Question

Total number of moles for the reaction $2HI\leftrightharpoons H_2+I_2$ if $\alpha$ is degree of

dissociation is

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

Answer 1

The correct option is A

2

Initially $2HI\leftrightharpoons H_2+I_2$

Moles of $HI=2,H_2=0,I_2=0$

At equilibrium,
Moles of $HI=2-\alpha ,H_2=\frac{\alpha }{2},I_2=\frac{\alpha }{2}$
Total moles = $2-\alpha +\frac{\alpha }{2}+\frac{\alpha }{2}=2$