Use the following data
2 H₂(g) + O₂(g) $\rightarrow$ 2 H₂O(g) $\Delta$ H_rxn° = -483.6 kJ/mol_rxn
2 H₂(g) + O₂(g) $\rightarrow$ 2 H₂O(l) $\Delta$ H_rxn° = -571.6 kJ/mol_rxn
To calculate $\Delta$ H_rxn° for the reaction
H₂O(l) $\rightarrow$ H₂O(g)

Question

Use the following data
2 H₂(g) + O₂(g)

\rightarrow

2 H₂O(g)

\Delta

H_rxn° = -483.6 kJ/mol_rxn
2 H₂(g) + O₂(g)

\rightarrow

2 H₂O(l)

\Delta

H_rxn° = -571.6 kJ/mol_rxn
To calculate

\Delta

H_rxn° for the reaction
H₂O(l)

\rightarrow

H₂O(g)

Quizplus · Accepted Answer

The correct answer is found by subtracting the enthalpy change of the reaction forming liquid water from the enthalpy change of the reaction forming gaseous water: -483.6 kJ/mol - (-571.6 kJ/mol) = 88 kJ/mol for 2 moles of water, or 44.0 kJ/mol for 1 mole of water.

Use the Following Data 2 H2(g) + O2(g) →\rightarrow→ 2 H2O(g) &Emsp;

Use the Following Data
2 H₂(g) + O₂(g) $\rightarrow$ 2 H₂O(g) &Emsp;