Question

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

Answer 1

The ΔHrxn for the given reaction is 241.8 kJ/mol.

Step-by-step explanation:

The enthalpy change (ΔHrxn) for the reaction H2O(g) → H2(g) + ½ O2(g) can be calculated by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products. Since the enthalpy of formation for H2O(g) is known, we can use standard enthalpy of formation values for H2(g) and O2(g) to calculate the enthalpy change. The balanced equation is:

2H2O(g) → 2H2(g) + O2(g)

The standard enthalpy of formation for H2O(g) is -241.8 kJ/mol, the standard enthalpy of formation for H2(g) is 0 kJ/mol, and the standard enthalpy of formation for O2(g) is 0 kJ/mol.

Therefore, the ΔHrxn for the reaction is:

ΔHrxn = 2×0 kJ/mol + 0 kJ/mol - (-241.8 kJ/mol)

= 241.8 kJ/mol