question_answer 1)

At 700 K, equilibrium constant for the reaction: \[{{H}_{2}}(g)+{{I}_{2}}(g)\rightleftharpoons 2HI(g)\] is 54.8. If \[0.5mol\,{{L}^{-1}}\]of\[HI\left( g \right)\]is present at equilibrium at700 K. What is the concentration of \[{{H}_{2}}(g)\]and \[{{I}_{2}}(g)\]assuming that we initially started with \[HI\left( g \right)\]and allowed it to reach equilibrium at 700 K?

Answer:

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