Question

If the pressure of $N_2+H_2$ mixture in stoichiometric proportions in a closed apparatus is 100 atm and 20% of the mixture reacts for the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$, then the pressure at the same temperature would be-

• A. 100
• B. 90
• C. 85
• D. 80

Answer 1

The correct option is B 90

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
$p_{N_2}+p_{H_2}=P+3P=100atm$
$\Rightarrow P=25atm$
20% mixture reacted means $\alpha =0.2$ $\text{Amount reacted}=x=C\alpha =P\alpha$
$\text{Amount reacted}=0.2P=0.2\times 25=5atm$
$\begin{array}{cccccc}& N_2\left(g\right)& +& 3H_2\left(g\right)& \rightleftharpoons & 2N{H}_3\left(g\right)\\ \text{At initial}& 25& & 75& & 0\\ \text{At eqb.}& (25-x)& & (75-3x)& & +2x\end{array}$
$\text{Total pressure}=(25-x)+(75-3x)+2x$
$=25-x+75-3x+2x$
$=25-5+75-15+10$
$=20+60+10=90$