Rate constant k of a reaction varies with temperature according to the equation
logk = constant `-E_a/(2.303R)(1/T)`
where `E_a` is the energy of activation 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 the energy of activation for this reaction

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given equation
The relationship between the rate constant \( k \) and temperature \( T \) is given by:
\[ \log k = \text{constant} - \frac{E_a}{2.303R} \left(\frac{1}{T}\right) \]
where:
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