Question

For a gaseous phase reaction $A+2B\rightleftharpoons {AB}_2$, $K_C=0.3475L^2{mole}^{-2}$ at 200${}^{o}C$. When 2 moles of $B$ are mixed with one mole of $A$, what total pressure is required to convert 60% of $A$ in ${AB}_2$?

• A. $P=1.8atm$
• B. $P=63atm$
• C. $P=181.5atm$
• D. $P=34atm$

Answer 1

The correct option is C $P=181.5atm$

$A+2B\rightleftharpoons {AB}_2$

Initial moles	1	2	0
Moles at equilibrium	$(1-x)$	$(2-2x)$	$x$

Total moles at equilibrium =$1-x+2-2x+x=3-2x$
Let pressure at equilibrium be $P$,
Now,
$P_{{AB}_2}^{\prime }=\left[\frac{x}{3-2x}\right]P,P_A^{\prime }=\left[\frac{1-x}{3-2x}\right]P,P_B^{\prime }=\left[\frac{2-2x}{3-2x}\right]P$

$\therefore K_P=\frac{x.P}{(3-2x).P\frac{(1-x)}{(3-2x)}.P^2\frac{{(2-2x)}^2}{{(3-2x)}^2}}$

$K_P=\frac{x.({3-2x)}^2}{P^2(1-x)({2-2x)}^2}$

Alternate to derive $K_P$ or equation (i),

$\because$ $K_P=\frac{n_{{AB}_2}}{n_A\times {{(n}_B)}^2}\times {\left(\frac{P}{\sum n}\right)}^{△n}$

$K_P=\frac{x}{(1-x)({2-2x)}^2}\times {\left(\frac{P}{(3-2x)}\right)}^{-2}$

$=\frac{x({3-2x)}^2}{(1-x)({2-2x)}^2.P^2}$
Given that,$x=0.6$ and $△n=-2$
$\because K_P=K_C{RT}^{△n}$

$=0.3475\times {(0.0821\times 473)}^{-2}$

By equations (1) and (2),

$=0.3475\times {(0.0821\times 473)}^{-2}$

$=\frac{0.6({3-1.2)}^2}{P^2(1-0.6)({2-1.2)}^2}$

$=\frac{0.6\times ({1.8)}^2}{P^2\left(0.4\right)({0.8)}^2}$

$\therefore P=181.5atm$