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

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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)

Chemistry

1 answer:

Tems11 [2.7K] · Answer 1 · 2026-03-01T23:32:53+03:00

Response:

ΔH = -793.6 kJ

Reasoning:

The ΔH for this particular reaction can be calculated through Hess's law, which allows one to combine the ΔH values of the half-reactions to find the overall ΔH:

The half-reactions are as follows:

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 calculation involves summing (4) + 4×(3) - (2) - 2×(1) - 2×(5):

(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)

<pThus, ΔH is:

ΔH = -639.1 kJ -369.2 kJ -356.9 kJ +483.6 kJ +88.0 kJ

ΔH = -793.6 kJ

I trust this clarifies things!