• # question_answer The following reaction has attained equilibrium $CO(g)+2{{H}_{2}}(g)\rightleftharpoons C{{H}_{3}}OH(g),\,\Delta {{H}^{{}^\circ }}=-92.0kJmo{{l}^{-1}}$ What will happen if (i) volume of the reaction vessel is suddenly reduced to half? (ii) the partial pressure of hydrogen is suddenly doubled? (iii) an inert gas is added to the system?

${{K}_{c}}=\frac{[C{{H}_{3}}OH]}{[CO]\,{{[{{H}_{2}}]}^{2}}},\,\,{{K}_{p}}\,=\frac{{{p}_{C{{H}_{3}}OH}}}{{{p}_{CO}}\times p_{{{H}_{2}}}^{2}}$ (i) When volume of the vessel is reduced to half, the concentration of each reactant or product becomes double. Thus,${{Q}_{c}}=\frac{2[C{{H}_{3}}OH]}{2[CO]\times {{\{2[{{H}_{2}}]\}}^{2}}}=\frac{1}{4}{{K}_{c}}$ As ${{Q}_{c}}<{{K}_{c}},$ equilibrium will shift in the forward direction, producing more of  $C{{H}_{3}}OH$ to make ${{Q}_{c}}={{K}_{c}}$. (ii) ${{Q}_{p}}=\frac{{{p}_{C{{H}_{3}}OH}}}{{{p}_{CO}}}\times {{(2{{p}_{{{H}_{2}}}})}^{2}}=\frac{1}{4}{{K}_{p}}.$ Again, ${{Q}_{p}}<{{K}_{p}},$equilibrium will shift in the forward direction to make ${{Q}_{p}}={{K}_{p}}$. (iii) As volume remains constant, molar concentrations will not change. Hence, there is no effect on the state of equilibrium.
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