Question

Given the reaction

2 H2(g) + O2(g) → 2 H2O(g)
ΔH = –483.6 kJ

What is ΔHrxn for the following?
H2O(g) → H2(g) + ½ O2(g)
ΔH = ? kJ

Answer 1

The enthalpy change for the reaction H2O(g) → H2(g) + ½O2(g) is +241.8 kJ.

In the given reaction: 2H2(g) + O2(g) → 2H2O(g), the enthalpy change (ΔH) is -483.6 kJ.

The question asks for the enthalpy change for the reverse reaction: H2O(g) → H2(g) + ½O2(g).

To find this, we can apply Hess's Law.

We can reverse the given reaction and change the sign of the enthalpy change, which would give us: 2H2O(g) → 2H2(g) + O2(g) with ΔH = +483.6 kJ.

Since we need the enthalpy change for 1 mole of H2O, we divide the enthalpy change by 2, resulting in a ΔHrxn of +241.8 kJ.