong three-dimensional bonding among carbon atoms in diamond.

(a) The rate of the reaction is extremely slow because of the relatively small value of ΔG°ΔG° for the reaction. (b) The entropy of the system decreases because the carbon atoms in graphite are less ordered than those in diamond. (c) The reaction has an extremely large activation energy due to strong three-dimensional bonding among carbon atoms in diamond. (d) The reaction does not occur because it is not thermodynamically favorable.

(d) The amount of energy released when the product bonds form is much greater than the amount of energy needed to break the reactant bonds.

(a) The total number of gaseous product molecules is less than the total number of gaseous reactant molecules, thus ΔS<0. (b) The total number of gaseous product molecules is greater than the total number of gaseous reactant molecules, thus ΔS>0. (c) The amount of energy released when the product bonds form is much less than the amount of energy needed to break the reactant bonds. (d) The amount of energy released when the product bonds form is much greater than the amount of energy needed to break the reactant bonds.

(b) Voltage decreases but remains above zero

(a) Voltage increases (b) Voltage decreases but remains above zero (c) Voltage becomes zero and remains at zero (d) No change in voltage occurs (e) Direction of voltage change cannot be predicted without additional information

(d) No change in voltage occurs

(a) Voltage increases (b) Voltage decreases but remains above zero (c) Voltage becomes zero and remains at zero (d) No change in voltage occurs (e) Direction of voltage change cannot be predicted without additional information

(d) has a slow rate at 298K because the activation energy is relatively high

(a) is only thermodynamically favorable in the presence of a catalyst (b) occurs with an increase in entropy because O2(g) is a product (c) is reversible and H2O2(aq) is produced almost as fast as it decomposes (d) has a slow rate at 298K because the activation energy is relatively high

(d) ΔG<0⁢ because although entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles, energy is released as the bond between the H and Cl atoms forms.

(a) ΔG>0⁢ because energy is released as the bond between the H and Cl atoms forms, and entropy increases because the number of gaseous product particles is less than the number of gaseous reactant particles. (b) ΔG>0⁢ because energy is absorbed as the bond between the H and Cl atoms forms, and entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles. (c) ΔG<0⁢ because although energy is absorbed as the bond between the H and Cl atoms forms, entropy increases because the number of gaseous product particles is less than the number of gaseous reactant particles. (d) ΔG<0⁢ because although entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles, energy is released as the bond between the H and Cl atoms forms.

(c) ΔG° < 0 and Keq > 1

(a) ΔG° > 0 and Keq > 1 (b) ΔG° > 0 and Keq < 1 (c) ΔG° < 0 and Keq > 1 (d) ΔG° < 0 and Keq < 1

(a) ΔG°<0 and TΔSº is smaller in magnitude than ΔHº.

(a) ΔG°<0 and TΔSº is smaller in magnitude than ΔHº. (b) ΔG°<0 and TΔSº is larger in magnitude than ΔHº. (c) ΔG°>0 and TΔSº is smaller in magnitude than ΔHº. (d) ΔG°>0 and TΔSº is larger in magnitude than ΔHº.

(b) ΔG°=−RT(lnK<0 because K>1

(a) ΔG°=−RTlnK>0 because K>1 (b) ΔG°=−RT(lnK<0 because K>1 (c) ΔG°=ΔH°−TΔS°<0 because ΔH°<0 and ΔS°>0 (d) ΔG°=ΔH°−TΔS°>0 because ΔH°<0 and ΔS°<0

(a) ΔH = Positive; ΔS = Positive

(a) ΔH = Positive; ΔS = Positive (b) ΔH = Positive; ΔS = Negative (c) ΔH = Positive; ΔS = Equals to zero (d) ΔH = Negative; ΔS = Positive (e) ΔH = Negative; ΔS = Negative

(b) ΔS°rxn is negative because the number of molecules in the gas phase decreases as the reaction proceeds.

(a) ΔS°rxn is negative because the number of N2O4 molecules increases as the reaction proceeds. (b) ΔS°rxn is negative because the number of molecules in the gas phase decreases as the reaction proceeds. (c) ΔS°rxn is positive because the number of N2O4 molecules increases as the reaction proceeds. (d) ΔS°rxn is positive because the number of molecules in the gas phase decreases as the reaction proceeds.