Question

Calculate the standard enthalpy of formation of solid AlCl₃ from the following data:

2 Al(s) + 6 HCl(aq) → 2 AlCl₃(aq) + 3 H₂(g)                  ΔH^o = -1048 kJ

HCl(aq) → HCl(g)                  ΔH^o = 75 kJ

Cl₂(g) + H₂(g) → 2 HCl(g)                      ΔH^o = -186 kJ

AlCl₃(s) → AlCl₃(aq)                                                       ΔH^o = -323 kJ

Answer 1

Reaction 1 : 2 Al (s) + 6 HCl (aq) 2 AlCl₃ (aq) + 3 H₂ (g) : H^o₁ = -1048 kJ

Reaction 2 : HCl (aq) HCl (g)                                         : H^o₂ = 75 kJ

Reaction 3 : Cl₂ (g) + H₂ (g) 2 HCl (g)                            : H^o₃ = -186 kJ

Reaction 4 : AlCl₃ (s) AlCl₃ (aq)                                    : H^o₄ = -323 kJ

The equation for formation of solid AlCl₃ is

Al (s) + 3/2 Cl₂ (g) AlCl₃ (s)

We have to combine the given four reactions in such a way so as to obtain the equation for formation of solid AlCl₃. Then same operations can be performed on their respective enthalpies to get the standard enthalpy of formation.

Consider the equation : (0.5 * reaction 1) - (3 * reaction 2) + (1.5 * reaction 3) - (reaction 4)

This equation gives the equation for formation of solid AlCl₃

H^o_f AlCl₃ = (0.5 * H^o₁) - (3 * H^o₂) + (1.5 * H^o₃) - (H^o₄)

H^o_f AlCl₃ = (0.5 * -1048 kJ) - (3 * 75 kJ) + (1.5 * -186 kJ) - (-323 kJ)

H^o_f AlCl₃ = -524 kJ - 225 kJ - 279 kJ + 323 kJ

H^o_f AlCl₃ = -705 kJ