For a gaseous reaction
`A(g) + 3 B(g) rightarrow 3C(g) + 3D(g)`
`Delta 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 solve the problem, we need to calculate the change in enthalpy (ΔH) for the given reaction using the relationship between ΔH and ΔE. The reaction is:

\[ A(g) + 3B(g) \rightarrow 3C(g) + 3D(g) \]

Given:
- ΔE = 27 kcal
- Temperature (T) = 37°C = 310 K (after converting to Kelvin)
- R = 2 cal K\(^{-1}\) mol\(^{-1}\)

### Step-by-Step Solution:

1. **Calculate ΔnG (Change in moles of gas)**:
- Count the moles of gaseous products and reactants.
- Products: 3 moles of C + 3 moles of D = 6 moles
- Reactants: 1 mole of A + 3 moles of B = 4 moles
- Therefore, ΔnG = moles of products - moles of reactants = 6 - 4 = 2.

2. **Use the relationship between ΔH and ΔE**:
- The formula is:
\[ \Delta H = \Delta E + \Delta n_G \cdot R \cdot T \]

3. **Substitute the known values into the equation**:
- ΔE = 27 kcal
- ΔnG = 2
- R = 2 cal K\(^{-1}\) mol\(^{-1}\) (convert to kcal: R = 0.002 kcal K\(^{-1}\) mol\(^{-1}\))
- T = 310 K

4. **Calculate the term ΔnG \cdot R \cdot T**:
- Convert R to kcal:
\[ R = 2 \text{ cal K}^{-1} \text{ mol}^{-1} = 0.002 \text{ kcal K}^{-1} \text{ mol}^{-1} \]
- Now calculate:
\[ \Delta n_G \cdot R \cdot T = 2 \cdot 0.002 \cdot 310 \]
\[ = 0.004 \cdot 310 = 1.24 \text{ kcal} \]

5. **Now substitute back into the ΔH equation**:
\[ \Delta H = 27 \text{ kcal} + 1.24 \text{ kcal} = 28.24 \text{ kcal} \]

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