Exam 1B, Spring 2001

Answer

First, A balanced reaction must be written:

2 HN₃(aq) + H₂O₂(aq) → 2 H₂O(l) + 3 N₂(g)

a. The rate of loss of hydrogen peroxide is given directly from the definition of a rate:

To the correct significant figures, this is 1×10^–3 M s^–1

b. The rate of gain of water is found from the reaction stoichiometry:

Rate of water gain = –2×Rate of loss of hydrogen peroxide = –(2)(–1.2×10^–3) = 2.4×10^–3 M s^–1

Again, to the correct number of significant figures, this is 2×10^–3 M s^–1

Answer

There are several ways that this problem could be answered. If graph paper were available, then plotting the concentration vs. time data in the manner appropriate for zero, first, and second order reactions and finding the linear plot would work. However, since no graph paper was provided, another method is needed.

The method of initial rates can be used by calculating the initial rate from the first two data points in each experiment. In general, the rate law for this reaction is given by Rate = k[NO₂]^m.

The initial rates for each experiment are:

Experiment 1:

Experiment 2:

The method of initial rates compares the two experimental rates:

Using m = 2 gives rate constants of k = 0.48 M^–1 s^–1 and k = 0.42 M^–1 s^–1, k_ave = 0.45 M^–1 s^–1.

An easier method is to use the half-lives. In each experiment the half-life is easily found:

Experiment 1 t_½ = 1.85 s and in Experiment 2 t_½ = 3.70 s. Since the half-life increases with decreasing initial concentration, the reaction must be second order. Then the rate constant can be found from

Giving an average rate constant, k_ave = 0.54 M^–1 s^–1.

Answer

Convert each of the temperatures to Kelvin and then use the two-point form of the Arrhenius equation.

T₁ = 25 + 273 = 298 K, k₁ = 8.8×10^–5 M^–1 s^–1

T₂ = 35 + 273 = 308 K, k₂ = 2.8×10^–4 M^–1 s^–1

E_a = 88000 J/mol = 88 kJ/mol

The total order of the reaction must be 2, based on the units of the rate constant.

Answer

Balance each reaction and then find the mass action expression.

a. 2 CO₂(g)→← 2 CO(g) + O₂(g)

b. 2 CH₄(g) + O₂(g)→← 2 CH₃OH(l)

c. CaCO₃(s)→← CaO(s) + CO₂(g)

K_c = [CO₂]_e

Answer

CdS(s) + H₂O(l) →← HS^–(aq) + Cd²⁺(aq) + OH^–(aq) K_c = 8×10^–28

H₂S(aq) + H₂O(l) →← H₃O⁺(aq) + HS^–(aq) K_c = 1.0×10^–7

H₃O⁺(aq) + OH^–(aq) →← 2 H₂O(l) K_c = 1.0×10¹⁴

Add reactions to give the target reaction:Multiply K_c values

CdS(s) + 2 H₃O⁺(aq) →← Cd²⁺(aq) + H₂S(aq) + 2 H₂O(l) K_c = (8×10^–28)(1.0×10⁷)(1.0×10¹⁴) = 8×10^–7

Answer

First, write the balanced reaction and add heat:

N₂(g) + O₂(g) + heat →← 2 NO(g)

a. When nitrogen monoxide is added the reaction shifts LEFT, towards REACTANTS, to remove the added nitrogen monoxide.

b. When the pressure is increased there is NO SHIFT because there are the same number of moles of gases on each side of the equation.

c. When the temperature is increased the reaction shifts RIGHT, towards PRODUCTS, to remove the added heat (towards the endothermic direction).

Answer

H₂(g) + I₂(g) →← 2 HI(g)

Since Δn = 0, K_p = K_c

ExperimentK_c

1

2

3

Experiment 3 is clearly in error, so the average equilibrium constant is K_c = 59.57