Question

In: Chemistry

Consider these reactions, where M represents a generic metal 1. 2M(s)+6HCl(aq)------>2MCl3(aq)+3H2(g) Delta H1=-749.0 KJ 2. HCl(g)------->HCl(aq)...

Consider these reactions, where M represents a generic metal

1. 2M(s)+6HCl(aq)------>2MCl3(aq)+3H2(g) Delta H1=-749.0 KJ

2. HCl(g)------->HCl(aq) Delta H2=-74.8 KJ

3. H2(g)+Cl2(g)------->2HCl(g) Delta H3=-1845.0 KJ

4. MCl3(s)-------->MCl3(aq) Delta H4=-298.0 KJ

Use the information above to determine the enthalpy of the following reaction.

2M(s)+3Cl2(g)----->2MCl3(s) Delta H=???

Expert Solution

2M(s)+6HCl(aq) -----> 2MCl₃(aq)+3H₂(g) : ΔH₁ = -749.0 KJ ----(1)

HCl(g) -----> HCl(aq) ; ΔH₂ = -74.8 KJ ----(2)

H₂(g)+Cl₂(g) -----> 2HCl(g) ; ΔH₃ = -1845.0 KJ ----(3)

MCl₃(s) ------> MCl₃(aq) : ΔH₄ = -298.0 KJ ----(4)

Use the information above to determine the enthalpy of the following reaction.

2M(s)+3Cl₂(g) -----> 2MCl₃(s) ΔH= ? ----(5)

Equ(5) can be obtained from remaining four equations as follows:

Eqn(5) = Eqn(1) + [6xEqn(2)]+[3xEqn(3) ] + [2xreverse of Eqn(4)]

ΔH= ΔH₁ +(6xΔH₂ ) + (3xΔH₃ ) + (2x(-ΔH₄ )

= -749 + (6x(-74.8)) + (3x(-1845)) + (2x(-(-298))) kJ

= -6136.8 kJ

queen_honey_blossom answered 4 months ago

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