Rate constant k of a reaction varies with temperature according to equation:
`log k=` constant ` - (E_a)/( 2.303 R).(1)/(T )`
What is the activation energy for the reaction. When a graph is plotted for log k versus `1/T` a straight line with a slope-6670 K is obtained. Calculate energy of activation for this reaction (R=8.314 `JK^(-1) mol^(-)` 1)

Explanation

To find the activation energy (Ea) for the reaction given the slope of the plot of log k versus 1/T, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given equation**:
The relationship between the rate constant (k) and temperature (T) is given by:
\[ \log k = \text{constant} - \frac{E_a}{2.303 R} \cdot \frac{1}{T} \]
Here, \( R \) is the universal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\)), and \( E_a \) is the activation energy.

2. **Understand the slope of the graph**:
When plotting log k against \( \frac{1}{T} \), the slope of the line is given by:
\[ \text{slope} = -\frac{E_a}{2.303 R} \]
From the problem, we know that the slope is -6670 K.

3. **Set up the equation using the slope**:
We can equate the slope to the expression involving \( E_a \):
\[ -\frac{E_a}{2.303 R} = -6670 \]

4. **Remove the negative signs**:
By multiplying both sides by -1, we have:
\[ \frac{E_a}{2.303 R} = 6670 \]

5. **Rearrange to solve for \( E_a \)**:
Multiply both sides by \( 2.303 R \):
\[ E_a = 6670 \cdot 2.303 R \]

6. **Substitute the value of \( R \)**:
Substitute \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \):
\[ E_a = 6670 \cdot 2.303 \cdot 8.314 \]

7. **Calculate \( E_a \)**:
First, calculate \( 2.303 \cdot 8.314 \):
\[ 2.303 \cdot 8.314 \approx 19.188 \]
Now, multiply this by 6670:
\[ E_a \approx 6670 \cdot 19.188 \approx 127711.4 \, \text{J} \]

8. **Convert to kJ if necessary**:
To convert joules to kilojoules, divide by 1000:
\[ E_a \approx 127.7114 \, \text{kJ/mol} \]

### Final Answer:
The activation energy \( E_a \) for the reaction is approximately **127.71 kJ/mol**.