Question

If for $2A_{2}B\left(g\right)\rightleftharpoons 2A_2\left(g\right)+B_2\left(g\right)$ $K_p=$ total pressure (at equilibrium) and starting the dissociation from $4$ moles of $A_{2}B$ then__________.

• A. degree of dissociation of $A_{2}B$ will be $\left(2/3\right)$
• B. total number of moles at equilibrium will be $\left(14/3\right)$
• C. at equilibrium the number of moles of $A_{2}B$ are not equal to the number of moles of $B_2$
• D. at equilibrium the number of moles of $A_{2}B$ are equal to the number of moles of $A_2$

Answer 1

Answer: (A)

degree of dissociation of $A_{2}B$ will be $\left(2/3\right)$

$2A_{2}B\left(g\right)\rightleftharpoons 2A_2\left(g\right)+B_2\left(g\right)K_P=$ Total pressure

$t=0$ $4$ $0$ $0$

$t=t_{eq}$ $4-x$ $x$ $x/2$

Let total pressure $=P$

$P^{B_2}=\frac{x/2}{4+x/2}.P=\frac{x}{8+x}\times P$

$P^{A_2}=\frac{x}{4+x/2}.P=\frac{2x}{8+x}\times P$

$P^{A_{2}B}=\frac{4-x}{4+x/2}.P=\frac{8-2x}{8+x}\times P$

$K_P=P=\frac{(\frac{x}{8+x}\times P)\left(\frac{{4x}^2{P}^2}{{(8+x)}^2}\right)}{P^2\times 4\times \frac{{(4-x)}^2}{{(8+x)}^2}}=\frac{x^2}{{(4-x)}^2}.\frac{x}{(8+x)}.P$

$\Rightarrow \frac{x^3}{{(4-x)}^2(8+x)}=1$

$x^3=(16+x^2-8x)(8+x)=128+{8x}^2-64x+16x+x^3-{8x}^2$

$x^3=x^3+128-48x$

$x=\frac{128}{48}=8/3$

$\alpha =x/4=2/3$

Hence $\alpha =2/3=$ degree of dissociation