Equilibrium - Result Question 19
####20. If $\alpha$ is the fraction of $HI$ dissociated at equilibrium in the reaction, $2 HI(g) \rightarrow H_2$ $(g)+I_2(g)$, starting with 2 moles of $HI$, the total number of moles of reactants and products at equilibrium are
[1996]
(a) $2+2 \alpha$
(b) 2
(c) $1+\alpha$
(d) $2-\alpha$
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Solution:
- (b) According to equation
Total moles at equilibrium $=2-2 \alpha+\alpha+\alpha=2$ mole
