Listed in the table are forward and reverse rate constants for the reaction
`2NO(g)hArrN_(2)(g)+O_(2)(g)`
`{:(,Temperature(K),k_(f)(M^(-1)s^(-1)),k_(b)(M^(-1)s^(-1))),(,1400,0.29,1.1xx10^(-6)),(,1500,1.3,1.4xx10^(-5)):}`
Select the correct statement

Options

A. Reaction is exothermic and value of eqilibrium constatnt`(K_(eq))` at 1400 K is `3.79xx10^(-6)`
B. Reaction is endothermic and value of `k_(eq)` 1400 K is `2.63xx10^(5)`
C. Reaction is exothermic and value of `k_(eq)` 1400 K is `2.63xx10^(5)`
D. Reaction is endothermic and value of `k_(eq)` 1500 K is `9.28xx10^(4)`

Answer

To solve the problem, we need to determine whether the reaction is exothermic or endothermic based on the given rate constants at two different temperatures, and also calculate the equilibrium constant (K_eq) for the reaction at those temperatures.

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction given is:
\[ 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \]

2. **Understand the Rate Constants**: We have the forward rate constant (k_f) and the reverse rate constant (k_b) for two temperatures:
- At 1400 K: \( k_f = 0.29 \, M^{-1}s^{-1} \) and \( k_b = 1.1 \times 10^{-6} \, M^{-1}s^{-1} \)
- At 1500 K: \( k_f = 1.3 \, M^{-1}s^{-1} \) and \( k_b = 1.4 \times 10^{-5} \, M^{-1}s^{-1} \)

3. **Calculate the Equilibrium Constant (K_eq)**:
- The equilibrium constant is calculated using the formula:
\[ K_{eq} = \frac{k_f}{k_b} \]

4. **Calculate K_eq at 1400 K**:
\[ K_{eq} (1400 K) = \frac{0.29}{1.1 \times 10^{-6}} = 263636.36 \approx 2.63 \times 10^5 \]

5. **Calculate K_eq at 1500 K**:
\[ K_{eq} (1500 K) = \frac{1.3}{1.4 \times 10^{-5}} = 92857.14 \approx 9.29 \times 10^4 \]

6. **Analyze the Change in K_eq**:
- At 1400 K, \( K_{eq} \approx 2.63 \times 10^5 \)
- At 1500 K, \( K_{eq} \approx 9.29 \times 10^4 \)

We observe that as the temperature increases from 1400 K to 1500 K, the value of \( K_{eq} \) decreases.

7. **Determine the Type of Reaction**:
- For an exothermic reaction, increasing the temperature causes the equilibrium constant to decrease. This is consistent with Le Chatelier's principle, which states that if a system at equilibrium is subjected to a change in temperature, the equilibrium will shift in the direction that counteracts the change.
- Since \( K_{eq} \) decreased with an increase in temperature, we conclude that the reaction is **exothermic**.

### Conclusion:
- The reaction is exothermic.
- The equilibrium constant at 1400 K is approximately \( 2.63 \times 10^5 \).