Heat of reaction `A(s) + B(g) rarr 2C(g)` is 40 kJ at 300 K and constant volume. Hence, heat of reaction at constant pressure and at 300 K is,

Options

A. 42.5 kJ
B. 37.5 kJ
C. `40.0 kJ`
D. `30.0 kJ`

Explanation

To solve the problem of finding the heat of reaction at constant pressure for the reaction \( A(s) + B(g) \rightarrow 2C(g) \) given that the heat of reaction at constant volume is 40 kJ at 300 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Heat of reaction at constant volume (\( \Delta U \)) = 40 kJ
- Temperature (\( T \)) = 300 K
- Gas constant (\( R \)) = 8.314 J/(mol·K)

2. **Determine \( \Delta N_g \):**
- Calculate the change in the number of moles of gaseous products and reactants (\( \Delta N_g \)).
- In the reaction, there are 2 moles of \( C(g) \) produced and 1 mole of \( B(g) \) consumed.
- Therefore, \( \Delta N_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - 1 = 1 \).

3. **Use the Relationship Between \( \Delta H \) and \( \Delta U \):**
- The relationship at constant temperature is given by:
\[ \Delta H = \Delta U + \Delta N_g \cdot R \cdot T \]
- Substitute the known values into the equation:
\[ \Delta H = 40 \text{ kJ} + (1) \cdot (8.314 \text{ J/(mol·K)}) \cdot (300 \text{ K}) \]

4. **Convert Units:**
- Convert \( R \) from J to kJ for consistency:
\[ R = 8.314 \text{ J/(mol·K)} = 0.008314 \text{ kJ/(mol·K)} \]
- Now calculate \( \Delta N_g \cdot R \cdot T \):
\[ \Delta N_g \cdot R \cdot T = 1 \cdot 0.008314 \text{ kJ/(mol·K)} \cdot 300 \text{ K} = 2.4942 \text{ kJ} \]

5. **Calculate \( \Delta H \):**
- Now substitute back into the equation for \( \Delta H \):
\[ \Delta H = 40 \text{ kJ} + 2.4942 \text{ kJ} = 42.4942 \text{ kJ} \]

6. **Final Answer:**
- Rounding to three significant figures, the heat of reaction at constant pressure is:
\[ \Delta H \approx 42.5 \text{ kJ} \]

### Conclusion:
The heat of reaction at constant pressure and at 300 K is approximately **42.5 kJ**.