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 find the heat of reaction at constant pressure (ΔH) given the heat of reaction at constant volume (ΔU), we can use the following relationship:

\[ \Delta U = \Delta H - \Delta N_g RT \]

Where:
- ΔU = heat of reaction at constant volume
- ΔH = heat of reaction at constant pressure
- ΔN_g = change in the number of moles of gas (moles of products - moles of reactants)
- R = universal gas constant (8.314 J/(K·mol))
- T = temperature in Kelvin

### Step-by-Step Solution

1. **Identify the given values:**
- ΔU = 40 kJ = 40,000 J (since 1 kJ = 1000 J)
- Temperature (T) = 300 K
- R = 8.314 J/(K·mol)

2. **Determine ΔN_g:**
- From the reaction \( A(s) + B(g) \rightarrow 2C(g) \):
- Moles of gaseous products (C) = 2
- Moles of gaseous reactants (B) = 1
- Therefore, ΔN_g = moles of products - moles of reactants = 2 - 1 = 1

3. **Substitute the values into the equation:**
- Rearranging the equation gives us:
\[ \Delta H = \Delta U + \Delta N_g RT \]
- Substitute the known values:
\[ \Delta H = 40,000 \, \text{J} + (1)(8.314 \, \text{J/(K·mol)})(300 \, \text{K}) \]

4. **Calculate the second term:**
- Calculate \( \Delta N_g RT \):
\[ \Delta N_g RT = 1 \times 8.314 \times 300 = 2494.2 \, \text{J} \]

5. **Calculate ΔH:**
- Now substitute this back into the equation for ΔH:
\[ \Delta H = 40,000 \, \text{J} + 2494.2 \, \text{J} = 42,494.2 \, \text{J} \]

6. **Convert ΔH to kJ:**
- Since we need the answer in kJ:
\[ \Delta H = \frac{42,494.2 \, \text{J}}{1000} = 42.4942 \, \text{kJ} \]

7. **Round the answer:**
- Rounding to three significant figures gives:
\[ \Delta H \approx 42.5 \, \text{kJ} \]

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