Question 9 of 37advanced🔧 ApplyNumerical3 marks

The activation energy for the reaction 2HI(g)→H2+I2(g)2 \text{HI(g)} \rightarrow \text{H}_2 + \text{I}_2 (\text{g}) is 209.5 kJ mol−1^{-1} at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Correct Answer

The fraction of molecules having energy equal to or greater than activation energy is 1.47×101.47 \times 10⁻¹⁹.

Exercise: Intext Questions | Q: 3.9 | (Page 24)
For More Understanding

Explanation

This question tests understanding of the Arrhenius equation and the Boltzmann distribution. The fraction of molecules with energy ≥\geq activation energy is given by the exponential term e^(-Ea/RT) from the Arrhenius equation k = Ae^(-Ea/RT). This represents the probability that a molecule possesses sufficient energy to overcome the activation barrier at a given temperature.

Solution Steps

  1. Step 1: Write the formula for fraction of molecules having energy ≥\geq activation energy: Fraction = e^(-Ea/RT)

  2. Step 2: Convert activation energy to J mol⁻¹: Ea = 209.5 kJ mol⁻¹ = 209.5×10209.5 \times 10³ J mol⁻¹ = 209500 J mol⁻¹

  3. Step 3: Substitute values: R = 8.314 J mol⁻¹ K⁻¹, T = 581 K

  4. Step 4: Calculate Ea/RT = 209500 / (8.314×5818.314 \times 581) = 209500 / 4830.434 = 43.37

  5. Step 5: Calculate fraction = e^(-43.37) = 1.47×101.47 \times 10⁻¹⁹

PreviousNext