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 need to use the relationship between activation energy (Ea) and temperature (T) for two different conditions. The formula we will use is derived from the Arrhenius equation, which states that the rate of a reaction is proportional to the exponential of the negative activation energy divided by the product of the gas constant and temperature.

The relationship can be expressed as:

\[ \frac{E_{A1}}{T_1} = \frac{E_{A2}}{T_2} \]

Where:
- \(E_{A1}\) = Activation energy of the uncatalyzed reaction
- \(T_1\) = Temperature of the catalyzed reaction (in Kelvin)
- \(E_{A2}\) = Activation energy of the catalyzed reaction
- \(T_2\) = Temperature of the uncatalyzed reaction (in Kelvin)

### Step-by-Step Solution:

1. **Identify the values given in the problem:**
- \(E_{A1} = 20 \, \text{kJ/mol} = 20000 \, \text{J/mol}\) (convert kJ to J)
- \(E_{A2} = 10 \, \text{kJ/mol} = 10000 \, \text{J/mol}\)
- \(T_1 = 27^\circ C = 300 \, \text{K}\) (convert Celsius to Kelvin by adding 273)

2. **Set up the equation using the relationship:**
\[ \frac{20000 \, \text{J/mol}}{300 \, \text{K}} = \frac{10000 \, \text{J/mol}}{T_2} \]

3. **Cross-multiply to solve for \(T_2\):**
\[ 20000 \cdot T_2 = 10000 \cdot 300 \]

4. **Calculate the right side:**
\[ 10000 \cdot 300 = 3000000 \]

5. **Now, divide both sides by 20000 to find \(T_2\):**
\[ T_2 = \frac{3000000}{20000} = 150 \, \text{K} \]

6. **Convert \(T_2\) back to Celsius:**
\[ T_2 = 150 \, \text{K} - 273 = -123^\circ C \]

7. **Final answer:**
The temperature at which the uncatalyzed reaction will have the same rate as that of the catalyzed reaction at 27°C is approximately **-123°C**.