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(C) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with a smaller number of gaseous moles. For

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The equilibrium will shift in the backward direction.

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(C) According to Le Chatelier's principle,an increase in pressure shifts the equilibrium towards the side with a smaller number of gaseous moles. For the given reaction: $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$. Number of gaseous moles on the reactant side = $1$ $(H_2O)$. Number of gaseous moles on the product side = $1 + 1 = 2$ ($CO$ and $H_2$). Since the product side has more gaseous moles,an increase in pressure will shift the equilibrium to the side with fewer gaseous moles,which is the reactant side (backward direction).

How will an increase in pressure affect the following equilibrium? $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$

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