For a gaseous reaction
`A(g) + 3 B(g) rightarrow 3C(g) + 3D(g)`
`Delat E` is 27 kcal at `37^(@)C`. Assuming `R = 2 cal K^(-1) mol^(-1)` the value of `Delta H` for the above reaction will be

Options

A. `15.8 kcal`
B. `28.2 kcal`
C. `25.7 kcal`
D. `20.0 kcal`

Answer

To find the value of ΔH for the given gaseous reaction, we will use the relationship between the change in internal energy (ΔE) and the change in enthalpy (ΔH). The formula we will use is:

\[ \Delta H = \Delta E + P \Delta V \]

### Step-by-Step Solution:

1. **Identify Given Values**:
- ΔE = 27 kcal
- Temperature (T) = 37°C = 310 K (since 37 + 273 = 310)
- R = 2 cal K⁻¹ mol⁻¹

2. **Convert ΔE to Calories**:
- Since ΔE is given in kcal, we convert it to calories:
\[ \Delta E = 27 \text{ kcal} = 27 \times 1000 \text{ cal} = 27000 \text{ cal} \]

3. **Determine Δn_g (Change in Moles of Gas)**:
- From the reaction:
\[ A(g) + 3B(g) \rightarrow 3C(g) + 3D(g) \]
- Moles of gaseous products = 3 (from C) + 3 (from D) = 6
- Moles of gaseous reactants = 1 (from A) + 3 (from B) = 4
- Therefore,
\[ \Delta n_g = \text{Moles of products} - \text{Moles of reactants} = 6 - 4 = 2 \]

4. **Calculate PΔV**:
- Using the ideal gas law, we know that:
\[ P \Delta V = \Delta n_g \cdot R \cdot T \]
- Substitute the values:
\[ P \Delta V = 2 \cdot (2 \text{ cal K}^{-1} \text{ mol}^{-1}) \cdot (310 \text{ K}) = 2 \cdot 2 \cdot 310 = 1240 \text{ cal} \]

5. **Calculate ΔH**:
- Now substitute ΔE and PΔV into the ΔH equation:
\[ \Delta H = \Delta E + P \Delta V = 27000 \text{ cal} + 1240 \text{ cal} = 28240 \text{ cal} \]

6. **Convert ΔH back to kcal**:
- To convert calories back to kilocalories:
\[ \Delta H = \frac{28240 \text{ cal}}{1000} = 28.24 \text{ kcal} \]

### Final Answer:
\[ \Delta H = 28.24 \text{ kcal} \]

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