A catalyst lowers the activation energy of a reaction from 20 kJ mol^(-1) to 10 kJ mol^(-1). The temperature at which uncatalysed reaction will have same rate as that of catalysed at 27°C is

Options

A. 327°C
B. 283°C
C. 445°C
D. 600°C

Answer

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

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

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

### Step-by-Step Solution:

1. **Identify the Activation Energies**:
- The activation energy without the catalyst (\( E_a \)) is 20 kJ/mol.
- The activation energy with the catalyst (\( E_a' \)) is 10 kJ/mol.

2. **Convert Activation Energies to Same Units**:
- Convert kJ/mol to J/mol for consistency:
- \( E_a = 20 \text{ kJ/mol} = 20000 \text{ J/mol} \)
- \( E_a' = 10 \text{ kJ/mol} = 10000 \text{ J/mol} \)

3. **Set Up the Ratio of Activation Energies and Temperatures**:
- According to the Arrhenius equation, we can set up the ratio:
\[ \frac{E_a}{E_a'} = \frac{T'}{T} \]
Where \( T' \) is the temperature for the uncatalyzed reaction and \( T \) is the temperature for the catalyzed reaction (27°C).

4. **Convert the Given Temperature to Kelvin**:
- The temperature \( T \) at which the catalyzed reaction occurs is:
\[ T = 27°C = 27 + 273 = 300 \text{ K} \]

5. **Substitute the Values into the Ratio**:
\[ \frac{20000}{10000} = \frac{T'}{300} \]
Simplifying the left side gives:
\[ 2 = \frac{T'}{300} \]

6. **Solve for \( T' \)**:
- Multiply both sides by 300:
\[ T' = 2 \times 300 = 600 \text{ K} \]

7. **Convert \( T' \) Back to Celsius**:
- To convert Kelvin back to Celsius:
\[ T' = 600 \text{ K} - 273 = 327°C \]

### Final Answer:
The temperature at which the uncatalyzed reaction will have the same rate as that of the catalyzed reaction at 27°C is **327°C**.