Answer:
ΔH = -793,6 kJ
Explanation:
It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:
If half-reactions are:
1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ
2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ
3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ
4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ
5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ
The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:
(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ
+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ
-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ
-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ
-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ
= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>
Where ΔH is:
ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ
<em>ΔH = -793,6 kJ</em>
I hope it helps!
