The activation energy for the reaction:
`2AB rarr A_(2)+B_(2)(g)`
is `159.7 kJ mol^(-1)` at `500 K`. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
(Given: `2.3 xx 8.314 J K^(-1) mol^(-1) xx 500 K = 9561.1 J mol^(-1)`)

Explanation

To calculate the fraction of molecules of reactants having energy equal to or greater than the activation energy for the reaction \(2AB \rightarrow A_2 + B_2\), we will use the Arrhenius equation. The step-by-step solution is as follows:

### Step 1: Understand the given data
- Activation energy (\(E_a\)) = 159.7 kJ/mol
- Temperature (\(T\)) = 500 K
- Gas constant (\(R\)) = 8.314 J/(K·mol)

### Step 2: Convert activation energy to J/mol
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