The aqueous oxidation of iodide ion by hydrogen peroxide at follows the overall…

CHEMISTRY · Multi-part · AP · Chemistry · Kinetics · Kinetics

Problem

The aqueous oxidation of iodide ion by hydrogen peroxide at

298\,\text{K}

follows the overall stoichiometry

H_2O_2 + 2\,I^- + 2\,H^+ \to I_2 + 2\,H_2O

. A simplified sequence of elementary steps consistent with many experiments is: Step 1

H_2O_2 + I^- \rightleftharpoons HOI + OH^-

(fast,

K_{eq} \approx 1.0 \times 10^{-8}

298\,\text{K}

) Step 2

HOI + I^- + H^+ \to I_2 + H_2O

(slow) Additional studies at

350\,\text{K}

reveal that the observed rate law changes from

\text{rate} = k[H_2O_2][I^-][H^+]

(at

298\,\text{K}

) to

\text{rate} = k'[H_2O_2]^1[H^+]^1

(at

350\,\text{K}

, very low

[I^-]

order). Answer the following questions. (1) Part 1: (a) Using the elementary steps, write the rate law expected at a temperature where the equilibrium constant for Step 1 is very small. Justify your form of the rate law by referencing the species present immediately before the slow step. (2) Part 2: (b) Explain qualitatively why the observed reaction order with respect to

I^-

decreases markedly when the temperature is raised to

350\,\text{K}

. (3) Part 3: (c) The enthalpy change for Step 1 is positive (endothermic). Relate this information to the temperature dependence of the equilibrium constant and show how it accounts for the change in the apparent rate law. (4) Part 4: (d) Predict how the initial rate at

350\,\text{K}

would respond to doubling

I^-

while leaving

H_2O_2

and

H^+

unchanged. Provide reasoning based on your discussion in parts (a)–(c).

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