`K_p` for a reaction at `25^@C` is 10 atm. The activation energy for forward and reverse reactions are 12 and 20 kJ/mol respectively . The `K_c` for the reaction at `40^@C` will be

Options

A. `4.33 xx 10^(-1)` M
B. `3.33 xx 10^(-2)` M
C. `3.33 xx 10^(-10)`M
D. `4.33 xx 10^(-2)` M

Explanation

To find the \( K_c \) for the reaction at \( 40^\circ C \), we will follow these steps:

### Step 1: Understand the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c \cdot R \cdot T^{\Delta n} \]
where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas.

### Step 2: Calculate \( \Delta H \)
The change in enthalpy \( \Delta H \) can be calculated using the activation energies:
\[ \Delta H = E_f - E_r \]
where:
- \( E_f \) is the activation energy for the forward reaction (12 kJ/mol),
- \( E_r \) is the activation energy for the reverse reaction (20 kJ/mol).

Calculating \( \Delta H \):
\[ \Delta H = 12 \, \text{kJ/mol} - 20 \, \text{kJ/mol} = -8 \, \text{kJ/mol} \]

### Step 3: Convert \( \Delta H \) to J/mol
Since we will use the gas constant \( R \) in J/(K·mol), we convert \( \Delta H \):
\[ \Delta H = -8 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -8000 \, \text{J/mol} \]

### Step 4: Use the Van't Hoff equation to find \( K_c \) at \( 40^\circ C \)
The Van't Hoff equation relates the change in equilibrium constant with temperature:
\[ \ln \left( \frac{K_{c2}}{K_{c1}} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
where:
- \( K_{c1} \) is the equilibrium constant at \( T_1 = 25^\circ C = 298 \, K \),
- \( K_{c2} \) is the equilibrium constant at \( T_2 = 40^\circ C = 313 \, K \).

### Step 5: Calculate \( K_c \) at \( 25^\circ C \)
From the given \( K_p = 10 \, \text{atm} \) at \( 25^\circ C \), we first need to find \( K_c \):
Assuming \( \Delta n = 0 \) (no change in moles of gas), the equation simplifies to:
\[ K_p = K_c \cdot R \cdot T \]
So,
\[ K_c = \frac{K_p}{R \cdot T} \]
Substituting the values:
\[ K_c = \frac{10 \, \text{atm}}{0.0821 \, \text{L·atm/(K·mol)} \cdot 298 \, K} = \frac{10}{24.4758} \approx 0.408 \, \text{mol/L} \]

### Step 6: Substitute values into the Van't Hoff equation
Now we can substitute the values into the Van't Hoff equation:
\[ \ln \left( \frac{K_{c2}}{0.408} \right) = -\frac{-8000}{8.314} \left( \frac{1}{313} - \frac{1}{298} \right) \]

Calculating the right side:
1. Calculate \( \frac{1}{313} - \frac{1}{298} \):
\[ \frac{1}{313} - \frac{1}{298} \approx -0.0005 \]
2. Calculate \( -\frac{-8000}{8.314} \):
\[ \approx 962.5 \]

Thus,
\[ \ln \left( \frac{K_{c2}}{0.408} \right) \approx 962.5 \times (-0.0005) \approx -0.48125 \]

### Step 7: Solve for \( K_{c2} \)
Exponentiating both sides:
\[ \frac{K_{c2}}{0.408} = e^{-0.48125} \]
Calculating \( e^{-0.48125} \approx 0.619 \):
\[ K_{c2} = 0.408 \times 0.619 \approx 0.253 \]

### Final Answer
The \( K_c \) for the reaction at \( 40^\circ C \) is approximately \( 0.253 \, \text{mol/L} \).

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