$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)$
The above equilibrium when subjected to pressure:

Remains unaffected

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Proceeds in the backward direction

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Proceeds in the forward direction

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None of the above

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Solution

The correct option is C Proceeds in the backward direction
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)$
The above equilibrium when subjected to pressure proceeds in the backward direction.
The number of moles of gaseous products (2) are more than the number of moles of gaseous reactants (1). When pressure is increased, the equilibrium shifts in a direction where there are less number of moles of gaseous species. This decreases total pressure. This is in accordance with Le-Chatelier's principle.

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Similar questions

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:

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:
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$

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

Q. For the reaction,

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

, by increasing pressure following equilibrium :

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:

Q. Consider the water gas equilibrium reaction,

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

Which of the following statements is true at equilibrium?

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C(s)+H2O(g)⇌CO(g)+H2(g)The above equilibrium when subjected to pressure:

$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g)$
The above equilibrium when subjected to pressure: