A hydrogenation reaction is carried out at `500 K`. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is `400 K`. Calculate the activation energy of the reaction if the catalyst lowers the activation barrier by `20 kJ mol^(-1)`.

Options

A. 110 kJ/mol
B. 10 kJ/mol
C. 100 kJ/mol
D. 1000 kJ/mol

Explanation

To solve the problem, we will use the Arrhenius equation, which relates the rate constant of a reaction to its activation energy and temperature. The equation is given by:

\[ k = A e^{-\frac{E_a}{RT}} \]

where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor (Arrhenius constant)
- \( E_a \) = activation energy
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin

### Step 1: Write the Arrhenius equation for both temperatures

For the reaction at \( 500 \, K \):
\[ k_{500} = A e^{-\frac{E_a}{R \cdot 500}} \]

For the reaction at \( 400 \, K \) (with the catalyst):
\[ k_{400} = A e^{-\frac{E_a - 20 \, \text{kJ}}{R \cdot 400}} \]

### Step 2: Since both reactions proceed at the same rate, equate the two rate constants

Since \( k_{500} = k_{400} \), we can set the two equations equal to each other:
\[ A e^{-\frac{E_a}{R \cdot 500}} = A e^{-\frac{E_a - 20 \, \text{kJ}}{R \cdot 400}} \]

### Step 3: Cancel \( A \) from both sides

Assuming \( A \) is not zero, we can cancel it out:
\[ e^{-\frac{E_a}{R \cdot 500}} = e^{-\frac{E_a - 20 \, \text{kJ}}{R \cdot 400}} \]

### Step 4: Take the natural logarithm of both sides

Taking the natural logarithm gives:
\[ -\frac{E_a}{R \cdot 500} = -\frac{E_a - 20 \, \text{kJ}}{R \cdot 400} \]

### Step 5: Rearranging the equation

Multiplying both sides by \( -R \):
\[ \frac{E_a}{500} = \frac{E_a - 20 \, \text{kJ}}{400} \]

### Step 6: Cross-multiply to eliminate the fractions

Cross-multiplying gives:
\[ 400 E_a = 500 (E_a - 20 \, \text{kJ}) \]

### Step 7: Expand and rearrange

Expanding the right side:
\[ 400 E_a = 500 E_a - 10000 \, \text{kJ} \]

Rearranging gives:
\[ 10000 \, \text{kJ} = 500 E_a - 400 E_a \]
\[ 10000 \, \text{kJ} = 100 E_a \]

### Step 8: Solve for \( E_a \)

Dividing both sides by 100:
\[ E_a = 100 \, \text{kJ/mol} \]

### Conclusion

The activation energy of the reaction is \( 100 \, \text{kJ/mol} \).

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