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(B) The given equilibrium reaction is $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$.According to Le Chatelier's principle,when pressur

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Backward direction

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(B) The given equilibrium reaction is $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$.According to Le Chatelier's principle,when pressure is increased,the equilibrium shifts in the direction that decreases the total number of moles of gaseous species.In this reaction,the number of moles of gaseous reactants is $1$ $(H_2O_{(g)})$ and the number of moles of gaseous products is $2$ $(CO_{(g)} + H_{2(g)})$.Since the number of gaseous moles on the product side $(2)$ is greater than on the reactant side $(1)$,an increase in pressure will shift the equilibrium towards the side with fewer gaseous moles,which is the reactant side.Therefore,the equilibrium will shift in the backward direction.

If pressure increases,what is the effect on the given equilibrium $C_{(s)} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + H_{2(g)}$?

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