I found the following reactions including the Energy-Delta all in Wikipedia and in at least one more source each:

(Delta with "-" are exothermic reaction which is generating heat)

Steam Reformation: $$\ce{CH4 + H2O <=> 3H2 + CO}\quad\Delta H=+206\ \mathrm{kJ/mol}\tag1$$ Water gas-shift: $$\ce{CO + H2O <=> CO2 + H2}\quad\Delta H=-42\ \mathrm{kJ/mol}\tag2$$ Hydrogen heat of combustion: $$\ce{2H2 + O2 <=> 2H2O}\quad\Delta H=-572\ \mathrm{kJ/mol}\tag3$$ Methane heat of combustion: $$\ce{CH4 + 2O2 <=> CO2 + 2H2O}\quad\Delta H=-891\ \mathrm{kJ/mol}\tag4$$

Now, if I add to (1) $\ce{H2O + 2O2}$ I may do (1),(2), and (3) in a row: $$\ce{(1) CH4 +2H2O +2O2 <=> 3H2 + CO +H2O +2O2}\quad\Delta H=+206\ \mathrm{kJ/mol}$$ $$\ce{(2) <=> 4H2 + CO2 +2O2}\quad\Delta H=(+206-42)\ \mathrm{kJ/mol}$$ $$\ce{(3) <=> 4H2O + CO2}\quad\Delta H=(+206-42+2\times(-572))\ \mathrm{kJ/mol}$$

Subtracting $\ce{2H2O}$ on both ends of the equation I get:

$$\ce{CH4 +2O2 <=> 2H2O + CO2}\quad\Delta H=-980\ \mathrm{kJ/mol}\tag5$$

This almost looks like the reactions in (4) only with a $\Delta H$ which is $89\ \mathrm{kJ/mol}$ different.

From my understanding of the conservation of energy that shall not be possible.

What do I miss?

  • 1
    $\begingroup$ Make sure to account for whether H2O is liquid or gas $\endgroup$
    – Andrew
    Oct 1, 2022 at 20:03
  • $\begingroup$ @Andrew As I understood it the energy difference between liquid or gas shall only make a difference for the lower heating value (lhv). The above number are all regarding the higher heating value (hhv). Or not? Additionally the vaporation energy of water is 40,66 kj/mol. If that is relevant for 2 water molecules than there still would be an error of 8kj/mol. $\endgroup$
    – Thomas R
    Oct 1, 2022 at 20:33
  • $\begingroup$ @Andrew The only possible though improbable liquid hydrogen is in (1) the Water gas-shift which takes place between 90-230°. Even this would just meen a correction of 40.66 kj/mol. Though even if thats relevant there still would be an error of 48 kj/mol. $\endgroup$
    – Thomas R
    Oct 1, 2022 at 21:04
  • $\begingroup$ The heat of vaporization of water at 25 C is typically reported as approximately 44 kJ/mol. I'm not sure where you're getting 40.66. The fact that your error is approx. 2 x 44 is not a coincidence. $\endgroup$
    – Andrew
    Oct 1, 2022 at 22:28
  • 1
    $\begingroup$ $\Delta H_r$ values are temperature dependent. You can only add them in the way that you are trying to do if they are all from the same temperature. The values you are using appear to be the standard enthalpy changes for these reactions, which by definition are the values for 25 C. Whether or not you understand what standard enthalpy changes are, it is relatively easy to find the state of H2O in the reactions as defined for the values you have given. It is not the same in all of them. $\endgroup$
    – Andrew
    Oct 2, 2022 at 7:46

1 Answer 1


Let's look at the 4 main reactions that we need to consider here. The reactions are as follows: Steam reformation, Water gas-shift, Combustion of hydrogen and the combustion of methane.

Similar to what is given in the question, I've listed the reactions with the states involved. [1]

\begin{align} \ce{CH4(g) + H2O(g) &-> 3H2(g) + CO (g)} &\Delta H^°_{298} = \pu{205.9 kJ/mol} \label{steam-reform} \tag{1} \\ \ce{CO(g) + H2O(g) &-> CO2(g) + H2(g)} &\Delta H^°_{298} = \pu{-41 kJ/mol} \label{water-gas} \tag{2} \\ \ce{2H2(g) + O2(g) &-> 2H2O(l)} &\Delta H^\circ_{298} = \pu{-572 kJ/mol} \label{hydrogen} \tag{3} \\ \ce{CH4(g) + 2O2(g) &-> CO2(g) + 2H2O(l)} &\Delta H^\circ_{298} = \pu{-891 kJ/mol} \label{methane} \tag{4} \end{align}

Here, the goal is to get to \eqref{methane} from the other three equations. The path taken in the OP's solution produces the following result.

$$\ce{CH4(g) + 2H2O(g) + CO(g) + 4H2 + 2O2 -> 4H2(g) + CO(g) + CO2(g) + 4H2O(l)} $$

Simplifying the reaction by removing common factors, we get,

$$\ce{CH4(g) + 2H2O(g) + 2O2 -> CO2(g) + 4H2O(l)}$$

The $\Delta H^\circ_{298}$ value for the above mentioned reaction is correct to what the OP proposes and rightfully so. It comes out to be $\pu{979.1 kJ/mol}$. However, the more observant would notice. This isn't \eqref{methane}. We're missing one reaction.

$$\ce{H2O(g) -> H2O(l)}$$

The value of $\Delta H^\circ_{298}$ for this reaction is observed to be $\pu{44 kJ/mol}$. Using this in the equation we got previously and eliminating some of the water, we get \eqref{methane} with the right value of $\Delta H^\circ_{298}$.


  1. Subramani, V.; Sharma, P.; Zhang, L.; Liu, K. Catalytic Steam Reforming Technology for the Production of Hydrogen and Syngas. Hydrogen and Syngas Production and Purification Technologies 2009, 14–126. DOI: 10.1002/9780470561256.ch2.
  • $\begingroup$ To make this answer more generally useful, it would be good to explain why H2O(g) is used in those two reactions rather than H2O(l) (the standard state at 25 C). The reason (in case you don't already know it) is that phase changes are generally the largest sources of differences in $\Delta_r H$ between different temperatures. Otherwise the differences arise only from the difference in heat capacities between reactants and products, which is typically fairly small. By using H2O(g) at 25 C, the resulting $\Delta H$ is a closer approximation of what it would be at T>100 C. $\endgroup$
    – Andrew
    Oct 3, 2022 at 13:14
  • $\begingroup$ I think your formula (4) is wrong. There shall be 2O2 and 2H2O. Apart from that I did not found a source yet, that the energy-values are for a temperature of 298° Kelvin. And apart from the formula (3) I did not find sources for the states (gas or liquid) for (1), (2), (4) $\endgroup$
    – Thomas R
    Oct 4, 2022 at 9:23
  • $\begingroup$ @ThomasR, the states for (4) come from the general definition of enthalpy of combustion. I'll add the source to the (1) and (2). It is behind a paywall though. $\endgroup$ Oct 4, 2022 at 12:07
  • $\begingroup$ @ThomasR - It's actually rare to find $\Delta H$ values for temperatures other than 298 K, since that's the standard. If a source does not specify a temperature, it's almost always 298. But if you want to be sure, check authoritative sources like NIST or scientific journals, which will specify temp or use the proper notation $\Delta H^\circ$, where the $^\circ$ symbol indicates standard conditions of 298 K unless otherwise indicated in a subscript. $\endgroup$
    – Andrew
    Oct 4, 2022 at 20:01

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