The rate of a reaction triples when temperature changes from `20^(@)C` to `50^(@)C`. Calculate the energy of activation.

Answer

To calculate the energy of activation (Ea) when the rate of a reaction triples as the temperature changes from 20°C to 50°C, we can follow these steps:

### Step 1: Convert temperatures from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]

- For \( T_1 = 20°C \):
\[ T_1 = 20 + 273 = 293 \, K \]

- For \( T_2 = 50°C \):
\[ T_2 = 50 + 273 = 323 \, K \]

### Step 2: Write the relationship between rate constants
Given that the rate triples, we can express the relationship between the rate constants at the two temperatures:
\[ k_2 = 3k_1 \]
Thus, we can write:
\[ \frac{k_2}{k_1} = 3 \]

### Step 3: Use the Arrhenius equation
The Arrhenius equation in logarithmic form is given by:
\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 4: Substitute known values
We know:
- \( R = 8.314 \, \text{J/mol·K} \)
- \( \frac{k_2}{k_1} = 3 \) implies \( \log(3) \)

Calculating \( \log(3) \):
\[ \log(3) \approx 0.477 \]

Now substituting into the equation:
\[ 0.477 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{293} - \frac{1}{323} \right) \]

### Step 5: Calculate the difference in reciprocals of temperatures
Calculating \( \frac{1}{293} - \frac{1}{323} \):
\[ \frac{1}{293} \approx 0.003412 \]
\[ \frac{1}{323} \approx 0.003096 \]
\[ \frac{1}{293} - \frac{1}{323} \approx 0.003412 - 0.003096 = 0.000316 \]

### Step 6: Rearranging the equation to solve for \( E_a \)
Rearranging gives:
\[ E_a = 0.477 \times 2.303 \times 8.314 \times \frac{1}{0.000316} \]

### Step 7: Calculate \( E_a \)
Calculating each part:
1. \( 2.303 \times 8.314 \approx 19.1 \)
2. \( E_a = 0.477 \times 19.1 \times \frac{1}{0.000316} \)

Calculating:
\[ E_a \approx 0.477 \times 19.1 \times 3165.5 \]
\[ E_a \approx 28,111.8 \, \text{J/mol} \]

### Step 8: Convert to kJ/mol
To convert \( E_a \) to kJ/mol:
\[ E_a \approx 28.1 \, \text{kJ/mol} \]

### Final Answer
The energy of activation \( E_a \) is approximately:
\[ E_a \approx 28.1 \, \text{kJ/mol} \]

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