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:

forward direction

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backward direction

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no effect

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none of these

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Solution

The correct option is B backward direction
$C_{(s)} + H_{2} O_{(g)} ⇌ C O_{(g)} + H_{2 (g)}$
According to Le-Chatelier's principle, the reactions in which number of moles of gaseous reactions are less than number of gaseous moles of products, such reactions are favored at low pressure.
$∴$ The equilibrium will shift in backward direction at high 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. For the reaction,

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

, by increasing pressure following equilibrium :

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:

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

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

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If pressure increases then its effect on given equilibriumC(s)+H2O(g)⇌CO(g)+H2(g)It is shifted in:

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: