$C_{(s)} + H_{2} O_{(g)} \to C O (g) + H_{2} (g)$
The pressure on the system is increased in the above reaction then:

[H_{2} O]

increase

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[H_{2}]

decrease

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[CO] decrease,

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all of the above

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Solution

The correct option is D all of the above
As C is in solid state, any change in its amount will not affect the equilibrium in the gas phase.
Since the number of moles of the products is more than that of the gaseous reactants, increase in pressure will shift the equilibrium towards the left.
So, $[H_{2} O]$ increase, $[H_{2}]$ decrease, $[C O]$ decrease,

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