`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^(-1)` respectively. The `K_(c)` for the reaction at `40^(@)C` will be:

Options

A. `4.33xx10^(-1) M`
B. `3.4xx10^(-2)M`
C. `3.4xx10^(-1)M`
D. `4.33xx10^(-2)M`

Answer

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

### Step 1: Calculate the change in enthalpy (\( \Delta H \))
The change in enthalpy can be calculated using the activation energies for the forward and reverse reactions.

\[ \Delta H = E_a(\text{forward}) - E_a(\text{reverse}) = 12 \, \text{kJ/mol} - 20 \, \text{kJ/mol} = -8 \, \text{kJ/mol} \]

### Step 2: Convert \( \Delta H \) to J/mol
Since the gas constant \( R \) is usually in J/(mol·K), we convert \( \Delta H \) to joules:

\[ \Delta H = -8 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -8000 \, \text{J/mol} \]

### Step 3: Calculate \( K_c \) at \( 25^\circ C \)
We know that \( K_p \) at \( 25^\circ C \) is given as \( 10 \, \text{atm} \). The relationship between \( K_p \) and \( K_c \) is given by:

\[ K_c = \frac{K_p}{RT} \]

Where:
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 25^\circ C = 298 \, \text{K} \)

Substituting the values:

\[ K_c = \frac{10}{0.0821 \times 298} = \frac{10}{24.4758} \approx 0.409 \, \text{mol/L} \]

### 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:

\[ \log \left( \frac{K_{c2}}{K_{c1}} \right) = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Where:
- \( K_{c1} = K_c \) at \( 25^\circ C \)
- \( K_{c2} = K_c \) at \( 40^\circ C \)
- \( T_1 = 298 \, \text{K} \)
- \( T_2 = 313 \, \text{K} \)

Substituting the values:

\[ \log \left( \frac{K_{c2}}{0.409} \right) = \frac{-8000}{2.303 \times 8.314} \left( \frac{1}{298} - \frac{1}{313} \right) \]

Calculating the right-hand side:

1. Calculate \( \frac{1}{298} - \frac{1}{313} \):

\[ \frac{1}{298} \approx 0.003356 \quad \text{and} \quad \frac{1}{313} \approx 0.003194 \]
\[ \frac{1}{298} - \frac{1}{313} \approx 0.000162 \]

2. Calculate \( \frac{-8000}{2.303 \times 8.314} \):

\[ \frac{-8000}{19.095} \approx -418.5 \]

3. Now substituting back:

\[ \log \left( \frac{K_{c2}}{0.409} \right) = -418.5 \times 0.000162 \approx -0.0677 \]

### Step 5: Solve for \( K_{c2} \)
Taking antilog:

\[ \frac{K_{c2}}{0.409} = 10^{-0.0677} \approx 0.85 \]

Thus,

\[ K_{c2} = 0.85 \times 0.409 \approx 0.348 \, \text{mol/L} \]

### Final Answer
The \( K_c \) for the reaction at \( 40^\circ C \) is approximately \( 0.348 \, \text{mol/L} \).

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