I understand this is a simple question (I’m in high school), but for some reason the numbers just are not working for me.

Given: $$\Delta H^\circ_\text{c} = -1301\ \mathrm{kJ}\ \text{for}\ \ce{C2H2(g)}$$ $$\Delta H^\circ_\text{f} = -394\ \mathrm{kJ}\ \text{for}\ \ce{CO2(g)}$$ $$\Delta H^\circ_\text{f} = -242\ \mathrm{kJ}\ \text{for}\ \ce{H2O(g)}$$ Calculate $\Delta H_\text{f}$ for $\ce{C2H2(g)}$.

My process: $$(1) : \ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(g) }\quad \Delta H^\circ_\text{r} = -1301\ \mathrm{kJ}$$ $$(2) : \ce{C(s) + O2(g) -> CO2(g)} \quad \Delta H^\circ_\text{r} = -394\ \mathrm{kJ}$$ $$(3) : \ce{H2(g) + 1/2O2(g) -> H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$

$$\text{Target:}\ \ce{2C(s) + H2(g) -> C2H2(g)}$$

After some experimentation I arrive at:

$$-1\times(1) + 2\times(2) + (3) = \text{Target}$$

As shown through: $$-1\times(1){:}\ \ce{2CO2(g) + H2O(g) -> C2H2(g) + 5/2 O2(g)} \quad \Delta H^\circ_\text{r} = 1301\ \mathrm{kJ}$$ $$2\times(2){:}\ \ce{2C(s) + 2O2(g) -> 2CO2(g)} \quad \Delta H^\circ_\text{r} = -788\ \mathrm{kJ}$$ $$(3){:}\ \ce{H2(g) + 1/2O2(g) -> H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$ $$\text{Sum:}\ \ce{2C(s) + H2(g) -> C2H2(g)} \quad \Delta H^\circ_\text{f} = 271\ \mathrm{kJ}$$

Now as far as I know, all the givens are accurate empirical values. But the empirical value for $\Delta H^\circ_\text{f}$ for $\ce{C2H2(g)}$ is $227\ \mathrm{kJ}$.

What’s even weirder is that this disparity $(271-227=44)$ is equal to the difference between standard enthalpies of formation for water in gas and liquid states $(-242 - -286 = 44)$.

Have I made an error somewhere, or is water formed in the liquid state during the combustion of acetylene?

  • $\begingroup$ The standard enthalpy of reaction involves the species in their standard states. So what is the standard state of water? $\endgroup$
    – bon
    Apr 11, 2015 at 18:38
  • $\begingroup$ Oh. The reason water is formed as a gas is a result of the enthalpy change, but if we're looking just at the product in standard conditions it's just a liquid? And if we were to wait long enough... I guess the water would return to liquid state as the heat returned to 25°C? $\endgroup$
    – MattDs17
    Apr 11, 2015 at 18:42
  • $\begingroup$ Welcome to chemistry.SE BTW! $\endgroup$
    – M.A.R.
    Apr 11, 2015 at 18:46
  • $\begingroup$ Could I get confirmation on my above conclusion? I think it makes sense, but I want to be sure. I guess my teacher has been using $\ce{H2O(g)}$ for simplicity's sake. $\endgroup$
    – MattDs17
    Apr 11, 2015 at 19:03
  • $\begingroup$ Yes that is correct $\endgroup$
    – bon
    Apr 11, 2015 at 19:18

1 Answer 1


Like many past chemistry.se questions, the answer to this one hinges on the presence of the $^\circ$.

$\Delta H^\circ_\text{c}$ is the $\Delta H$ value for a "standard" reaction where acetylene at 1 atm of pressure is combusted isothermally at 298 K with 1 atm of oxygen gas, forming 1 atm of $\ce{CO2}$ and liquid water (which is the most thermodynamically stable form of water at 1 atm and 298 K).

I think it's safe to say that no combusion of acetylene, ever in the history of the world, has occured under those conditions. But those "standard" conditions are the only conditions where the $\Delta H^{\circ}_f$ is guaranteed to be -227 kJ/mol.

By the way, in practical and engineering literatures, $\Delta H_c^{\circ}$ of a fuel is commonly called the higher heating value of the fuel. It corresponds to $\Delta H_{rxn}^{\circ}$ of a reaction like

$\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(l)}$

The lower heating value, conversely, corresponds to $\Delta H_{rxn}^{\circ}$ of reactions like

$\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(g)}$


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