`N_(2)+3H_(2) hArr 2NH_(3)`
Which of the following statements is correct if `N_(2)` added at equilibrium condition?

Options

A. The equilibrium will shift in the forward firection because according to the second law of thermodynamics, the entropy must increase in the direction of the spontaneous reaction.
B. The condition of equilibrium is `G_(N_(2)) +3G_(H_(2)) =2G_(NH_(3))`, where `G` is the Gibbs frre energy per mole of the gaseous species measured at partial pressure. The condition of equilibrium is unafected by the use of catlyst which increases the rate of both the forward and backward reactions to the same extent.
C. The catalyst will increase the rate of forward reaction by `alpha` and that of backward reaction by `beta`.
D. The catalyst will not alter the rate of either of the reactions.

Explanation

`N_(2)+3H_(2) hArr 2NH_(3)`
`DeltaG = sum Delta ("products") -sum G("reactants")`
`DeltaG = 2 xxG (NH_(3)) - [G(N_(2)) +[3 xx G(H_(2))]`
At equilibrium, `DeltaG = 0`
`0 = 2 xx G(NH_(3)) - [G(N_(2)) +{3 xx G(H_(2))}]`
`[G(N_(2))+ {3 xx G(H_(2))}] = 2 xx G(NH_(3))`
The condition of equilibrium is unaffected by the use of catalyst which increases the rate of both the forward and backward reactions so the same extent.