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 need to find the energy of activation (Ea) using the given slope from the graph of log k versus 1/T. The equation provided is:

\[ \log k = \text{constant} - \frac{E_a}{2.303R} \left( \frac{1}{T} \right) \]

### Step-by-Step Solution:

1. **Identify the Slope from the Graph**:
- From the problem, we know that the slope (m) of the graph of log k versus 1/T is given as -6670 K.
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