A) \[2Ag+2HCl+(O)\xrightarrow{{}}2AgCl+{{H}_{2}}O\]
B) \[N{{i}_{2}}{{O}_{2}}(s)+{{H}_{2}}O(l)+2{{e}^{-}}2NiO(s)+2O{{H}^{-}}\]
C) \[{{E}^{o}}=+0.40V\]
D) \[FeO(s)+{{H}_{2}}O(l)+2{{e}^{-}}Fe(s)+2O{{H}^{-}};\]
Correct Answer: C
Solution :
\[\underset{\begin{smallmatrix} p-\left( \frac{2p}{9}+\frac{p}{9} \right) \\ =\frac{6p}{9} \end{smallmatrix}}{\mathop{2NOBr(g)}}\,\underset{\frac{2p}{9}}{\mathop{2NO(g)}}\,+\underset{\frac{p}{9}}{\mathop{B{{r}_{2}}(g)}}\,\] From \[{{K}_{p}}=\frac{{{({{p}_{No}})}^{2}}\times ({{p}_{B{{r}_{2}}}})}{{{({{p}_{NOBr}})}^{2}}}\] \[=\frac{{{\left( \frac{2p}{9} \right)}^{2}}\left( \frac{p}{9} \right)}{{{\left( \frac{6p}{9} \right)}^{2}}}=\frac{p}{81}\] \[\therefore \] \[\frac{{{K}_{p}}}{p}=\frac{1}{81}\]You need to login to perform this action.
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