Question

Consider the reaction:
${Cl}_2\left(aq\right)+H_{2}S\left(aq\right)\to S\left(s\right)+2H^{+}\left(aq\right)+2{Cl}^{-}\left(aq\right)$

The rate equation for this reaction is
rate $=k\left[{Cl}_2\right]\left[H_{2}S\right]$

Which of these mechanisms is/are consistent with this rate equation?
A. ${Cl}_2+H_{2}S\to H^{+}+{Cl}^{-}+{Cl}^{+}+{HS}^{-}$ (slow)
${Cl}^{+}+{HS}^{-}\to H^{+}+{Cl}^{-}+S$ (fast)

B. $H_{2}S\rightleftharpoons H^{+}+{HS}^{-}$ (fast equilibrium)
${Cl}_2+{HS}^{-}\to 2{Cl}^{-}+H^{+}+S$ (Slow)

• A. $B$ only
• B. Both $A$ and $B$
• C. Neither $A$ nor $B$
• D. $A$ only

Answer 1

The correct option is D $A$ only

The slow step is the rate-determining step.

From the given rate law expression (Rate$=k\left[{Cl}_2\right]\left[H_{2}S\right]$), it can be said that the slow step should involve $1$ molecule of ${Cl}_2$ and $1$ molecule of $H_{2}S$.

If we consider mechanism (A) we find that $1$ molecule of ${Cl}_2$ and $1$ molecule of $H_{2}S$ participate in the reaction. Hence, the rate law expression will be

Rate$=k\left[{Cl}_2\right]\left[H_{2}S\right]$

If we consider mechanism (B) we find

Rate$=k\left[{Cl}_2\right]\left[{HS}^{-}\right]...\left(i\right)$

For equation

$H_{2}S\rightleftharpoons H^{+}+{HS}^{-}$

$K=\frac{\left[H^{+}\right]\left[{HS}^{-}\right]}{\left[H_{2}S\right]}$

or $\left[{HS}^{-}\right]=\frac{K\left[H_{2}S\right]}{\left[H^{+}\right]}$

Substituting this value in equation (i) we find

Rate, $r=k\left[C{l}_2\right]\frac{K\left[H_{2}S\right]}{\left[H^{+}\right]}=k^{\prime }\frac{\left[C{l}_2\right]\left[H_{2}S\right]}{\left[H^{+}\right]}$

This is different from given rate law expression Rate$=k\left[{Cl}_2\right]\left[H_{2}S\right]$.

Hence only, mechanism ($A$) is consistent with the given rate equation.