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Effective Equilibrium Constant

H2 and CO2 ...

Question

$H_{2}$ and $C O_{2}$ at initial pressure of 10 atm and 20 atm respectively react at $1000$ to form $C O$ and $H_{2} O$ to attain equilibrium : $H_{2} (g) + C O_{2} (g) ⇌ H_{2} O (g) + C O (g)$ . Calculate partial pressure of $H_{2} O$ (nearest integer) (atm) at equilibrium if $K_{p}$ is $1.60$ at $1000$ .

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Solution

Let $x$ atm be the equilibrium partial pressure of water vapour.
The equilibrium partial pressure of $C O$ will also be $x$ atm.
The equilibrium partial pressures of $H_{2}$ and $C O_{2}$ will be $10 - x$ atm and $20 - x$ atm respectively.

The equilibrium constant expression is $K_{p} = \frac{P_{H_{2} O} P_{C O}}{P_{H_{2}} P_{C O}}$

$1.60 = \frac{x \times x}{(10 - x) \times (20 - x)}$

$0.625 x^{2} = 200 - 10 x - 20 x + x^{2}$

$0.375 x^{2} - 30 x + 200 = 0$

$x = 72.7$ or $x = 7.34$

The value 72.7 is discarded as the equilibrium partial pressure of hydrogen cannot be negative.
[ $10 - x = 10 - 72.7 < 0$ ]

Hence, the partial pressure of water vapour at equilibrium is 7.34 atm.

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