# Computing enthalpy of formation of dissolved gas from its enthalpy of dissolution

Provided that I know the enthalpy of formation $$\Delta H_f^0(X_{g})$$ of a substance $$X$$ in gas phase and its enthalpy of dissolution $$\Delta H^0_{sol}(X)$$ in the same conditions, considering that the dissolution reaction is $$X_g \leftrightharpoons X_{aq}$$ is it legit to consider that the enthalpy of formation of that substance $$X$$ in aqueous phase would be $$\Delta H^0_f(X_{aq}) = \Delta H^0_{sol}(X) - \Delta H_f^0(X_{g})$$?

As a concrete exemple, considering that the enthalpy of formation of $$H_2$$ in gas phase is 0 kJ/mol and that its enthalpy of dissolution is approximately -4.15 kJ/mol (using the data provided here), can we conclude that the enthalpy of formation of $$H_2$$ in aqueous phase is approximately -4.15 kJ/mol?

• That seems correct. – Buck Thorn Aug 21 '19 at 17:13