For a gaseous reaction at 300K, `DeltaH- DeltaU = -4.98` kJ , assuming that `R = 8.3 JK^(-1) mol^(-1), Delta n_((g))` is

Options

A. 1
B. 2
C. `-2`
D. 0

Explanation

To solve the problem, we need to find the change in the number of moles of gas, denoted as \( \Delta n_g \), using the relationship between enthalpy change (\( \Delta H \)), internal energy change (\( \Delta U \)), and the ideal gas constant \( R \).

### Step-by-Step Solution:

1. **Understand the Relationship**:
The relationship between enthalpy change and internal energy change for a gaseous reaction is given by:
\[ \Delta H - \Delta U = \Delta n_g \cdot R \cdot T \]
where:
- \( \Delta H \) is the change in enthalpy,
- \( \Delta U \) is the change in internal energy,
- \( \Delta n_g \) is the change in the number of moles of gas,
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin.

2. **Substituting Known Values**:
From the problem, we know:
\[ \Delta H - \Delta U = -4.98 \text{ kJ} \]
Convert this value to Joules (since \( R \) is in Joules):
\[ -4.98 \text{ kJ} = -4980 \text{ J} \]

3. **Plugging in the Values**:
We can rearrange the equation to solve for \( \Delta n_g \):
\[ \Delta n_g = \frac{\Delta H - \Delta U}{R \cdot T} \]
Substituting the values:
- \( R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1} \)
- \( T = 300 \text{ K} \)
\[ \Delta n_g = \frac{-4980 \text{ J}}{8.3 \text{ J K}^{-1} \text{ mol}^{-1} \cdot 300 \text{ K}} \]

4. **Calculating \( \Delta n_g \)**:
Calculate the denominator:
\[ 8.3 \cdot 300 = 2490 \text{ J mol}^{-1} \]
Now, calculate \( \Delta n_g \):
\[ \Delta n_g = \frac{-4980}{2490} \approx -2.00 \text{ mol} \]

5. **Conclusion**:
The change in the number of moles of gas (\( \Delta n_g \)) is approximately -2.00 mol. This indicates a decrease in the number of moles of gas during the reaction.

### Final Answer:
\[ \Delta n_g \approx -2.00 \text{ mol} \]