Answer:
\[{{K}_{c}}=\frac{[C{{H}_{3}}OH]}{[CO]\,{{[{{H}_{2}}]}^{2}}},\,\,{{K}_{p}}\,=\frac{{{p}_{C{{H}_{3}}OH}}}{{{p}_{CO}}\times p_{{{H}_{2}}}^{2}}\]
(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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