Question

Hydrogen $\left(amoles\right)$ and iodine $\left(bmoles\right)$ react to give $2x$ moles of the $HI$ at equilibrium. The total number of moles at equilibrium is:

• A. $a+b+2x$
• B. $(a-b)+(b-2x)$
• C. $(a+b)$
• D. $a+b-x$

Answer 1

Answer: (C)

$(a+b)$

$H_2+I_2⟶2HI$

Initially,

No. of moles of $H_2=a\text{moles}$

No. of moles of $I_2=b\text{moles}$

No. of moles of $HI=0$

Reacted amount of-

$H_2=x\text{moles}$

$I_2=x\text{moles}$

$HI=2x\text{moles}$

$\therefore$ At equillibrium,

No. of moles of $H_2=a-x$

No. of moles of $I_2=b-x$

No. of moles of $HI=2x$

$\therefore$ Total no. of moles at equillibrium $=(a-x)+(b-x)+2x=a+b-2x+2x=a+b$

Hence, total no. of moles at equillibrium is $(a+b)$.