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 need to determine how many times the rate of a first-order reaction increases in the presence of a catalyst, given the activation energies in both cases.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Activation energy without catalyst (EA1) = 60 kJ/mol
- Activation energy with catalyst (EA2) = 50 kJ/mol
- Temperature (T) = 27°C = 300 K (since T(K) = T(°C) + 273)

2. **Convert Activation Energies to Joules:**
- EA1 = 60 kJ/mol = 60,000 J/mol
- EA2 = 50 kJ/mol = 50,000 J/mol

3. **Use the Arrhenius Equation:**
The ratio of the rate constants \( k_2 \) (with catalyst) and \( k_1 \) (without catalyst) can be expressed using the Arrhenius equation:
\[ \frac{k_2}{k_1} = e^{\frac{-(EA2 - EA1)}{RT}} \]
where \( R \) is the universal gas constant (8.314 J/(mol·K)).

4. **Calculate the Difference in Activation Energies:**
\[ EA1 - EA2 = 60,000 J/mol - 50,000 J/mol = 10,000 J/mol \]

5. **Substitute Values into the Equation:**
\[ \frac{k_2}{k_1} = e^{\frac{-10,000}{(8.314)(300)}} \]

6. **Calculate the Exponent:**
\[ \frac{-10,000}{(8.314)(300)} \approx -4.007 \]

7. **Calculate \( \frac{k_2}{k_1} \):**
\[ \frac{k_2}{k_1} = e^{-4.007} \approx 0.0183 \]

8. **Find the Rate Increase:**
Since \( \frac{k_2}{k_1} \) gives us the ratio of the rate constants, we can find how many times the rate increases:
\[ \text{Rate increase} = \frac{1}{0.0183} \approx 54.64 \]

### Final Answer:
The rate of reaction will increase approximately 55 times in the presence of the catalyst.

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