Question

For a gaseous phase reaction, $A+2B\rightleftharpoons A{B}_2{K}_C=0.3475litr{e}^{2}mo{l}^{-2}$ at ${200}^{\circ }C$. When $2$ mole of $B$ are mixed with one mole of $A$. What total pressure is required to convert $60\%$ of $A$ in $A{B}_2$?
Express answer as $10X$

Answer 1

$\underset{1(1-x)}{A}+\underset{0(2-2x)}{2B}\rightleftharpoons \underset{0x}{A{B}_2}$

Total moles at equilibrium=$1-x+2-2x+x=3-2x$

Let the pressure of equilibrium is $P$

$P_{A{B}_2}^1=\frac{x}{3-2x}\times P$

$P_A^1=\frac{1-x}{3-2x}\times P$

$P_B^1=\frac{2-2x}{3-2x}\times P$
$K_p=\frac{P_{A{B}_2}^1}{P_A^1\times (P_B^1{)}^2}$

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

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

Given that $x=\frac{60}{100}=0.6$ and $△n=-2$

$K_P=K_C{\left(RT\right)}^{△n}=0.3475\times (0.0821{)}^{-2}\times (473{)}^{-2}=2.3\times {10}^{-4}$
$K_p=2.3\times {10}^{-4}=\frac{x}{(1-x)\times (2-x{)}^2}\times \frac{P}{(3-2x{)}^2}$

$P=181.45atm$

$X=181.45$

$10X=1815$