Question

$n$ mole each of $H_{2}O,H_2$ and $O_2$ are mixed at a suitable high temperature to attain the equilibrium :
$2H_{2}O\rightleftharpoons 2H_2+O_2$
If $y$ mole of $H_{2}O$ are dissociated and the total pressure maintained is $P$, $K_p$ is :

• A. $\left[P(n+y/2){(n+y)}^2\right]/\left[(3n+y/2){(n-y)}^2\right]$
• B. $\left[P(n+3y/2){(n+2y)}^2\right]/\left[(3n+4y/2){(n-y)}^2\right]$
• C. $\left[P(n+y/2){(n+2y)}^2\right]/\left[(3n+2y/2){(2n-y)}^2\right]$
• D. None of these

Answer 1

Answer: (A)

$\left[P(n+y/2){(n+y)}^2\right]/\left[(3n+y/2){(n-y)}^2\right]$

Initial moles of water, hydrogen and oxygen are n each.

The equilibrium moles of water, hydrogen and oxygen are $n-y$ , $n+y$ and $n+0.5y$ respectively

Total number of moles at equilibrium are $n-y+n+y+n+0.5y=3n+0.5y$

The mole fractions of water, hydrogen and oxygen at equilibrium are $\frac{n-y}{3n+0.5y}$, $\frac{n+y}{3n+0.5y}$ and $\frac{n+0.5y}{3n+0.5y}$ respectively.

The expression for the equilibrium constant is $K_p=\frac{P_{H_2}^2\times P_{O_2}}{P_{H_{2}O}^2}$

$K_p=\frac{X_{H_2}^2{P}^2\times X_{O_2}P}{X_{H_{2}O}^2{P}^2}$

$K_p=P\times \frac{X_{H_2}^2\times X_{O_2}}{X_{H_{2}O}^2}$

Substitute values in the above equation

$K_P=P\times \frac{[\frac{n+y}{3n+0.5y}{]}^2\times \frac{n+0.5y}{3n+0.5y}}{[\frac{n-y}{3n+0.5y}{]}^2}$

$K_P=\left[P(n+y/2){(n+y)}^2\right]/\left[(3n+y/2){(n-y)}^2\right]$