The reaction `2NO_(2)(g) hArr N_(2)O_(4)(g)` is an exothermic equilibrium . This means that:

Options

A. equilibration of this gas mixture will be slower at high temperature
B. A mole of `N_(2)O_(4)`will occupy twice the volume of a mole of `NO_(2)` at the same temperature.
C. the equilibrium will move to the right if an equilibrium maxture is cooled
D. the postion of equilibrium will move to the left with increasing gas pressure

Answer

To solve the question regarding the exothermic equilibrium reaction \(2NO_2(g) \rightleftharpoons N_2O_4(g)\), we will analyze the implications of the reaction being exothermic and apply Le Chatelier's principle to determine the correct statement among the provided options.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction \(2NO_2(g) \rightleftharpoons N_2O_4(g)\) is an equilibrium reaction where two moles of nitrogen dioxide (\(NO_2\)) gas react to form one mole of dinitrogen tetroxide (\(N_2O_4\)) gas. Since the reaction is exothermic, it releases heat.

**Hint**: Remember that in an exothermic reaction, heat can be considered a product.

2. **Effect of Temperature on Equilibrium**:
According to Le Chatelier's principle, if the temperature of an exothermic reaction is decreased (the system is cooled), the equilibrium will shift to the right to produce more heat, favoring the formation of \(N_2O_4\).

**Hint**: Cooling an exothermic reaction favors the forward reaction.

3. **Analyzing the Options**:
- **Option A**: "Equilibrium of this gaseous mixture will be slower at high temperature."
- This statement is incorrect. While the equilibrium constant may change, the rate of the reaction does not necessarily slow down at higher temperatures.

- **Option B**: "A mole of \(N_2O_4\) will occupy twice the volume of a mole of \(NO_2\) at the same temperature."
- This statement is also incorrect. According to the ideal gas law, at the same temperature and pressure, the volume is directly proportional to the number of moles. Since 2 moles of \(NO_2\) produce 1 mole of \(N_2O_4\), the volume of \(NO_2\) will be greater.

- **Option C**: "The equilibrium will move to the right if the equilibrium mixture is cooled."
- This statement is correct. Cooling the system favors the formation of \(N_2O_4\) because it is an exothermic reaction.

- **Option D**: "The position of equilibrium will move to the left with increasing gas pressure."
- This statement is incorrect. Increasing pressure favors the side with fewer moles of gas. Since there are 2 moles of \(NO_2\) and 1 mole of \(N_2O_4\), increasing pressure will shift the equilibrium to the right.

4. **Conclusion**:
The only correct statement is Option C: "The equilibrium will move to the right if the equilibrium mixture is cooled."

### Final Answer:
The correct option is **C**: The equilibrium will move to the right if the equilibrium mixture is cooled.