Using enthalpies of formation and combustion to find formation enthalpy of Acetylene (Hess' Law)

Question

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?

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

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