The following reaction is reactant-favored at equilibrium at room temperature. COCl2(g) -> CO(g) + Cl2(g)

Will raising or lowering the temperature make it product-favored?

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To determine whether raising or lowering the temperature will make the reaction product-favored, we need to consider the reaction's enthalpy change (∆H) and the temperature dependence of the reaction quotient, Q.

The enthalpy change (∆H) can provide insight into whether the reaction is exothermic or endothermic.

- If ∆H < 0 (negative), the reaction is exothermic, meaning it releases heat.
- If ∆H > 0 (positive), the reaction is endothermic, meaning it absorbs heat.

In this case, we would need to examine experimental data or references to determine the enthalpy change (∆H) for the reaction COCl2(g) -> CO(g) + Cl2(g). However, let's assume it is exothermic for the purpose of this explanation.

At a lower temperature:
Lowering the temperature for an exothermic reaction causes the reaction to shift towards the products. This is because reducing the temperature decreases the kinetic energy of the molecules, and fewer reactant molecules have enough energy to overcome the activation energy barrier. Therefore, lowering the temperature makes the reaction product-favored in this case.

At a higher temperature:
Raising the temperature for an exothermic reaction causes the reaction to shift towards the reactants. Increasing the temperature provides more energy to the system, allowing a higher proportion of the reactant molecules to exceed the activation energy barrier. Consequently, raising the temperature makes the reaction reactant-favored.

It is important to note that this explanation assumes the reaction being exothermic. To obtain a definitive answer, it is necessary to consider the specific enthalpy change (∆H) for the reaction COCl2(g) -> CO(g) + Cl2(g) using experimental data or references.