Online Quiz - Chapter 10: Reaction Rates & Chemical Equilibrium

What will happen if CO2 is added to the reaction below? H2CO3(aq) ↔ CO2(g) + H2O(l) a) Equilibrium shifts toward the reactant. b) A change will occur, but we cannot predict what that change will be. c) Equilibrium shifts toward the products. d) No change will occur because the reaction is at equilibrium.

a) Equilibrium shifts toward the reactant. Rationale: Adding more CO2, one of the products, shifts the equilibrium to the left, since it speeds up the rate of the reverse reaction.

The formation of carbon dioxide from carbon and oxygen gas is exothermic but occurs very slowly. What statement about this chemical reaction is true? C(s) + O2(g) → CO2(g) a) The activation energy of the reaction is very high. b) The reaction absorbs heat in order to make CO2. c) Lowering the temperature of the reaction will increase the speed of the reaction. d) This is a decomposition reaction.

a) The activation energy of the reaction is very high. Rationale: Activation energy is the energy barrier that blocks reactions. The higher the barrier, the slower the reaction goes. One of the reasons catalysts can speed up reaction rates is that they can lower the activation energy (via a different reaction pathway like first forming an intermediate and then the products).

To speed up a reaction, you should ______. a) increase the amount of reactant and raise the temperature. b) add more reactant and lower the temperature. c) raise the temperature and reduce the amount of reactant. d) add a catalyst and reduce the temperature.

a) increase the amount of reactant and raise the temperature. Rationale: Three factors that speed up a reaction rate are: more reactants, high temperature, and catalysts.

The equilibrium constant for the production of carbon dioxide from carbon monoxide and oxygen is Kc = 2 x 10^11. This means that the reaction mixture at equilibrium is likely to consist of _______. a) mostly products. b) mostly starting materials. c) an equal mixture of products and reactants. d) twice as much product as starting material.

a) mostly products Rationale: Kc = [products] / [reactants] - if large Kc, then at equilibrium, there are more products than reactants - if small Kc, then at equilibrium, there are more reactants than products

Hemoglobin, represented by Hb in the equation below, binds to oxygen to form oxyhemoglobin, HbO2. Which action will increase the amount of oxyhemoglobin in a patient? Hb + O2 ↔ HbO2 a) Decrease the amount of oxygen in the patient's body. b) Increase the amount of oxygen in the patient's body. c) Remove hemoglobin from the patient's body. d) none of them.

b) Increase the amount of oxygen in the patient's body. Rationale: Increasing the amount of O2, one of the reactants, speeds up the forward reaction rate. Therefore more products, HbO2, are made.

Which changes will shift the reaction to the products in the exothermic reaction below? HF(aq) + H2O(l) ↔ F-(aq) + H3O+(aq) a) adding F- and raising the temperature b) adding HF and lowering the temperature c) adding H3O+ and lowering the temperature d) adding HF and raising the temperature

b) adding HF and lowering the temperature Rationale: Since it is an exothermic reaction, the equilibrium can be written as: HF(aq) + H2O(l) ↔ F-(aq) + H3O+(aq) + heat 1. Adding HF shifts the equilibrium to the products side, since more HF speeds up the forward reaction rate. 2. Lower the temperature does the same, because the reverse reaction rate slows down more than the forward rate.

Which two changes would increase the rate of this reaction? CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) a) adding O2 and H2O b) adding a catalyst and adding CO2 c) adding O2 and raising the temperature d) adding CH4 and lowering the temperature

c) adding O2 and raising the temperature Rationale: More reactants and higher temperature can speed up the rate of reactions.

Which of the following equilibrium constants indicates the reaction that gives the smallest amount of product? a) Kc = 5 × 10^-1 b) Kc = 5 × 10^0 c) Kc = 5 × 10^10 d) Kc = 5 × 10^-10

d) Kc = 5 × 10^-10 Rationale: Equilibrium constant (Kc) is the ratio of the amount of products to the amount of reactants when the equilibrium is reached - higher Kc = more products - lower Kc = more reactants