Q. Consider the following reaction equilibrium $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$ Initially, 1 mole of $N_{2}$ and 3 moles of $H_{2}$ are take in a 2 L flask. At equilibrium state if, the number of moles of $N_{2}$is 0.6, what is the total number of moles of all gases present in the flask?
A) 0.8
B) 1.6
C) 3.2
D) 6.4
Explanation
(c)
$N_{2}(g)\ \ \ \ \ \ \ + 3H_{2}(g)\ \ \ \ \ \ \ \ \ \ \rightleftharpoons 2NH_{3}(g)$
(Initially) 1 mole 3 mole 0 moles(2L)
(At equilibrium) 0.6 or $(3 - 3x)$ $2x$ moles
$(1 - x)$ moles moles
Number of moles of $N_{2} = 0.6 = 1 - x$
$\therefore$ $x = 1 - 0.6 = 0.4$
So, $3 - 3x = 3 - 3 \times 0.4 = 1.8$,
and $2x = 2 \times 0.4 = 0.8$
Therefore, the total number of moles at equilibrium.
$= (1 - x) + (3 - 3x) + 2x$
$= 0.6 + 1.8 + 0.8$
$= 3.2$ mol
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.