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

Options

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

Answer

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

### Step 1: Determine the change in enthalpy (\( \Delta H \))
Given the activation energies:
- Activation energy for the forward reaction (\( E_f \)) = 12 kJ/mol
- Activation energy for the reverse reaction (\( E_b \)) = 20 kJ/mol

The change in enthalpy (\( \Delta H \)) can be calculated as:
\[ \Delta H = E_f - E_b = 12 \, \text{kJ/mol} - 20 \, \text{kJ/mol} = -8 \, \text{kJ/mol} \]

### Step 2: Convert \( \Delta H \) to J/mol
Since the gas constant \( R \) is in J, we need to convert \( \Delta H \) from kJ to J:
\[ \Delta H = -8 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -8000 \, \text{J/mol} \]

### Step 3: Use the relationship between \( K_p \) and \( K_c \)
We know that:
\[ K_p = K_c \cdot R^{\Delta n} \cdot T \]
Where:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T \) is in Kelvin
- \( \Delta n \) is the change in moles of gas (assumed to be 1 in this case)

At \( 25^\circ C \) (which is \( 298 \, K \)), we have:
\[ K_p = 10 \, \text{atm} \]

Rearranging the equation gives:
\[ K_c = \frac{K_p}{R \cdot T} \]

### Step 4: Calculate \( K_c \) at \( 25^\circ C \)
Substituting the values:
\[ K_c = \frac{10}{0.0821 \cdot 298} \]
Calculating this:
\[ K_c = \frac{10}{24.4758} \approx 0.408 \, \text{mol/L} \]

### Step 5: 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:
\[ \log \frac{K_{c2}}{K_{c1}} = \frac{\Delta H}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:
- \( T_1 = 298 \, K \) (25°C)
- \( T_2 = 313 \, K \) (40°C)

Substituting the known values:
\[ \log \frac{K_{c2}}{0.408} = \frac{-8000}{2.303 \cdot 8.314} \left( \frac{1}{298} - \frac{1}{313} \right) \]

Calculating the right side:
\[ \log \frac{K_{c2}}{0.408} = \frac{-8000}{19.094} \left( \frac{1}{298} - \frac{1}{313} \right) \]
Calculating \( \left( \frac{1}{298} - \frac{1}{313} \right) \):
\[ \frac{1}{298} - \frac{1}{313} \approx 0.000168 \]
Now substituting this value:
\[ \log \frac{K_{c2}}{0.408} = -418.1 \cdot 0.000168 \approx -0.0702 \]

### Step 6: Solve for \( K_{c2} \)
\[ \frac{K_{c2}}{0.408} = 10^{-0.0702} \approx 0.849 \]
Thus:
\[ K_{c2} = 0.408 \cdot 0.849 \approx 0.346 \, \text{mol/L} \]

### Final Answer
The \( K_c \) for the reaction at \( 40^\circ C \) is approximately **0.346 mol/L**.

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