Question

Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 2) X(s)+2Cl2(g)⟶XCl4(s)ΔH2=+356.9 kJ 2) X(s)+2Cl2(g)⟶XCl4(s)ΔH2=+356.9 kJ 3) 12H2(g)+12Cl2(g)⟶HCl(g)ΔH3=−92.3 kJ 3) 12H2(g)+12Cl2(g)⟶HCl(g)ΔH3=−92.3 kJ 4) X(s)+O2(g)⟶XO2(s)ΔH4=−639.1 kJ 4) X(s)+O2(g)⟶XO2(s)ΔH4=−639.1 kJ 5) H2O(g)⟶H2O(l)ΔH5=−44.0 kJ 5) H2O(g)⟶H2O(l)ΔH5=−44.0 kJ what is the enthalpy, ΔH,ΔH, for this reaction? XCl4(s)+2H2O(l)⟶XO2(s)+4HCl(g)

Answer 1

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

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

ΔH = -793,6 kJ

I hope it helps!