An acidified solution of butanone reacts with iodine as shown - AQA - A-Level Chemistry - Question 1 - 2022 - Paper 2

Using the rate equation, rearrange for k:

$k = \frac{rate}{[CH₃CH₂COCH₃][I₂][H⁺]}$

Substituting in the values:

$k = \frac{1.45 \times 10^{-4}}{4.35 \times 0.00500 \times 0.825}$

Calculating gives: $k = 2.75 \times 10^{-2} \text{ mol}^{-2} \text{ dm}^{6} \text{ s}^{-1}$.

When concentrations are halved, the new initial rate is:

$rate' = k[\frac{1}{2}CH₃CH₂COCH₃][\frac{1}{2}I₂][\frac{1}{2}H⁺] = k \cdot \frac{1}{8}[CH₃CH₂COCH₃][I₂][H⁺]$

Thus,

$rate' = \frac{1.45 \times 10^{-4}}{8} = 1.81 \times 10^{-5} \text{ mol dm}^{-3} \text{s}^{-1}$.

A common observation is the disappearance of the purple color of iodine.

The graph shows a decreasing trend as the temperature increases. This indicates that as temperature rises, the value of 1/t decreases, suggesting an increased rate of reaction. The shape of the graph is exponential due to the Arrhenius equation relationship, where higher temperatures provide more energy for molecular collisions.

From the graph, extract the 1/t value at 35 °C, which corresponds to approximately:

$t = \frac{1}{value~at~35°C}$

For example, if 1/t = 0.03 s⁻¹, then: $t = 33.3 ~s$.

Using the equation,

$\ln(k_2/k_1) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Substituting in values:

1. Substitute rate constants and temperatures: $\ln(\frac{1.70 \times 10^{-4}}{1.55 \times 10^{-4}})$

2. Evaluate and rearrange for Eₐ, giving: $Eₐ = 69.57 ~kJ~mol^{-1}.$

The mechanism involves a nucleophilic addition:

1. The cyanide ion (CN⁻) attacks the carbonyl carbon of butanone.
2. This forms a tetrahedral intermediate.
3. Protonation occurs from dilute acid, resulting in the formation of cyanohydrin.