Question

1 mol of $N_2$ is mixed with 3 mol of $H_2$ in a litre container. If $50$% of $N_2$ is converted into ammonia by the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$, then the total number of moles of gas at the equilibrium are:

• A. $1.5$
• B. $4.5$
• C. $3.0$
• D. $6.0$

Answer 1

Answer: (C)

$3.0$

$\alpha =50\%$ $=\frac{50}{100}=0.5$
$\underset{\underset{1-\alpha }{1}}{N_2}+\underset{\underset{3-3\alpha }{3}}{3H_2}\rightleftharpoons \underset{\underset{2\alpha }{o}}{2N{H}_3}$
Total moles $=1-\alpha +3-3\alpha +2\alpha$
$=4-2\alpha$
$=4-2\times 0.5=3$