Question

Consider the reaction, Cl2 + H2S => 2 HCl + S, which is found to be first order in Cl2. Which step of the proposed mechanism must be slow in order to agree with this rate law? Cl2 => 2 Cl Cl + H2S => HCl + HS Cl + HS => HCl + S
A. only 3
B. only 2
C. 1
D. Either 2 or 3

Answer 1

Answer:

Second step of the mechanism is slow step.

Explanation:

The given chemical reaction is as follows

$Cl_{2}+H_{2}S\rightarrow 2HCl+S$

The mechanism of the reaction is as follows.

$Cl_{2}\rightarrow 2Cl$

$Cl+H_{2}S\rightarrow HCl+HS$

$Cl+HS\rightarrow HCl+S$

The rate of the reaction.

$rate=k[Cl_{2}]^\frac{1}{2}[H_{2}S]$

Therefore, Step -2 must be slow.