The heat of reaction for an endothermic reaction at constant volume in equilibrium is 1200 calories more than that at constant pressure at 300 k . Calculate the ratio of equilibrium constant `K_p and K_c` .

Options

A. `1.648 xx 10^(6)`
B. `1.648 xx 10^(3)`
C. `1.648 xx 10^(-3)`
D. `1.648 xx 10^(-6)`

Explanation

To solve the problem, we need to find the ratio of the equilibrium constants \( K_p \) and \( K_c \) for the given endothermic reaction. We know that the heat of reaction at constant volume is 1200 calories more than that at constant pressure at 300 K.

### Step-by-Step Solution:

1. **Understand the relationship between \( K_p \) and \( K_c \)**:
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c (RT)^{\Delta n} \]
where \( \Delta n \) is the change in the number of moles of gas during the reaction, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

2. **Identify the heat of reaction**:
We are given that the heat of reaction at constant volume (\( \Delta U \)) is 1200 calories more than that at constant pressure (\( \Delta H \)):
\[ \Delta U = \Delta H - 1200 \text{ calories} \]

3. **Use the relationship between \( \Delta U \) and \( \Delta H \)**:
At constant pressure, we can relate \( \Delta U \) and \( \Delta H \) using the equation:
\[ \Delta H = \Delta U + P \Delta V \]
For an ideal gas, \( P \Delta V = nRT \). Therefore, we can write:
\[ \Delta H - \Delta U = nRT \]
Substituting \( \Delta U \) from step 2:
\[ 1200 = nRT \]

4. **Calculate \( n \)**:
Rearranging the equation gives:
\[ n = \frac{1200}{RT} \]
We will use \( R = 2 \text{ cal/(mol K)} \) and \( T = 300 \text{ K} \):
\[ n = \frac{1200}{2 \times 300} = \frac{1200}{600} = 2 \]

5. **Substitute \( n \) into the \( K_p \) and \( K_c \) relationship**:
Now we can substitute \( \Delta n \) into the equation for \( K_p \):
\[ K_p = K_c (RT)^{\Delta n} \]
Here, \( \Delta n = -2 \):
\[ K_p = K_c (RT)^{-2} \]

6. **Calculate \( K_p/K_c \)**:
Rearranging gives us:
\[ \frac{K_p}{K_c} = (RT)^{-2} \]
Substituting \( R = 2 \text{ cal/(mol K)} \) and \( T = 300 \text{ K} \):
\[ \frac{K_p}{K_c} = \left(2 \times 300\right)^{-2} = (600)^{-2} = \frac{1}{360000} \approx 2.78 \times 10^{-6} \]

### Final Result:
The ratio of equilibrium constants \( \frac{K_p}{K_c} \) is approximately \( 2.78 \times 10^{-6} \).