9.

The activation energy for the reaction
2HI (g) → H₂ + I₂(g)
is 209.5 kJ mol^–1 at 581 k. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

2HI (g) → H + I₂(g)
Activation energy, E_a = 209.5 kJ mol⁻¹ 
Multiply by 1000 to convert in j 
E_a= 209500 J mol⁻¹ 
Temperature, T = 581 K
Gas constant, R = 8.314 JK⁻¹ mol⁻¹
According to Arhenious equation
K = A e ^–Ea/RT 
In this formula term e ^–Ea/RT represent the number of molecules which have energy equal or more than activation energy 
Number of molecules = e ^–Ea/RT 
Plug the values we get 
Number of molecules
$$\begin{gathered} ={\mathrm{e}}^{-\frac{209500}{8.314\times 581}} \\ = {\mathrm{e}}^{-43.4} \\ = \frac{1}{ {\mathrm{e}}^{43.4}} \end{gathered}$$

taking antilog of we get

1.47 x 10^-19