For the gas phase reaction `2NOhArrN_(2)+O_(2), DeltaH^(@)=-"43.5 kcal mole"^(-1)` Which one of the stateements below is true for `N_(2)(g)+O_(2)(g)hArr 2NO(g)`

Options

A. `K_(C)` in independent to T
B. `K_(C)` increases as T decreases
C. `K_(C)` deacreases as T decreases
D. `K_(C)` varies with addition of NO

Answer

To solve the question regarding the gas phase reaction \(2NO \rightleftharpoons N_2 + O_2\) with \(\Delta H^\circ = -43.5 \, \text{kcal/mol}\), we need to analyze the reverse reaction \(N_2 + O_2 \rightleftharpoons 2NO\) and determine the correct statement about the equilibrium constant \(K_c\) for this reaction.

### Step-by-Step Solution:

1. **Identify the Reaction and Its Enthalpy Change**:
- The given reaction is:
\[ 2NO \rightleftharpoons N_2 + O_2 \]
- The enthalpy change (\(\Delta H^\circ\)) for this reaction is \(-43.5 \, \text{kcal/mol}\), indicating that it is an exothermic reaction.

2. **Determine the Enthalpy Change for the Reverse Reaction**:
- For the reverse reaction:
\[ N_2 + O_2 \rightleftharpoons 2NO \]
- The enthalpy change will be the opposite of the forward reaction:
\[ \Delta H^\circ = +43.5 \, \text{kcal/mol} \]
- This indicates that the reverse reaction is endothermic.

3. **Analyze the Temperature Dependence of the Equilibrium Constant**:
- According to the van 't Hoff equation, the relationship between the equilibrium constant (\(K_c\)) and temperature is given by:
\[ \log \frac{K_2}{K_1} = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
- Here, \(K_1\) is the equilibrium constant at temperature \(T_1\) and \(K_2\) is at temperature \(T_2\).

4. **Consider the Effect of Increasing Temperature**:
- Since \(\Delta H\) for the reverse reaction is positive, increasing the temperature (\(T_2 > T_1\)) will lead to an increase in \(K_c\) for the endothermic reaction:
\[ K_2 > K_1 \]
- This indicates that the equilibrium shifts to the right (towards the products) with an increase in temperature.

5. **Evaluate the Given Statements**:
- **Option 1**: \(K_c\) is independent of \(T\) - Incorrect.
- **Option 2**: \(K_c\) increases as \(T\) decreases - Incorrect (they have a direct relationship).
- **Option 3**: \(K_c\) decreases as \(T\) decreases - Correct (since \(K_c\) increases with temperature for endothermic reactions).
- **Option 4**: \(K_c\) varies with the addition of \(NO\) - Incorrect (the equilibrium constant is only affected by temperature).

### Conclusion:
The correct statement is that \(K_c\) decreases as the temperature decreases.