# Comparing the enthalpy change between the formation of water in two states

The enthalpy of the reaction $$\ce{H2(g) + 1/2 O2 (g) -> H2O (g)}$$ is $$\Delta H_1$$ and that of $$\ce{H2(g) + 1/2 O2 (g) -> H2O (l)}$$ is $$\Delta H_2$$. Then,

a) $$\Delta H_1 < \Delta H_2$$
b) $$\Delta H_1 + \Delta H_2 = 0$$
c) $$\Delta H_1 > \Delta H_2$$
d) $$\Delta H_1 = \Delta H_2$$

According to me, in a reaction forming a gaseous product the pressure would be greater than the one forming a liquid product. Therefore the enthalpy change should be greater for the first reaction. However, the correct answer is a). Where am I going wrong?

• Hess' Law might prove to be your friend here. Welcome to Chem SE.. Commented Aug 9, 2022 at 4:20
• $\Delta H_2 + \Delta H_{\ce{H2O},\mathrm{evap}} = \Delta H_1$ Commented Aug 9, 2022 at 8:26
• In other words, (a) is wrong and (c) is right. State (g) is a hypothetical ideal gas state of water vapor at 25 C and 1 bar that has not condensed. Commented Aug 9, 2022 at 11:15