Question

If $\alpha$ is the reaction of HI dissociated at equilibrium in the reaction $2HI\to H_2+I_2$ then starting with 2 moles of HI, the total number of moles of reactants and products at equilibrium are:

• A. 1
• B. 2
• C. 1+$\alpha$
• D. 2+2$\alpha$

Answer 1

Answer: (D)

2+2$\alpha$

$2HI\rightleftharpoons H_2+I_2$

t=0, $200$

t=$t_{eq}$, $2(1-\alpha )2\alpha 2\alpha$

Hence, total number of moles of reactants and products at equillibrium are

$n_f=2-2\alpha +2\alpha +2\alpha =2+2\alpha$