Question

# iii) Calculate ΔH for the reaction 2 Al (s) + 3 Cl2 (g) → 2 AlCl3...

iii) Calculate ΔH for the reaction 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s) from the data

2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g) ΔH = -1049. kJ

HCl (g) → HCl (aq) ΔH = -74.8 kJ

H2 (g) + Cl2 (g) → 2 HCl (g) ΔH = -1845. kJ

AlCl3 (s) → AlCl3 (aq) ΔH = -323. kJ

We have to calculate ΔH for the reaction: 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)

It is given that:

2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g) ΔH = -1049. kJ -----------------------------------------------(1)

HCl (g) → HCl (aq) ΔH = -74.8 kJ -----------------------------------------------------------------------------------------(2)

H2 (g) + Cl2 (g) → 2 HCl (g) ΔH = -1845. kJ --------------------------------------------------------------------------(3)

AlCl3 (s) → AlCl3 (aq) ΔH = -323. kJ-------------------------------------------------------------------------------------(4)

We must calculate:

(1) + 6x (2) + 3x (3) - 2 x (4)

then we will get the required equation.

ΔH for this reaction is:

-1049. kJ + 6 x -74.8 kJ + 3 x -1845. kJ - (-323. kJ)

= -7035.8 + 323 kJ

= - 6172.8 kJ