Question

carbon dioxide and water react to form methane and oxygen, like this:
co₂(g)+2h₂o(g)→ch₄(g)+2o₂(g)
the reaction is endothermic. suppose a mixture of co₂, h₂o, ch₄ and o₂ 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.
perturbation change in composition shift in equilibrium
the temperature is raised. the pressure of o₂ will select to the right
to the left
(none)
the temperature is lowered. the pressure of h₂o will select to the right
to the left
(none)

Explanation:

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

For an endothermic reaction \(CO_{2}(g)+2H_{2}O(g)\to CH_{4}(g)+2O_{2}(g)\), heat is a reactant (\(\Delta H> 0\)). When the temperature is raised, according to Le - Chatelier's principle, the system will shift to consume the added heat.

Step2: Effect of raising temperature

Since heat is a reactant in an endothermic reaction, increasing the temperature shifts the equilibrium to the right. As the reaction proceeds to the right, more \(CH_{4}\) and \(O_{2}\) are formed. So the pressure of \(O_{2}\) (a product) will increase.

Step3: Effect of lowering temperature

When the temperature is lowered, the system will shift to produce heat. For an endothermic reaction, the equilibrium shifts to the left. As the reaction proceeds to the left, more \(CO_{2}\) and \(H_{2}O\) are formed. So the pressure of \(H_{2}O\) (a reactant) will increase.

Answer:

• When the temperature is raised: The pressure of \(O_{2}\) will increase and the equilibrium shifts to the right.
• When the temperature is lowered: The pressure of \(H_{2}O\) will increase and the equilibrium shifts to the left.