An acid reacts with glycerine to form complex and equilibrium is established. If the heat of reaction at constant volume for this reaction is 1200 cal more than at constant pressure and the temperature is 300 K, then which of the following expression is true?

Options

A. `K_(p) lt K_(c)`
B. `K_(p) gt K_(c)`
C. `K_(p) = K_(c)`
D. Can't be predicted

Explanation

To solve the problem, we need to analyze the relationship between the internal energy change (ΔE) and the enthalpy change (ΔH) for the reaction, and how they relate to the number of gaseous moles (ΔNg).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The heat of reaction at constant volume (ΔE) is 1200 cal more than at constant pressure (ΔH).
- Therefore, we can express this as:
\[ ΔE = ΔH + 1200 \text{ cal} \]

2. **Using the First Law of Thermodynamics**:
- According to the first law of thermodynamics, we have the relationship:
\[ ΔH = ΔE + ΔN_g RT \]
- Here, ΔN_g is the change in the number of gaseous moles, R is the gas constant, and T is the temperature.

3. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ ΔE - ΔH + ΔN_g RT = 0 \]
- Substituting ΔE from step 1 into this equation:
\[ (ΔH + 1200) - ΔH + ΔN_g RT = 0 \]
- This simplifies to:
\[ 1200 + ΔN_g RT = 0 \]

4. **Calculating ΔNg**:
- Rearranging gives us:
\[ ΔN_g RT = -1200 \]
- Now, substituting R = 8.314 J/(mol·K) and T = 300 K (converting calories to joules: 1 cal = 4.184 J):
\[ ΔN_g \cdot 8.314 \cdot 300 = -1200 \cdot 4.184 \]
- Calculating the right side:
\[ ΔN_g \cdot 2494.2 = -5020.8 \]
- Solving for ΔN_g:
\[ ΔN_g = \frac{-5020.8}{2494.2} \approx -2.01 \approx -2 \]

5. **Finding the Relationship Between Kp and Kc**:
- The relationship between Kp and Kc is given by:
\[ K_p = K_c R T^{ΔN_g} \]
- Substituting ΔN_g = -2, R = 0.0821 L·atm/(K·mol) (or 8.314 J/(K·mol)), and T = 300 K:
\[ K_p = K_c \cdot (0.0821 \cdot 300)^{-2} \]
- Calculating gives:
\[ K_p = K_c \cdot (24.63)^{-2} \]
- This implies:
\[ K_p = K_c \cdot (1.6 \times 10^{-7}) \]

6. **Conclusion**:
- Since Kp is a small fraction of Kc, we conclude that:
\[ K_c > K_p \]

### Final Answer:
The correct expression is:
\[ K_c > K_p \]