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The rate of decomposition of acetaldehyd...

The rate of decomposition of acetaldehyde into methane and CO in the presence of `I_(2)` at 800 K follows the rate law:
Rate `=k[CH_(3)CHO][I_(2)]`
The decomposition is believed to go by a two step mechanism:
`CH_(3)CHO + I_(2) to CH_(3)I + HI + CO`
`CH_(3)I + HI to CH_(4) + I_(2)`
What is the catalyst for the reaction ? Which of the two steps is a slower one ?

Text Solution

Verified by Experts

The correct Answer is:
`I_(2)` is catalyst; first step is slow.
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