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

Using the rate equation:

Substituting the values from Table 1:

$ext{1.45 imes 10^{-4}} = k (4.35)(0.00500)(0.825)$

Calculating:

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

Thus,

$k ≈ 1.60 imes 10^{-2} ext{ mol}^{-2} ext{ dm}^{6} ext{s}^{-1}$

If all concentrations are halved:

$ext{new rate} = k \left(\frac{[C_4H_8COC]}{2}\right) \left(\frac{[I_2]}{2}\right) \left(\frac{[H^+]}{2}\right)$

Substituting known values:

$\text{new rate} = k (2.175)(0.00250)(0.4125)$

Calculating:

$\text{new rate} = 3.63 \times 10^{-5} ext{ mol dm}^{-3} ext{s}^{-1}$.

A common observation is the disappearance of the brown color of iodine, indicating that all iodine has reacted.

The graph is exponential, showing that as the temperature increases, the value of 1/t decreases sharply. This indicates that reaction rate increases significantly with temperature, as more molecules have the required energy to react, leading to a faster completion of the reaction.

Referring to the graph, we estimate the value of 1/t at 35°C, which correlates to around 0.02 s⁻¹. Hence,

$t = \frac{1}{0.02} = 50 ext{ seconds}$.

Using the formula:

$\ln(k1/k2) = (Ea/R)(1/T2 - 1/T1)$ Substituting the known values:

$\ln(1.55 \times 10^{-5}/1.70 \times 10^{-4}) = (Ea/8.31)(1/333 - 1/303)$ Calculate the left side, then solve for Ea:

After calculations, we find:

$Ea ≈ 76.2 ext{ kJ mol}^{-1}$.

1. Nucleophilic Attack: The cyanide ion (CN⁻) attacks the carbonyl carbon of butanone, forming an intermediate.
2. Formation of Alkoxide: This intermediate becomes an alkoxide ion.
3. Protonation: After this, the alkoxide undergoes protonation by dilute acid to form the final alcohol product.