Chlorate(I) ions undergo the following reaction under aqueous conditions - CIE - A-Level Chemistry - Question 5 - 2019 - Paper 1

The rate equation can be expressed based on the determined orders:

$\text{rate} = k \cdot [\text{ClO}^-]^2 \cdot [\text{NH}_3]^2$

Using the data from experiment 1:

• Given:
  • Rate = 0.256 mol dm⁻³ s⁻¹
  • [ClO⁻] = 0.200 mol dm⁻³
  • [NH₃] = 0.100 mol dm⁻³

Substituting these values into the rate equation:

$0.256 = k \cdot (0.200)^2 \cdot (0.100)^2$

Solving for k:

$k = \frac{0.256}{(0.200)^2 \cdot (0.100)^2}$ $k = \frac{0.256}{0.0004} = 640 \, \text{mol}^{-3} \text{dm}^9 \text{s}^{-1}$

Thus, the rate constant is:

$k = 640 \, \text{dm}^6 \, \text{mol}^{-2} \, \text{s}^{-1}$

A typical graph for the temperature vs. rate constant, k, would show that as the temperature increases, the value of k also increases. This can be represented as a positively sloping curve on a graph with temperature on the x-axis and k on the y-axis.

To confirm that the reaction is first-order with respect to [I⁻], the concentration of [I⁻] can be plotted against time. A linear relationship would indicate a first-order reaction. If the plot is a straight line, then it confirms that the rate law follows the equation:

$\text{rate} = k \cdot [I^-]$

The overall equation can be deduced by adding the three steps of the mechanism:

1. $\text{ClO}^- + \text{H}_2\text{O} \rightarrow \text{HClO} + \text{OH}^-$
2. $\text{I}^- + \text{HClO} \rightarrow \text{H}_2\text{O} + \text{Cl}^-$
3. $\text{HClO} + \text{OH}^- \rightarrow \text{H}_2\text{O} + \text{I}^-$

After canceling intermediates and combining the reactants and products, the overall reaction becomes:

$\text{ClO}^- + 2\text{I}^- + 2\text{H}_2\text{O} \rightarrow \text{HClO} + 2\text{I}^- + 2\text{OH}^-$

Step 1 involves a redox reaction where the oxidation state of chlorine changes. In the transformation:

$\text{ClO}^- + \text{H}_2\text{O} \rightarrow \text{HClO} + \text{OH}^-$

Chlorine is reduced as it goes from +1 in ClO⁻ to +0 in HClO, indicating a gain of electrons. Similarly, the hydroxide ion shows a change, confirming the redox nature of this step.