The activation energy for the reaction, `2Hi(g) to H_(2)(g) + I_(2)(g)` is 209.5 kJ `moli^(-1)` at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.

Explanation

To solve the problem of calculating the fraction of molecules of reactants having energy equal to or greater than the activation energy for the reaction \(2 \text{HI}(g) \rightarrow \text{H}_2(g) + \text{I}_2(g)\), we will use the Arrhenius equation. Here's a step-by-step solution:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation relates the rate constant \(k\) of a reaction to the temperature \(T\) and the activation energy \(E_a\):
\[ k = A e^{-\frac{E_a}{RT}} \]
Where:
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