Question

At temperature $T$, a compound ${AB}_2\left(g\right)$ dissociates according to the reaction $2{AB}_2\left(g\right)\iff 2AB\left(g\right)+B_2\left(g\right)$ with a degree of dissociation, $x$, which is small compared with unity. Deduce the expression for $K_p$ in terms of $x$ and the total pressure $P$.

• A. $\frac{P{x}^3}{2}$
• B. $\frac{P{x}^2}{3}$
• C. $\frac{P{x}^3}{3}$
• D. $\frac{P{x}^2}{2}$

Answer 1

Answer: (A)

$\frac{P{x}^3}{2}$

${2AB}_{2\left(g\right)}\leftrightharpoons 2A{B}_{\left(g\right)}+B_{2\left(g\right)}$ [Total Pressure p]
at=0 1mole 0 0
$t=eq$ $1-xmole$ $xmole$ $\frac{x}{2}$ moles
Total moles , --> $1-x+x+$$\frac{x}{2}$=1+$\frac{x}{2}$=$\frac{2+x}{2}$
partial pressure of $P_{A{B}_2}=X_{A{B}_2}\times P$
$\frac{2(1-X)P}{2-X}$
$P_{AB}$=$\frac{2XP}{2-X}$ ,

$P_{B_2}$=$\frac{XP}{2+X}$
$K_P=\frac{P_{AB}^2\times P_{B_2}}{P_{A{B}_2}^2}=\frac{\frac{4x^2{p}^2}{{(2+x)}^2}\times \frac{xP}{2+x}}{\frac{4{(1-x)}^2{P}^x}{{(2+x)}^2}}$
$\frac{{Px}^3}{(2+x){(1-x)}^2}as1>>>>>x$
$K_P=\frac{{Px}^3}{2\times 1}=\frac{{Px}^3}{2}$