Answer:
Reciprocating the reaction:\[\underset{At\,equilibrium}{\overset{{}}{\mathop{{}}}}\,\underset{0.5M}{\mathop{2HI(g)}}\,\rightleftharpoons
\underset{x}{\mathop{{{H}_{2}}(g)}}\,+\underset{x}{\mathop{{{I}_{2}}(g)}}\,;{{K}_{c}}=\frac{1}{54.8};\]
\[{{K}_{c}}=\frac{[{{H}_{2}}][{{I}_{2}}]}{{{[HI]}^{2}}}\]
\[\frac{1}{54.8}=\frac{x\times
x}{{{(0.5)}^{2}}}\]
\[\frac{1}{54.8}=\frac{x\times
x}{{{(0.5)}^{2}}}\]
\[x=0.0675M\]
\[\therefore \] \[[{{H}_{2}}]=[{{I}_{2}}]=0.0675M\]
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