2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ𝐻1=−947.0 kJ HCl(g)⟶HCl(aq) Δ𝐻2=−74.8 kJ H2(g)+Cl2(g)⟶2HCl(g) Δ𝐻3=−1845.0 kJ MCl3(s)⟶MCl3(aq) Δ𝐻4=−256.0 kJ 2M(s)+3Cl2(g)⟶2MCl3(s)

Question

Consider these reactions, where M represents a generic metal.

1. 
   
2. 
   
3. 
   
4. 
   

Use the given information to determine the enthalpy of the reaction

Answer 1

HESS' LAW

2M(s) + 3Cl2(g) ===> 2MCl3(s) TARGET EQUATION

GIven:

(1) 2M(s) + 6HCl(aq) ==> 2MCl3(aq) + 3H2(g) ... -947.0 kJ

(2) HCl(g) ==> HCl(aq) ... -74.8 kJ

(3) H2(g) + Cl2(g) ==> 2HCl(g) ... -1845.0 kJ

(4) MCl3(s) ==> MCl3(aq) ... -256.0 kJ

Procedure:

copy (1): 2M(s) + 6HCl(aq) ==> 2MCl3(aq) + 3H2(g) ... ∆H = -947.0 kJ

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

copy (2)x6: 6HCl(g) ==> 6HCl(aq) ... ∆H = -448.8 kJ

reverse(4)x2: 2MCl3(aq) ==> 2MCl3(s) ... ∆H = +512.0 kJ

Add up these equations canceling and combining like terms to obtain the following:

2M(s) + 3Cl2(g) ==> 2MCl3(s) ... TARGET EQUATION

∆H = -947 + -5535 + -449 + 512

∆H = -6419 kJ