Question

A reaction carried out by 1 mol of N${}_2$ and $3$ mol of H${}_2$ shows at equilibrium the mole fraction of NH${}_3$ as $0.012$ at 500${}^o$C and $10$ atm pressure. report the pressure at which mole % of $NH$${}_3$ in equilibrium mixture is increased to $10.4$ (in atm)

• A. $32$
• B. $50$
• C. $105$
• D. $179$

Answer 1

The correct option is C $105$

$N_2+{3H}_2\rightleftharpoons {2NH}_3$
1 0 0 Moles before reaction
$(1-x)$ $(3-3x)$ $2x$ Moles at equilibrium
Given mole fraction of ${NH}_3$=$0.012$ at P = $10$ atm
$\therefore \frac{2x}{(4-2x)}=0.012\Rightarrow x=0.0237$
$\therefore K_P=\frac{\left(P_{{NH}_3}^{\prime }\right)}{P_{N_2}^{\prime }\times {\left(P_{H_2}^{\prime }\right)}^3}=\frac{{\left[\frac{2x.P}{(4-2x)}\right]}^2}{\frac{(1-x).P}{(4-2x)}{\left[\frac{(3-3x)P}{(4-2x)}\right]}^3}$
$K_P=\frac{{4x}^2{(4-2x)}^2}{(1-x){(3-3x)}^2{p}^2}$
$=\frac{4\left(0.0237\right){{}^2[4-2(0.0237\left)\right]}^2}{(1-0.0237)[3-3{\left(0.0237\right)}^3]\times 100}$
$K_P=1.431\times {10}^{-5}{atm}^{-2}$
Let mole % of ${NH}_3$ in equilibrium mixture be increased to 10.4 at pressure $P$.
$\therefore \frac{2x}{(4-2x)}=\frac{10.4}{100}orx=0.1884$
Now again using equation (i)
$K_P=\frac{4x^2{(4-2x)}^2}{(1-x){(3-3x)}^3.P^2}$
$1.431\times {10}^{-5}=\frac{[4\times {\left(0.1884\right)}^2][4-2({0.1884)}^2]}{[1-0.1884]{[3-3(0.1884\left)\right]}^3\times P^2}$
$P=105.041atm$