Question

The rate of decomposition of aldehyde $\left( {C{H_3}CHO} \right)$into $C{H_4}$ and $CO$ in presence of ${I_2}$ at $800K$ follows the rate law: $r = K\left[ {C{H_3}CHO} \right]\left[ {{I_2}} \right]$. The decomposition is believed to go to the two step mechanism:
$C{H_3}CHO + {I_2} \to C{H_3}I + HI + CO$
$C{H_3}I + HI \to C{H_4} + {I_2}$
What is the catalyst for the reaction? Which of the steps is the slower one?
A.$HI$, 1st step
B.${I_2}$, 1st step
C.$HI$, 2nd step
D.${I_2}$, 2nd step

Answer 1

To answer this question, you must recall the rate law. For a multi- step reaction, the rate law expression is written for the slowest reaction which is also known as the Rate determining step or RDS.

Complete answer:
A catalyst is a substance which speeds up a reaction and is not consumed in the overall reaction. On course of the reaction Iodine is used up in the first substance but is formed again in the second step. Overall it is not consumed in the reaction and thus, ${I_2}$ is the catalyst.
We know that the rate law expression contains the concentrations of the reactants of the rate determining step. Since the expression contains the concentrations of acetaldehyde and iodine which are present in the first reaction, hence, the first reaction is the slower step

Thus, the correct answer is B.

Additional information:
When we study a chemical reaction, it is important for us to consider other factors other than the chemical properties of the reactants, namely, the conditions under which the reaction takes place, the mechanism by which the reaction proceeds, the equilibrium toward which it is moving, and the rate at which it is occurring.

Note:
The rate law is an experimentally determined expression that can be used to predict the relationship between the rate of the reaction and the concentrations of reactants. For elementary reactions, the rate equation is generally derived using the first principles given by the collision theory. The rate equation of a reaction with a multi-step mechanism cannot be calculated simply using the stoichiometric coefficients of the overall reaction.