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

`([CH_3OH])/([CO][H_2]^2)` and `K_p=(p_(CH_3OH))/(p_(CO)xxp_(H_2^2))`
(i) When volume of the vessel is reduced to half, the concentration of the species becomes double .
`Q_c=(2[CH_3OH])/(2[CO](2[H_2I])^2)=1/4K_c`
As `Q lt K_c`, equilibrium will shift in the forward direction producing more of `CH_3OH`.
(ii)When `pH_2` is made double , then
`Q_p=(p_(CH_3OH))/(p_(CO)xx (2p_(H_2))^2)=1/4K_p`
As `Q_c lt K_p` equilibrium will shift in the forward direction.
(iii) As volume remains constant , molar concentration will not change . Hence, no effect on equilibrium state.