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Types of Catalysis

The rate of d...

Question

The rate of decomposition of aldehyde $(C H_{3} C H O)$ into $C H_{4}$ and $C O$ in presence of $I_{2}$ at 800 K follows the rate law: $r = K [C H_{3} C H O] [I_{2}]$ . The decomposition is believed to go to the two step mechanism:
$C H_{3} C H O + I_{2} \to C H_{3} I + H I + C O$
$C H_{3} I + H I \to C H_{4} + I_{2}$
What is the catalyst for the reaction? Which of the two steps is the slower one?

H I, I^{s t} s t e p

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I_{2}, I^{s t} s t e p

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H I, I I^{s t} s t e p

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I_{2}, I I^{s t} s t e p

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Solution

The correct option is D $I_{2}, I^{s t} s t e p$
Iodine is consumed in the first step and formed in the second step. Hence, iodine is the catalyst. The rate law expression contains the concentrations of acetaldehyde and iodine which appear in the first step. Hence, the first step is the slower step. It is the rate determining step.

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The equilibrium constant for the given reaction $H_{2} + I_{2} ⇌ 2 H I$ is correctly

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$[H I] (M)$	Rate of decomposition of $H I (m o l L^{- 1} s^{- 1})$
$0.1$	$2.75 \times 10^{- 8}$
$0.2$	$11 \times 10^{- 8}$
$0.3$	$24.75 \times 10^{- 8}$

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g), K_{c} = 47.6

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