Question

methane and water react to form hydrogen and carbon monoxide, like this:

mathrm{ch}_{4}(g)+mathrm{h}_{2} mathrm{o}(g)
ightarrow 3 mathrm{h}_{2}(g)+mathrm{co}(g)

the reaction is endothermic. suppose a mixture of ( mathrm{ch}_{4}, mathrm{h}_{2} mathrm{o}, mathrm{h}_{2} ) and ( mathrm{co} ) has come to equilibrium in a closed reaction vessel. predict what change, if any, the perturbations in the table below will cause in the composition of the mixture in the vessel. also decide whether the equilibrium shifts to the right or left.

Explanation:

Step1: Le - Chatelier's principle for temperature change in endothermic reactions

For an endothermic reaction \(CH_{4}(g)+H_{2}O(g)\to3H_{2}(g)+CO(g)\), heat is absorbed in the forward reaction. When the temperature is raised, according to Le - Chatelier's principle, the system will try to absorb the added heat. So, the equilibrium shifts in the direction of the endothermic (forward) reaction.
As the forward reaction occurs, \(CH_{4}\) is consumed. So, the pressure of \(CH_{4}\) (a reactant) will decrease. And the equilibrium shifts to the right.

Step2: Le - Chatelier's principle for temperature decrease in endothermic reactions

When the temperature is lowered, the system will try to release heat to counteract the change. For an endothermic reaction, the reverse reaction is exothermic. So, the equilibrium shifts in the reverse (exothermic) direction.
As the reverse reaction occurs, \(H_{2}\) (a product) is consumed. So, the pressure of \(H_{2}\) will decrease. And the equilibrium shifts to the left.

Answer:

perturbation	change in composition	shift in equilibrium
The temperature is lowered	The pressure of \(H_{2}\) will decrease	to the left