Question

consider the chemical equation in equilibrium.

ch₄(g) + h₂o(g) ⇌ co(g) + 3h₂(g)

what will happen to the equilibrium of this reaction if the pressure is increased?

the equilibrium will shift to the left to favor the reverse reaction.
the equilibrium will shift to the right to favor the forward reaction.
the equilibrium will not be affected by changing the pressure.
the equilibrium will not be reestablished after this kind of stress.

Explanation:

To determine the effect of pressure on the equilibrium of the reaction $\ce{CH_{4}(g) + H_{2}O(g)
ightleftharpoons CO(g) + 3H_{2}(g)}$, we use Le Chatelier's principle. First, we calculate the number of moles of gaseous reactants and products. On the reactant side, we have $1$ mole of $\ce{CH_{4}(g)}$ and $1$ mole of $\ce{H_{2}O(g)}$, so the total number of moles of gaseous reactants, $n_{reactants}=1 + 1=2$. On the product side, we have $1$ mole of $\ce{CO(g)}$ and $3$ moles of $\ce{H_{2}(g)}$, so the total number of moles of gaseous products, $n_{products}=1+3 = 4$. When pressure is increased, the equilibrium shifts in the direction that produces fewer moles of gas to counteract the pressure increase. Since the number of moles of gaseous reactants ($2$) is less than the number of moles of gaseous products ($4$), the equilibrium will shift to the left (towards the reactants) to favor the reverse reaction, which reduces the total number of moles of gas.

Answer:

A. The equilibrium will shift to the left to favor the reverse reaction.