The rate temperature changes from 300K to 310K. Activation energy of such a reaction will be `(R = 8.314 JK^(-1) mol^(-1) " and " log 2 = 0.3010)`

Options

A. `48.6 kJ mol^(-1)`
B. `58.5 kJ mol^(-1)`
C. `60.5 kJ mol^(-1)`
D. `53.6 kJ mol^(-1)`

Explanation

To find the activation energy (Ea) for the reaction when the temperature changes from 300 K to 310 K, we can use the Arrhenius equation in the form of the logarithmic relationship:

\[ \log \frac{k_2}{k_1} = -\frac{E_a}{2.303R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

### Step 1: Identify the known values
- \(T_1 = 300 \, K\)
- \(T_2 = 310 \, K\)
- \(R = 8.314 \, J \, K^{-1} \, mol^{-1}\)
- The temperature coefficient states that when the temperature increases by 10 K, the rate constant doubles, thus \(k_2 = 2k_1\).

### Step 2: Calculate \(\frac{k_2}{k_1}\)
Since the rate constant doubles when the temperature increases by 10 K:
\[ \frac{k_2}{k_1} = 2 \]

### Step 3: Substitute values into the equation
Substituting the values into the logarithmic equation:
\[ \log 2 = -\frac{E_a}{2.303 \times 8.314} \left( \frac{1}{310} - \frac{1}{300} \right) \]

### Step 4: Calculate \(\frac{1}{T_2} - \frac{1}{T_1}\)
Calculating the difference:
\[ \frac{1}{310} - \frac{1}{300} = \frac{300 - 310}{310 \times 300} = \frac{-10}{93000} = -\frac{1}{9300} \]

### Step 5: Substitute \(\log 2\) and calculate \(E_a\)
Using \(\log 2 = 0.3010\):
\[ 0.3010 = -\frac{E_a}{2.303 \times 8.314} \left(-\frac{1}{9300}\right) \]

This simplifies to:
\[ 0.3010 = \frac{E_a}{2.303 \times 8.314 \times 9300} \]

### Step 6: Solve for \(E_a\)
Rearranging gives:
\[ E_a = 0.3010 \times 2.303 \times 8.314 \times 9300 \]

Calculating the right-hand side:
1. Calculate \(2.303 \times 8.314 \approx 19.156\)
2. Then calculate \(19.156 \times 9300 \approx 178,000\)
3. Finally, multiply by \(0.3010\):
\[ E_a \approx 0.3010 \times 178,000 \approx 53598 \, J/mol \]

### Step 7: Convert to kilojoules
To convert to kilojoules:
\[ E_a \approx \frac{53598}{1000} \approx 53.598 \, kJ/mol \]

### Final Answer
The activation energy \(E_a\) is approximately:
\[ \boxed{53.6 \, kJ/mol} \]