For the reaction, $C (s) + H_{2} O (g) ⇌ H_{2} (g) + C O (g)$ , by increasing pressure following equilibrium :

remain unaffected

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proceed in backward direction

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proceed in forward direction

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cannot be predicted

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Solution

The correct option is B proceed in backward direction
$C (s) + H_{2} O (g) ⇌ H_{2} (g) + C O (g)$ by increasing pressure following equilibrium proceeds in backward direction.
The number of moles of gaseous reactants are smaller than the number of moles of gaseous products. When pressure is increased, the reaction will shift in a direction in which less number of moles of species are present. This nullifies the effect of increased pressure.

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Similar questions

Q. Predict the effect upon the following reaction equilibria of increased pressure.

(1) H_{2} O (g) ⇌ H_{2} (g) + \frac{1}{2} O_{2} (g) (2) C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g) (3) C (s) + H_{2} O (g) ⇌ H_{2} (g) + C O (g)

Q. If pressure increases then its effect on given equilibrium

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)

It is shifted in:

Calculate the equilibrium constant ( $K_{p}$ ) for the reaction, $C (s) + C O_{2} (g) ⇌ 2 C O (g)$ , at $1300 K$ from the following data:
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$

$H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g); K_{p} (1300 K) = 0.623 a t m$

C_{(s)} + H_{2} O_{(g)} \to C O (g) + H_{2} (g)

The pressure on the system is increased in the above reaction then:

C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)

The above equilibrium when subjected to pressure:

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For the reaction, C(s)+H2O(g)⇌H2(g)+CO(g), by increasing pressure following equilibrium :

For the reaction, $C (s) + H_{2} O (g) ⇌ H_{2} (g) + C O (g)$ , by increasing pressure following equilibrium :