`N_(2)+3H_(2)hArr2NH_(3) , K=4xx10^(6) "at" 298K` and `K=41 "at" 400 K`
Which statement is correct?

Options

A. If `N_(2)` is added at equilibrium 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 equilibrium is `2DeltaG_(NH_(3))=3DeltaG_(H_(2)+DeltaG_(N_(2))` where `G` 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 `400K` 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 \( K = 4 \times 10^6 \) at 298 K and \( K = 41 \) at 400 K, we will analyze each statement provided to determine which one is correct.

### Step-by-Step Solution:

1. **Understanding the Reaction and Equilibrium Constants**:
- The reaction is \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \).
- At 298 K, the equilibrium constant \( K = 4 \times 10^6 \) indicates that the formation of ammonia is highly favored at this temperature.
- At 400 K, \( K = 41 \) indicates that the equilibrium is less favorable for ammonia formation compared to 298 K.

2. **Analyzing the First Statement**:
- The first statement claims that adding \( N_2 \) at equilibrium will shift the equilibrium to the forward direction.
- According to Le Chatelier's principle, adding more reactant (in this case, \( N_2 \)) will shift the equilibrium towards the products (forward direction).
- This statement is **correct**.

3. **Analyzing the Second Statement**:
- The second statement discusses the condition for equilibrium and the relationship between Gibbs free energy (\( \Delta G \)).
- At equilibrium, \( \Delta G = 0 \), which means the free energy of the reactants equals the free energy of the products.
- The statement is **correct**.

4. **Analyzing the Third Statement**:
- This statement claims that adding a catalyst does not change \( K_p \) but changes \( \Delta G \).
- While it is true that a catalyst does not change the equilibrium constant \( K \), it does not change \( \Delta G \) either. Therefore, this statement is **incorrect**.

5. **Analyzing the Fourth Statement**:
- The fourth statement suggests that at 400 K, the addition of a catalyst will increase the forward reaction rate by two times and the reverse reaction rate by 1.7 times.
- A catalyst increases the rate of both the forward and reverse reactions equally, thus not changing the equilibrium position or the ratio of rates. This statement is also **incorrect**.

### Conclusion:
After analyzing all the statements, we find that the first and second statements are correct. However, since the question asks for which statement is correct and does not specify how many, we conclude that:

- The **correct statements** are:
- **First Statement**: Adding \( N_2 \) at equilibrium shifts the equilibrium to the forward direction.
- **Second Statement**: The condition for equilibrium holds true as \( \Delta G = 0 \).

### Final Answer:
The correct statements are the first and second ones.

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