The activiation energy for the reaction `2HI(g) to H_(2) + I_(2)(g)` is `209.5 KJ mol^(-1)` at 581 K.Calculate the farction of molecules of reactants having energy equal to or greater than activation energy ?

Answer

To calculate the fraction of molecules of reactants having energy equal to or greater than the activation energy for the reaction \(2HI(g) \rightarrow H_2(g) + I_2(g)\) with an activation energy (\(E_a\)) of \(209.5 \, \text{kJ mol}^{-1}\) at a temperature (\(T\)) of \(581 \, \text{K}\), we can use the Arrhenius equation.

### Step-by-Step Solution:

1. **Write the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \(k\) is the rate constant,
- \(A\) is the pre-exponential factor,
- \(E_a\) is the activation energy,
- \(R\) is the universal gas constant, and
- \(T\) is the absolute temperature in Kelvin.

2. **Identify the Variables**:
- Activation energy, \(E_a = 209.5 \, \text{kJ mol}^{-1} = 209500 \, \text{J mol}^{-1}\) (since \(1 \, \text{kJ} = 1000 \, \text{J}\))
- Universal gas constant, \(R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}\)
- Temperature, \(T = 581 \, \text{K}\)

3. **Calculate the Exponential Factor**:
We need to calculate the term \(e^{-\frac{E_a}{RT}}\):
\[ -\frac{E_a}{RT} = -\frac{209500}{8.314 \times 581} \]

4. **Calculate \(RT\)**:
\[ RT = 8.314 \times 581 = 4832.354 \, \text{J mol}^{-1} \]

5. **Calculate \(-\frac{E_a}{RT}\)**:
\[ -\frac{E_a}{RT} = -\frac{209500}{4832.354} \approx -43.4 \]

6. **Calculate the Fraction of Molecules**:
Now, we can find the fraction of molecules having energy equal to or greater than the activation energy:
\[ \text{Fraction} = e^{-43.4} \]

7. **Calculate \(e^{-43.4}\)**:
Using a calculator or mathematical software:
\[ e^{-43.4} \approx 1.47 \times 10^{-19} \]

### Final Answer:
The fraction of molecules of reactants having energy equal to or greater than the activation energy is approximately:
\[ \text{Fraction} \approx 1.47 \times 10^{-19} \]