Calculate the ΔH for the reaction: 2 Al (s) + 3 Cl2 (g) --> 2AlCl3 (s) from the following data.

Question

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

2) HCl (g) --> HCl (aq) ΔH= -74.8 kJ/Mole

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

4) AlCl3 (s) --> AlCl3 (aq) ΔH= -323 kJ/Mole

Answer 1

Hess' Law

2Al(s) + 3Cl2(g) ==> 2AlCl3(s) ... TARGET EQUATION

GIVEN:

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

2) HCl (g) --> HCl (aq) ΔH= -74.8 kJ/Mole

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

4) AlCl3 (s) --> AlCl3 (aq) ΔH= -323 kJ/Mole

PROCESS:

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

copy (3) & x3: 3H2(g) + 3Cl2(g)=> 6HCl(g) ... ∆H = -5535 kJ

reverse (4) & x2: 2AlCl3(aq) => 2AlCl3(s) ... ∆H = +646 kJ

copy (2) x 6: 6HCl(g) ==> 6HCl(aq) ... ∆H = -449 kJ

Add them up and combine or cancel as appropriate:

2Al(s) +6HCl(aq) +3H2(g) +3Cl2(g) + 2AlCl3(aq) + 6HCl(g) =>2AlCl3(aq) + 3H2(g) +6HCl(g) +2AlCl3(s) +6HCl(aq)

This leaves us with...

2Al(s) + 3Cl2(g) ==> 2AlCl3(s) which is our TARGET EQUATION

∆H = -1049 + (-5535) + 646 + (-449) = -6387 kJ