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BITSAT 2016 Paper

Section: Chemistry

Question : 73 of 150

Marks: +1, -0

\mathrm{Cl}_2(aq)+\mathrm{H}_2\mathrm{S}(aq)\rightarrow\mathrm{S}(s)+2\mathrm{H}^+(aq)+2\mathrm{Cl}^-(aq)

=k[\mathrm{Cl}_2][\mathrm{H}_2\mathrm{S}]

\mathrm{Cl}_2+\mathrm{H}_2\mathrm{S}\rightarrow\mathrm{H}^++\mathrm{Cl}^-+\mathrm{Cl}^++\mathrm{HS}^-

\mathrm{Cl}^++\mathrm{HS}^-\rightarrow\mathrm{H}^++\mathrm{Cl}^-+\mathrm{S}

\mathrm{H}_2\mathrm{S}\rightleftharpoons\mathrm{H}^++\mathrm{HS}^-

\mathrm{Cl}_2+\mathrm{HS}^-\rightarrow2\mathrm{Cl}^-+\mathrm{H}^++\mathrm{S}

(Slow)

Solution:

Since the slow step is the rate determining step hence if we consider option (A) we find

Rate

=k[\mathrm{Cl}_2][\mathrm{H}_2\mathrm{S}]

Now if we consider option (B) we find

Rate

=k[\mathrm{Cl}_2][\mathrm{HS}^-]

....(i)

For equation,

\mathrm{H}_2\mathrm{S}\rightleftharpoons\mathrm{H}^++\mathrm{HS}^-

K=\frac{[\mathrm{H}^+][\mathrm{HS}^-]}{\mathrm{H}_2\mathrm{S}}

[\mathrm{HS}^-]=\frac{K[\mathrm{H}_2\mathrm{S}]}{\mathrm{H}^+}

Substituting this value in equation (i) we find

Rate

=\frac{k[\mathrm{Cl}_2]K[\mathrm{H}_2\mathrm{S}]}{\mathrm{H}^+}=\frac{k'[\mathrm{Cl}_2][\mathrm{H}_2\mathrm{S}]}{[\mathrm{H}^+]}

Thus slow step should involve 1 molecule of

\mathrm{Cl}_2

and 1 molecule of

\mathrm{H}_2\mathrm{S}

hence only , mechanism (A) is consistent with the given rate equation

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