The following reaction has attained equilibrium
`CO(g)+2H_(2)(g)hArrCH_(3)OH(g)`. `DeltaH^(@)=-92.0KJmol^(-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 at constant volume.

Explanation

`K_(C)=([CH_(3)OH])/([CO][H_(2)]^(2)) K_(P)=P_(CH_(3)OH)/P_(COxxP_(H_(2))`
(i) When volume of the vessel is reduced to half, the concentration of each reactant of product becomes double. Thus. ltbgt `Q_(C)=(2[CH_(3)OH])/(2[CO]xx{2|H_(2)|}^(2))=(1)/(2)K_(c)`
As `Q_(c)ltK_(c1)` equliibrium will shift in the forward direction producing more of `CH_(3)OH` to make`Q_(c)=K_(c)`
(ii) `Q_(P)=P_(CH_(3)OH)/P_(CO)xx(1)/(2P_(H_(2))^(2))=(1)/(4)K_(c)`
Again`Q_(P)ltK_(P1)` equilibrium will shift in the forward direction producing more of `CH_(3)OH "to make" Q_(P)=K_(P)`
(iii) As volume remains constant molar concentrations will not change. Hence there is not not effect on the state of equilibrium.