Question

If $\alpha$ is the fraction of HI dissociated at equilibrium in the reaction, ${ 2HI(g) <=> H_2(g)+I_2(g)}$ starting with the 2 moles of HI, then the total number of moles of reactants and products at equilibrium are

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

Answer 1

Answer: (B)

2

Answer 2

$2HI(g) \rightleftharpoons H_2(g)+I_2(g)$
In initial \hspace2mm \, 2 \, mol \hspace2mm 0 \, mol \hspace2mm 0 \, mol
At equilibrium \hspace2mm (2-2\alpha) mol \hspace2mm \alpha \, mol \hspace2mm \alpha \, mol
So, at equilibrium total moles
$=2-2\alpha+\alpha+\alpha$
$=2-2\alpha+2\alpha=2$