The ratio `K_(p)` to `K_( c)` of a reaction is 24.63 L atm `mol^(-1)` at `27^(@)` C. If heat of reaction at constant pressure is 98.8 kcal, what is the heat of reaction (in kcal) at constant volume?

Answer

To solve the problem, we need to find the heat of reaction at constant volume (\( \Delta U \)) given the heat of reaction at constant pressure (\( \Delta H \)) and the ratio of \( K_p \) to \( K_c \).

### Step-by-Step Solution:

1. **Identify Given Data**:
- \( \frac{K_p}{K_c} = 24.63 \, \text{L atm mol}^{-1} \)
- \( \Delta H = 98.8 \, \text{kcal} \)
- Temperature \( T = 27^\circ C = 300 \, \text{K} \)
- Ideal gas constant \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)

2. **Use the Relationship Between \( K_p \) and \( K_c \)**:
The relationship between \( K_p \) and \( K_c \) is given by:
\[ \frac{K_p}{K_c} = R T^{\Delta N} \]
where \( \Delta N \) is the change in the number of moles of gas (moles of products - moles of reactants).

3. **Calculate \( \Delta N \)**:
Rearranging the formula gives:
\[ \Delta N = \frac{\log \left( \frac{K_p}{K_c} \right)}{\log (RT)} \]
First, calculate \( RT \):
\[ RT = 0.0821 \times 300 = 24.63 \]
Now, take the logarithm:
\[ \log \left( \frac{K_p}{K_c} \right) = \log(24.63) \approx 1.391 \]
\[ \log(24.63) = \log(600^{\Delta N}) \implies \Delta N \log(600) \approx 1.391 \]
\[ \log(600) \approx 2.778 \]
Now, solving for \( \Delta N \):
\[ \Delta N = \frac{1.391}{2.778} \approx 0.501 \]

4. **Use the Relationship Between \( \Delta H \) and \( \Delta U \)**:
The relationship between the heat of reaction at constant pressure and constant volume is given by:
\[ \Delta H = \Delta U + \Delta N \cdot R \cdot T \]
Rearranging gives:
\[ \Delta U = \Delta H - \Delta N \cdot R \cdot T \]

5. **Substitute Values**:
Convert \( \Delta H \) to calories:
\[ \Delta H = 98.8 \, \text{kcal} = 98800 \, \text{cal} \]
Now substitute \( \Delta N \), \( R \), and \( T \):
\[ \Delta U = 98800 - (0.501) \cdot (2) \cdot (300) \]
\[ = 98800 - (0.501 \cdot 600) = 98800 - 300.6 = 98500.4 \, \text{cal} \]
Converting back to kcal:
\[ \Delta U \approx 98.5004 \, \text{kcal} \]

### Final Answer:
The heat of reaction at constant volume is approximately \( \Delta U \approx 98.50 \, \text{kcal} \).