`N_(2) + 3H_(2) hArr 2NH_(3)" "K=4xx 10^(6)"at"298`
`" "K=41 "at" 400 k`
Which statements is correct?

Options

A. If `N_(2)` is added at equlibrium condition, the equilibrium will shift to the forward direction because according to `II^(nd)` law of thermodynamics the entropy must increases in the direction of spontaneous reaction .
B. The condition for equlibrium is `2DeltaG_(NH_(3)) = 3DeltaG_(N_(2)) + DeltaG_(N_(2))` where G is Gibbs free energy per mole of the gaseous species measured at that partial pressure.
C. Addition of catalyst does not change `K_(p)` but changes `DeltaH`.
D. At 400 K addition of catalyst will increase forward reaction by `2` times while reverse reaction rate will be changed by 1.7 times.

Explanation

To solve the question regarding the equilibrium reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) with given equilibrium constants at different temperatures, we will analyze the statements provided and determine which one is correct.

### Step-by-Step Solution:

1. **Understanding the Reaction and Equilibrium Constant (K)**:
The reaction given is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
The equilibrium constant \( K \) is given as:
- \( K = 4 \times 10^6 \) at \( 298 \, K \)
- \( K = 41 \) at \( 400 \, K \)

2. **Analyzing the Effect of Adding \( N_2 \)**:
According to Le Chatelier's Principle, if we add more \( N_2 \) at equilibrium, the system will shift to the right to produce more \( NH_3 \). Thus, the statement that "if \( N_2 \) is added at equilibrium, the equilibrium will shift forward" is true based on Le Chatelier's Principle.

3. **Evaluating the Condition for Equilibrium**:
The condition for equilibrium involves the Gibbs free energy change. At equilibrium, the change in Gibbs free energy (\( \Delta G \)) is zero. The statement regarding \( 2 \Delta G_{NH_3} = 3 \Delta G_{N_2} + \Delta G_{N_2} \) does not make sense in the context of equilibrium, as it misrepresents the relationship between the Gibbs free energies of the reactants and products. Therefore, this statement is incorrect.

4. **Impact of a Catalyst on \( K_p \)**:
A catalyst speeds up the rate of both the forward and reverse reactions equally but does not change the equilibrium constant \( K_p \). The equilibrium constant is only affected by temperature changes, not by the presence of a catalyst. Thus, the statement that "addition of catalyst does not change \( K_p \) but changes \( \Delta H \)" is incorrect because a catalyst does not change \( \Delta H \) either; it only affects the rate of reaction.

5. **Rate Changes with a Catalyst**:
The last statement suggests that at \( 400 \, K \), the rate of the forward reaction increases by 2 times while the reverse reaction changes by 1.7 times. This statement is misleading because a catalyst does not change the rates of reactions in such a manner; it affects both rates equally, maintaining the ratio defined by the equilibrium constant.

### Conclusion:
After analyzing all statements, the correct statement is:
- **Addition of catalyst does not change \( K_p \)**.

### Final Answer:
The correct statement is: **Addition of catalyst does not change \( K_p \)**.