Question
If $$\alpha $$ is the fraction of $$HI$$ dissociated at equilibrium in the reaction, $$2HI\left( g \right) \rightleftharpoons {H_2}\left( g \right) + {I_2}\left( g \right)$$ 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 :
$$2$$
Solution :
\[\underset{\begin{smallmatrix}
\\
\text{In initial}
\\
\text{At}\,\,\text{equilibrium}
\end{smallmatrix}}{\mathop{{}}}\,\underset{\begin{smallmatrix}
2\,\,mol \\
\left( 2-2\alpha \right)mol
\end{smallmatrix}}{\mathop{2HI\left( g \right)}}\,\,\rightleftharpoons \underset{\begin{smallmatrix}
0\,\,mol \\
\alpha \,mol
\end{smallmatrix}}{\mathop{{{H}_{2}}\left( g \right)}}\,+\underset{\begin{smallmatrix}
0\,\,mol \\
\alpha \,mol
\end{smallmatrix}}{\mathop{{{I}_{2}}\left( g \right)}}\,\]
$$\eqalign{
& {\text{So, at equilibrium total moles}} \cr
& = 2 - 2\alpha + \alpha + \alpha \cr
& = 2 - 2\alpha + 2\alpha \cr
& = 2 \cr} $$