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?

  • 1
    $\begingroup$ Hess' Law might prove to be your friend here. Welcome to Chem SE.. $\endgroup$ Commented Aug 9, 2022 at 4:20
  • $\begingroup$ $\Delta H_2 + \Delta H_{\ce{H2O},\mathrm{evap}} = \Delta H_1$ $\endgroup$
    – Poutnik
    Commented Aug 9, 2022 at 8:26
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    $\begingroup$ 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. $\endgroup$ Commented Aug 9, 2022 at 11:15

2 Answers 2


Well in order to convert H2O into liquid we need to lower the energy of system more than H2O of gaseous state (energy of gas is more than liquids and liquids have energy more than solids).

That means more have to be released in order to make it liquid.


Since enthalpy is Q at constant pressure the PV factor can be ignored [this is for the questioner]. The difference is the heat of condensation of the vapor that is released. Therefore the reaction to give liquid water is more negative than that to give gas. I looked it up the heat of formation of water vapor is -57.8 Kcal for liquid water -68.3. Now the problem is to define what is meant by larger? the less negative number or the greater release of heat. As usual test makers do not try to answer their own questions! [I notice that in the comments this dilemma was not resolved. altho Chet tried]. I personally deplore questions that make knowing sign conventions more important than knowing chemistry.


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