Answer:
The given reaction is:
\[{{N}_{2}}{{O}_{4}}(g)\]
\[\rightleftharpoons \]
\[2N{{O}_{2}}(g)\]
\[{{t}_{initial}}\]
\[1\]
\[0\]
\[{{t}_{equilibrium}}\]
\[1-0.5\]
\[2\times 0.5\]
Total mole =
\[0.5+1=\,1.5\]
Mole fraction
\[\frac{0.5}{1.5}\]
\[\frac{1}{1.5}\]
Partial pressure
\[\frac{0.5}{1.5}\times 1\]
\[\frac{1}{1.5}\times 1\]
(Partial
pressure of gas = Mole fraction of gas \[\times \]Total pressure)
\[{{K}_{p}}=\frac{{{({{p}_{N{{O}_{2}}}})}^{2}}}{{{p}_{{{N}_{2}}{{O}_{5}}}}}=\frac{{{\left(
\frac{1}{1.5} \right)}^{2}}}{\left( \frac{0.5}{1.5} \right)}\]
\[=\frac{1}{0.5\times
1.5}=1.33\]
\[\Delta
{{G}^{{}^\circ }}=-2.303RT{{\log }_{10}}{{K}_{p}}\]
\[=-2.303\times
8.314\times 333\log 1.33\]
\[=-789.67kJ\,mo{{l}^{-1}}\]
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