The following reaction attains equlibrium at high temperature
`N_(2)(g)+2H_(2)O(g)+heathArr2NO(g)+2H_(2)(g)`
The concentration of NO(g)is affected by

Options

A. increasing the nitrogen concentration
B. decreasing the nitrogen concentration
C. Compressing the reaction mixture
D. None of these

Explanation

To determine how the concentration of NO(g) is affected in the given equilibrium reaction:

\[ N_2(g) + 2H_2O(g) + \text{heat} \rightleftharpoons 2NO(g) + 2H_2(g) \]

we will analyze the effects of various changes based on Le Chatelier's principle.

### Step 1: Identify the nature of the reaction
The reaction is endothermic, as indicated by the presence of "heat" on the reactant side. This means that heat is absorbed during the reaction.

**Hint:** Remember that in endothermic reactions, increasing temperature shifts the equilibrium to the right (towards products).

### Step 2: Effect of increasing N2 concentration
If we increase the concentration of \( N_2 \), according to Le Chatelier's principle, the equilibrium will shift to the right to counteract the change. This will result in an increase in the concentration of \( NO \).

**Hint:** Increasing the concentration of a reactant shifts the equilibrium toward the products.

### Step 3: Effect of decreasing N2 concentration
Conversely, if we decrease the concentration of \( N_2 \), the equilibrium will shift to the left to compensate for the change, leading to a decrease in the concentration of \( NO \).

**Hint:** Decreasing the concentration of a reactant shifts the equilibrium toward the reactants.

### Step 4: Effect of compressing the reaction mixture
Compressing the reaction mixture decreases the volume. According to Le Chatelier's principle, the equilibrium will shift toward the side with fewer moles of gas.

In this reaction:
- Reactants: \( N_2(g) + 2H_2O(g) \) = 3 moles of gas
- Products: \( 2NO(g) + 2H_2(g) \) = 4 moles of gas

Since there are more moles of gas on the product side (4 moles) than on the reactant side (3 moles), compressing the mixture will shift the equilibrium to the left, resulting in a decrease in the concentration of \( NO \).

**Hint:** The side with fewer moles of gas is favored when the volume is decreased.

### Conclusion
- Increasing \( N_2 \) concentration increases \( NO \) concentration (Option A is correct).
- Decreasing \( N_2 \) concentration decreases \( NO \) concentration (Option B is correct).
- Compressing the reaction mixture decreases \( NO \) concentration (Option C is incorrect).

Thus, the concentration of \( NO(g) \) is affected by options A and B, while option C does not lead to an increase in \( NO \) concentration.

### Final Answer
The concentration of \( NO(g) \) is affected by:
- A) Increasing \( N_2 \) concentration (increases \( NO \))
- B) Decreasing \( N_2 \) concentration (decreases \( NO \))