Consider these reactions, where M represents a generic metal. Δ𝐻= ???KJ

Question

2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ𝐻1=−575.0 kJ

HCl(g)⟶HCl(aq)  Δ𝐻2=−74.8 kJ

H2(g)+Cl2(g)⟶2HCl(g) Δ𝐻3=−1845.0 kJ

MCl3(s)⟶MCl3(aq)  Δ𝐻4=−222.0 kJ

Use the given information to determine the enthalpy of the reaction

2M(s)+3Cl2(g)⟶2MCl3(s)

Answer 1

This is an example of the use of Hess' Law where we can use known reactions to find the value of an unknown (target) reaction. Our target reaction is 2M(s) + 3Cl₂(g) ==> 2MCl₃(s) ... TARGET EQUATION

Known reactions:

(1) 2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)Δ𝐻1=−575.0 kJ

(2) HCl(g)⟶HCl(aq)  Δ𝐻2=−74.8 kJ

(3) H2(g)+Cl2(g)⟶2HCl(g) Δ𝐻3=−1845.0 kJ

(4) MCl3(s)⟶MCl3(aq)  Δ𝐻4=−222.0 kJ

Procedure:

copy(1): 2M(s) + 6HCl(aq) ===> 2MCl3(aq) + 3H2(g) ... ∆H = -575 kJ

(2)x6: 6HCl(g) =========> 6HCl(aq) ... ∆H = 6x-74.8 = -449 kJ

(3)x3: 3H2(g) + 3Cl2(g) ==> 6HCl(g) ... ∆H = 3x-1845 = 5535 kJ

reverse(4)x2: 2MCl3(aq) ==> 2MCl3(s) ... ∆H = +2x220 = +440 kJ

Add reactions and cancel and combine common elements to obtain...

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

And add ∆H values to get the answer

-575 + (-449) + (-5535) + 440 = -6119 kJ