OCR, A Level Chemistry, 2023, H432/01 Periodic table, elements and physical chemistry. Section B. Questions 16 to 18.

Bromine, Br2, can be produced by the reaction:

5Br^–(aq) + BrO₃^–(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)

A student investigates the rate of this reaction by carrying out four experiments at the same temperature. The student’s results are shown below.

Explain how the reaction orders can be determined from the student’s results, and determine the rate equation and rate constant for this reaction.

To measure reaction rates, we change the concentration of one reactant while keeping the others constant. This helps isolate the effect of that specific reactant. What we are looking for is:

• If doubling the concentration of a reactant doubles the rate, the reaction is first order with respect to that reactant (m = 1).
• If doubling the concentration causes the rate to quadruple, the reaction is second order (m = 2)
• If changing the concentration doesn’t affect the rate, the reaction is zero order (m = 0).

Often, a quick visual comparison of the numbers helps identify the correct values. However, if the pattern isn’t immediately obvious, a more systematic and numerical approach helps. I will explain here in both ways. 

The rate equation for this reaction can be expressed as:

Rate = k [Br^–]^a [BrO₃^–]^b [H⁺]^c

Where k is the rate constant and a, b, and c are the reaction orders for Br^–, BrO₃^– and H⁺ respectively. To solve the problem, we select experiments where we can simplify some of the terms in the rate equation and calculate the relevant reaction orders.

By comparing experiment 1 and experiment 2, we see that in experiment 2, the student has trebled the concentration of Br⁻, while keeping BrO₃⁻ and H⁺ constant. The initial rate also trebles, which suggests that the reaction rate is directly proportional to [Br⁻]. This means the reaction order with respect to Br⁻ is 1 (a = 1). Or using a more systematic approach, if we divide the results from experiment 2 by those from experiment 1:

$$
\small{\frac{\text{Experiment 2}}{\text{Experiment 1}}:}
\frac{7.56 \times 10^{-4}}{2.52 \times 10^{-4}} =
\frac{k (6.00 \times 10^{-2})^a (1.20 \times 10^{-1})^b (8.00 \times 10^{-2})^c}
{k (2.00 \times 10^{-2})^a (1.20 \times 10^{-1})^b (8.00 \times 10^{-2})^c}
$$

Operating and simplifying, we get:

$$
\small{\frac{7.56 \times 10^{-4}}{2.52 \times 10^{-4}} =
\frac{(6.00 \times 10^{-2})^a}{(2.00 \times 10^{-2})^a}}
$$

3 = 3^a

Meaning that a = 1.

If we compare experiment 1 with experiment 3, we notice that in experiment 3, the initial concentration of Br⁻ has been doubled, while the concentration of BrO₃⁻ has been halved. Despite these changes, the initial rate of reaction remains the same. Since we already know that the reaction order with respect to Br⁻ is 1 (a = 1), we can conclude that the reaction order for BrO₃⁻ is also 1 (b = 1). Alternatively, we could confirm this using a more systematic method:

$$
\small{\frac{\text{Experiment 3}}{\text{Experiment 1}}:
\frac{2.52 \times 10^{-4}}{2.52 \times 10^{-4}} =
\frac{k (4.00 \times 10^{-2})^1 (6.00 \times 10^{-2})^b (8.00 \times 10^{-2})^c}
{k (2.00 \times 10^{-2})^1 (1.20 \times 10^{-1})^b (8.00 \times 10^{-2})^c}
}
$$

Simplifying:

$$
\small{1 = \frac{4.00 \times 10^{-2}}{2.00 \times 10^{-2}} \left( \frac{6.00 \times 10^{-2}}{1.20 \times 10^{-1}} \right)^b}
$$

1 = 2 × 0.5^b

0.5 = 0.5^b

This means that b = 1

The only reaction order left is c. The initial concentration of H⁺ has only changed in experiment 4. Using the information that we have (including a = b = 1), experiment 3 and experiment 4:

$$\small
\frac{\text{Experiment 3}}{\text{Experiment 4}}:
\frac{2.52 \times 10^{-4}}{3.15 \times 10^{-3}} =
\frac{k \, (4.00 \times 10^{-2})^1 \, (6.00 \times 10^{-2})^1 \, (8.00 \times 10^{-2})^c}{k \, (2.00 \times 10^{-2})^1 \, (6.00 \times 10^{-2})^1 \, (4.00 \times 10^{-1})^c}
$$

Simplifying:

$$\small
0.08 = 2 \times \left(\frac{8.00 \times 10^{-2}}{4.00 \times 10^{-1}}\right)^c
$$

0.08 = 2 × 0.2^c

0.04 = 0.2^c

Taking logarithms:

log (0.04) = c log (0.2)

c = log (0.04) / log (0.2) = 2

Then, c = 2

With the reaction orders, we can calculate the reaction constant by using the data from any of the experiments. Taking experiment 1:

Rate = k [Br^–]¹ [BrO₃^–]¹ [H⁺]²

2.52×10^-4 = k (2.00×10^-2) (1.20×10^-1) (8.00×10^-2)²

$$\small
k = \frac{2.52 \times 10^{-4}}{0.02 \times 0.12 \times (0.08)^2} = 16.40625
$$

To calculate the units of k, we use the rate equation, making k the subject and replacing the relevant units:

$$\small
k = \frac{\text{Rate}}{[ \text{Br}^-]^1 [ \text{BrO}_3^-]^1 [ \text{H}^+]^2} =
\frac{\text{mol} \, \text{dm}^{-3} \, \text{s}^{-1}}{(\text{mol} \, \text{dm}^{-3})^1 (\text{mol} \, \text{dm}^{-3})^1 (\text{mol} \, \text{dm}^{-3})^2} = \text{mol}^{-3} \, \text{dm}^9 \, \text{s}^{-1}
$$

Then, k = 16.4 mol^-3 dm⁹ s^-1

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