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 find the heat of reaction at constant volume (ΔU) given the heat of reaction at constant pressure (ΔH) and the ratio of Kp to Kc, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- ΔH (heat of reaction at constant pressure) = 98.8 kcal
- Kp/Kc = 24.63 L atm mol^(-1)
- Temperature (T) = 27°C = 300 K (since 27 + 273 = 300)
- Universal gas constant (R) = 0.0821 L atm K^(-1) mol^(-1)

2. **Use the Relationship Between Kp and Kc:**
The relationship between Kp and Kc is given by the equation:
\[ K_p = K_c \cdot R \cdot T^{\Delta N_g} \]
Rearranging gives:
\[ \frac{K_p}{K_c} = R \cdot T^{\Delta N_g} \]
We can substitute the known values into this equation.

3. **Substituting Values:**
Substitute Kp/Kc, R, and T into the equation:
\[ 24.63 = R \cdot T^{\Delta N_g} \]
\[ 24.63 = 0.0821 \cdot 300^{\Delta N_g} \]

4. **Solve for ΔN_g:**
Rearranging gives:
\[ 24.63 = 24.63 \cdot \Delta N_g \]
Dividing both sides by 24.63:
\[ 1 = 0.0821 \cdot 300^{\Delta N_g} \]
Solving for ΔN_g:
\[ \Delta N_g = 1 \]

5. **Use the Thermodynamic Relation:**
The relationship between ΔH and ΔU is given by:
\[ \Delta H = \Delta U + \Delta N_g \cdot R \cdot T \]
Rearranging gives:
\[ \Delta U = \Delta H - \Delta N_g \cdot R \cdot T \]

6. **Substituting Known Values:**
Substitute ΔH, ΔN_g, R, and T into the equation:
\[ \Delta U = 98.8 \, \text{kcal} - (1) \cdot (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \cdot (300 \, \text{K}) \]
Convert ΔH into calories for consistency:
\[ 98.8 \, \text{kcal} = 98800 \, \text{cal} \]
Now calculate:
\[ \Delta U = 98800 - (0.0821 \cdot 300) \]
\[ \Delta U = 98800 - 2463 \]
\[ \Delta U = 96337 \, \text{cal} \]

7. **Convert Back to Kcal:**
\[ \Delta U = \frac{96337}{1000} \, \text{kcal} = 96.34 \, \text{kcal} \]

### Final Answer:
The heat of reaction at constant volume (ΔU) is approximately **96.34 kcal**.