Question

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241.8 kJ 2 ) X ( s ) + 2 Cl 2 ( g ) ⟶ XCl 4 ( s ) Δ H 2 = + 361.7 kJ 3 ) 1 2 H 2 ( g ) + 1 2 Cl 2 ( g ) ⟶ HCl ( g ) Δ H 3 = − 92.3 kJ 4 ) X ( s ) + O 2 ( g ) ⟶ XO 2 ( s ) Δ H 4 = − 607.9 kJ 5 ) H 2 O ( g ) ⟶ H 2 O ( l ) Δ H 5 = − 44.0 kJ what is the enthalpy, Δ H , for this reaction? XCl 4 ( s ) + 2 H 2 O ( l ) ⟶ XO 2 ( s ) + 4 HCl ( g )

Answer 1

Answer:

-767,2kJ

Explanation:

It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ

The sum of (4) - (2) produce:

6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ

(6) + 4×(3):

7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ

(7) - 2×(1):

8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ

(8) - 2×(5):

9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= -767,2kJ

I hope it helps!