Q. 56 g of nitrogen and 8 g of hydrogen gas heated in a closed vessel. At equilibrium, 34 g of ammonia are present. The equilibrium number of moles of nitrogen, hydrogen and ammonia are, respectively
A) 1, 1, 2
B) 1, 2, 2
C) 2, 1, 1
D) 2, 2, 1
Explanation
(a)
${\ \ \ \ \ N}_{2}\ + \ \ \ \ \ \ \ \ \ \ 3H_{2} \rightleftharpoons 2{NH}_{3}$
56 g 8 g 0 g
$( = 2\ mol)( = 4\ mol)\ (0\ mol)$ initially
$(2 - 1)(4 - 3)\ \ \ \ 34\ g$ at equilibrium
$= 1\ \ \ \ \ \ \ \ \ \ \ \ = 1\ \ \ \ \ \ ( = 2\ mol)$
According to Eq.(1) 2 moles of ammonia are present a produce 2 moles of $NH_{3},$ we need 1 mole of $N_{2}$ and 3 of $H_{2}$, hence $2 - 1 = 1$ mole of $N_{2}$ and $4 - 3\text{=}1$ mole are present at equilibrium in vessel
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.