A) \[\frac{1}{64}\]
B) 16
C) \[\frac{1}{8}\]
D) \[\frac{1}{16}\]
Correct Answer: A
Solution :
For \[HI\rightleftharpoons \frac{1}{2}{{H}_{2}}+\frac{1}{2}{{I}_{2}};\,\,{{K}_{{{c}_{1}}}}=8\] Reverse given eqn. and multiply by 2. \[\therefore \,\]For \[{{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI;{{K}_{{{C}_{2}}}}={{\left( \frac{1}{{{K}_{{{c}_{1}}}}} \right)}^{2}}\] \[={{\left( \frac{1}{8} \right)}^{2}}=\frac{1}{64}\]You need to login to perform this action.
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