A) \[\frac{1}{16}\]
B) \[\frac{1}{64}\]
C) 16
D) \[\frac{1}{8}\]
Correct Answer: B
Solution :
\[HI(g)\frac{1}{2}{{H}_{2}}(g)+\frac{1}{2}{{I}_{2}}(g)\] \[{{K}_{1}}=\frac{{{[{{H}_{2}}]}^{1/2}}{{[{{I}_{2}}]}^{1/2}}}{[HI]}\] \[{{H}_{2}}(g)+{{I}_{2}}(g)2HI(g)\] \[{{K}_{2}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}\] ?(ii) From Eqs (i) and (ii) \[K_{1}^{2}=\frac{1}{{{K}_{2}}}\] \[\because \] \[\therefore \]\[{{K}_{2}}=\frac{1}{K_{1}^{2}}=\frac{1}{{{8}^{2}}}=\frac{1}{64}\]
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