Question

4AlCl₃(s)+3O₂ (g)→2Al₂O₃ (s)+6Cl₂ (g);∆H=-529.0 kJ

Determine ∆H for the following thermochemical equation.
Cl₂ (g)+Al₂O₃ (s)→AlCl₃ (s)+O₂ (g)
A. +264.5 kJ
B. +529.0 kJ
C. +88.2 kJ
D. +176.3 kJ
E. -176.3 kJ

Answer 1

To find the enthalpy change for the reverse reaction, one must take the given enthalpy change for the forward reaction, divide it by the stoichiometric coefficients if necessary, and reverse the sign. For the reaction given, the enthalpy change for the desired reverse reaction is +264.5 kJ.

Step-by-step explanation:

The student is asking how to determine the enthalpy change (ΔH) of the reverse reaction given the enthalpy change for the forward reaction. In this case, the enthalpy change for the following reaction is being sought:

Cl₂ (g) + Al₂O₃ (s) → AlCl₃ (s) + O₂ (g)

Given the enthalpy change for the forward reaction:

4AlCl₃(s)+3O₂ (g) → 2Al₂O₃ (s)+6Cl₂ (g); ΔH = -529.0 kJ

The reverse reaction would have an equal magnitude but opposite sign for ΔH since reversing a reaction simply changes the sign of the enthalpy change. However, we must consider the stoichiometry of the original equation. The given reaction involves halving the stoichiometric coefficients of the products in the original equation. By doing this, the enthalpy change must also be halved. Therefore:

ΔH for the reverse reaction = 529.0 kJ / 2 = +264.5 kJ

This is the enthalpy change for the reverse reaction, which corresponds to Choice A in the given options.

The correct option is A,