| (I) \[PC{{l}_{5}}(g)\rightleftarrows PC{{l}_{3}}(g)+C{{l}_{2}}(g)\] |
| (II) \[2HI(g)\rightleftarrows {{H}_{2}}(g)+{{I}_{2}}(g)\] |
| (III) \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftarrows 2N{{H}_{3}}(g)\] |
| Extent of the reactions taking place is |
A) I > II > III
B) I < II < III
C) II < III < I
D) III < I < II
Correct Answer: B
Solution :
[b] Extent of reaction can be calculated by the value of \[{{K}_{C}}\]. The higher value of \[{{K}_{C}}\], the larger the extent of reactions. \[PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)\] \[\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{0.01\times 0.01}{0.01}=0.01\] (II) \[2HI\rightleftharpoons {{H}_{2}}+{{I}_{2}},{{K}_{C}}=\frac{[{{H}_{2}}][{{I}_{2}}]}{{{[HI]}^{2}}}\] \[=\frac{(0.01)(0.01)}{{{(0.01)}^{2}}}=1\] (III) \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\] \[{{K}_{C}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}=\frac{{{(0.01)}^{2}}}{{{(0.01)}^{4}}}=\frac{1}{0.01\times 0.01}=10000\] Extent of reaction I<II<III.
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