The activation energy for a first order reaction is 60 kJ `mol^(-1)` in the absence of a catalyst and 50 kJ `mol^(-1)` in the presence of a catalyst. How many times will the rate of reaction grow in the presence of the catalyst if the reaction proceeds at `27^@ `C ?

Explanation

To solve the problem, we will use the Arrhenius equation and the relationship between the rate constants of a reaction in the presence and absence of a catalyst.

### Step-by-step Solution:

1. **Identify Given Data**:
- Activation energy without catalyst (Ea1) = 60 kJ/mol
- Activation energy with catalyst (Ea2) = 50 kJ/mol
- Temperature (T) = 27°C = 300 K (after converting to Kelvin)
- Gas constant (R) = 8.314 J/(mol·K)

2. **Convert Activation Energies to Joules**:
- Ea1 = 60 kJ/mol = 60,000 J/mol
- Ea2 = 50 kJ/mol = 50,000 J/mol

3. **Apply the Arrhenius Equation**:
The relationship between the rate constants (k1 for the reaction without catalyst and k2 for the reaction with catalyst) can be expressed as:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a1 - E_a2}{2.303 \cdot R \cdot T} \]

4. **Substitute Values into the Equation**:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{60,000 - 50,000}{2.303 \cdot 8.314 \cdot 300} \]

5. **Calculate the Numerator**:
\[ 60,000 - 50,000 = 10,000 \text{ J/mol} \]

6. **Calculate the Denominator**:
\[ 2.303 \cdot 8.314 \cdot 300 \approx 5737.74 \]

7. **Calculate the Logarithm**:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{10,000}{5737.74} \approx 1.743 \]

8. **Find the Antilog**:
To find the ratio of the rate constants:
\[ \frac{k_2}{k_1} = 10^{1.743} \approx 55.06 \]

9. **Conclusion**:
The rate of reaction will grow approximately 55.06 times in the presence of the catalyst.