Question

Consider the reaction:
 $ C{l_2}(aq) + {H_2}S(aq) \to S(s) + 2{H^ + }(aq) + 2C{l^ - }(aq) $
The rate equation for this reaction is $ Rate = k[C{l_2}][{H_2}S] $
Which of these mechanisms is/are consistent with this rate equation?
(A) $ C{l_2} + {H_2}S \to {H^ + } + C{l^ - } + C{l^ + } + H{S^ - }\;{\text{ }}\;{\text{ }}\left( {slow} \right) $
 $ C{l^ + } + H{S^ - } \to {H^ + } + C{l^ - } + S\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\;\;{\text{ }}\;{\text{ }}\;\left( {fast} \right) $
(B) $ {H_2}S \rightleftharpoons {H^ + } + H{S^ - }\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;\;\;{\text{ }}\;\left( {fast{\text{ }}equilibrium} \right) $
 $ C{l_2} + H{S^ - } \to 2C{l^ - } + {H^ + } + S\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\left( {slow} \right) $
(a) B only
(b) Both A and B
(c) Neither A nor B
(d) A only

Answer 1

Hint :The rate equation of any reaction is dependent on the rate-determining step. In any given reaction the slowest step will be the rate-determining step of the reaction. That step of the reaction is the one on which the rate equation is being written and hence we have to eliminate the options based on these concepts.

Complete Step By Step Answer:
By carefully analyzing both A and B mechanisms we can see that they both contain one step in which the reaction proceeds very slowly. This means that that is the rate-determining step.
Let us first look at the (A) mechanism first.
According to this mechanism the rate determining step is the first step represented by;
 $ \Rightarrow C{l_2} + {H_2}S \to {H^ + } + C{l^ - } + C{l^ + } + H{S^ - }\;{\text{ }}\;{\text{ }}\left( {slow} \right) $
This means that according to this reaction we can say that the rate of the reaction is:
 $ \Rightarrow Rate = k[C{l_2}][{H_2}S] $
Thus we can say that this mechanism is consistent with the rate equation.
Now let us take a look at the second mechanism (B).
According to that mechanism the rate determining step or the slowest step is:
 $ \Rightarrow C{l_2} + H{S^ - } \to 2C{l^ - } + {H^ + } + S\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\left( {slow} \right) $
This means that the rate equation will written based on this reaction which will be as follows:
 $ \Rightarrow Rate = \;k[{H_2}S{\text{ }}]{[C{I_2}\left] {\text{ }} \right[{H^ + }]^{ - 1}} $
This means that this mechanism doesn’t give the correct rate expression of the given reaction.
Thus we can say that the correct answer for this question will be option (d).

Note :
While writing the rate expression of a given reaction we express it in terms of the reactants. It is also important to note that the rate expression for any given reaction can only be determined experimentally. This means that the rate law expression has nothing to do with the balanced chemical equation or the stoichiometry.