▶️ Answer/Explanation
(a)(i)
Answer: The initial rate is determined from the tangent at \( t = 0 \). From the graph, the slope is approximately \(-0.0125 \, \text{mol dm}^{-3} \, \text{s}^{-1}\).
Explanation: The initial rate is the change in concentration of \( S_2O_8^{2-} \) per unit time at \( t = 0 \). The tangent at \( t = 0 \) gives a slope of \(-0.0125\), so the rate is \( 0.0125 \, \text{mol dm}^{-3} \, \text{s}^{-1} \).
(a)(ii)
Answer: A large excess of \( I^- \) ensures its concentration remains approximately constant, making the reaction pseudo-first order with respect to \( S_2O_8^{2-} \). The half-life then depends only on \( k \).
Explanation: Since \( [I^-] \) is constant, the rate equation simplifies to \( \text{rate} = k’ [S_2O_8^{2-}] \), where \( k’ = k[I^-] \). The half-life of a first-order reaction is \( t_{1/2} = \frac{\ln 2}{k’} \), allowing \( k \) to be determined.
(b)
Answer: \[ Fe^{2+} + S_2O_8^{2-} \rightarrow Fe^{3+} + 2SO_4^{2-} \] \[ Fe^{3+} + I^- \rightarrow Fe^{2+} + \frac{1}{2} I_2 \]
Explanation: \( Fe^{2+} \) is oxidised to \( Fe^{3+} \) by \( S_2O_8^{2-} \), and \( Fe^{3+} \) is reduced back to \( Fe^{2+} \) by \( I^- \), regenerating the catalyst.
(c)
Answer: An increase in temperature increases the rate constant \( k \) (Arrhenius equation) and thus the rate of reaction.
Explanation: The Arrhenius equation \( k = Ae^{-E_a/RT} \) shows that \( k \) increases with temperature. Since rate \( = k[S_2O_8^{2-}][I^-] \), the rate also increases.
(d)
Answer: The half-life \( t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{1.58 \times 10^{-2}} \approx 43.9 \, \text{s} \).
Explanation: For a first-order reaction, \( t_{1/2} = \frac{\ln 2}{k} \). Substituting \( k = 1.58 \times 10^{-2} \, \text{s}^{-1} \) gives \( t_{1/2} \approx 43.9 \, \text{s} \).
(e)
Answer: Step 1 (slow, rate-determining): \( NO + Br_2 \rightarrow NOBr_2 \)
Step 2 (fast): \( NOBr_2 + NO \rightarrow 2NOBr \)
Explanation: The rate-determining step involves one NO and one \( Br_2 \) molecule, matching the observed rate law. The second step is fast and does not affect the rate.
