AP Classroom Practice Questions

C12H22O11 + H2O → 2C6H12O6 The hydrolysis of sucrose is represented by the chemical equation above. This reaction is extremely slow in aqueous solution. However, when sucrase is added as shown in the diagram, the rate of the reaction is about 6,000,000 times faster. Based on this information, what best explains the large increase in the rate of hydrolysis that occurs with the addition of sucrase?

The reaction proceeds through a different reaction path with a lower activation energy in which the sucrase-sucrose complex is formed as an intermediate. Catalysts can increase the rate of a reaction by providing a different reaction path with a lower activation energy. The diagram is consistent with a 2-step mechanism in which the first step produces the sucrase-sucrose complex as an intermediate that is used up in the second step.

X → products Pure substance X decomposes according to the equation above. Which of the following graphs indicates that the rate of decomposition is second order in X ?

X-Axis: Time Y-Axis: 1/[X] Line: Straight, increasing slope

2HO2(g)→H2O2(g)+O2(g) The reaction represented by the chemical equation shown above occurs in Earth's atmosphere. In an experiment, [HO2] was monitored over time and the data plotted as shown in the following graph. Based on the information, which of the following is the rate law expression for the reaction?

Rate=k[HO2]^2

2 N2O5(g) → 4 NO2(g) + O2(g) A sample of N2O5 was placed in an evacuated container, and the reaction represented above occurred. The value of PN2O5, the partial pressure of N2O5(g), was measured during the reaction and recorded in the table below.

The decomposition of N2O5 is a first-order reaction. This option is correct. A plot of ln[A] versus t is a straight line.

The following questions relate to the below information. XY2 → X + Y2 The equation above represents the decomposition of a compound XY2. The diagram below shows two reaction profiles (path one and path two) for the decomposition of XY2. http://turlockapchem.weebly.com/uploads/8/4/3/6/8436628/unit_10_practice_test_solutions.pdf (p. 5) What most likely accounts for the difference between reaction path one and reaction path two?

The presence of a catalyst in path two

2 NO(g) + O2(g) → 2 NO2(g) Consider the following mechanism for the reaction represented above. Step 1: 2 NO ⇄ N2O2 (fast reversible) Step 2: N2O2 + O2 → 2 NO2 (slow) What is the rate law consistent with the mechanism?

The rate law that is consistent with the mechanism is rate = k[NO]^2[O2].

Step 1:H2(g)+ICl(g)→HI(g)+HCl(g) (slow) Step 2:HI(g)+ICl(g)→HCl(g)+I2(g) (fast) What is the overall chemical equation for the reaction and the rate law for elementary step 2?

H2(g)+2ICl(g) → 2HCl(g)+I2(g)H2(g)+2ICl(g)→2HCl(g)+I2(g) rate=k[HI][ICl]

2H2(g)+2NO(g)→N2(g)+2H2O(g) For the chemical reaction represented above, the following mechanism is proposed. Step 1:2NO(g)⇌N2O2(g)(fast equilibrium)Step 2:N2O2(g)+H2(g)→N2O(g)+H2O(g)(slow)Step 3:N2O(g)+H2(g)→N2(g)+H2O(g)(fast) What is the initial rate law expressions consistent with this proposed mechanism?

Rate = k[H2][NO]^2