How do I solve problems where I'm asked for $\Delta H^\circ_f$ and given $\Delta H^\circ_f$ values for equations with the involved elements? For example:

Find the $\Delta H^\circ_f$ of the following reaction: $\ce{C + O2 -> CO2}$, given:

$\ce{SrO + CO2 -> SrCO3}$ ; $\Delta H^\circ_f = \pu{-234 kJ}$

$\ce{2SrO -> 2Sr + O2}$ ; $\Delta H^\circ_f = \pu{1184 kJ}$

$\ce{2SrCO3 -> 2Sr + 2C + 3O2}$ ; $\Delta H^\circ_f = \pu{2440 kJ}$

I know that I'm somehow supposed to manipulate the equations to make them "add up" (adding straight down) to the target equation, but I'm not sure how, exactly, to achieve this.

Also, is there a different and/or better method for doing these?

I've made some progress on my own, using this About.com page. What I'm wondering now, is if elements are locked into compounds, can I split them up, as long as I keep them on the same side? In the example above, I've manipulated it to where I have $\ce{2CO2 + O2 -> 2C + 3O2}$; am I allowed to split the $\ce{2CO2}$ into $\ce{2C + 2O2}$?

I've made it as far as getting to $\ce{2C + 2O2 -> 2CO2}$. If you take these as molar ratios, it's the same as $\ce{C + O2 -> CO2}$, which is the target equation. What do I have to do to the enthalpies to go from the first to the second?

  • $\begingroup$ Hess's law isn't too difficult once you know what to look for. What may have caused you difficulty is 1.) either you are missing a reaction, or 2.) one of those listed reactions is incorrect. In order to properly use Hess's law, you need at least one reaction where $\ce{CO2}$ is present. $\endgroup$
    – user3735
    Commented Mar 13, 2014 at 3:35
  • $\begingroup$ @Jun-GooKwak Am I approaching it the right way with the manipulating equations method? $\endgroup$
    – evamvid
    Commented Mar 13, 2014 at 3:36
  • $\begingroup$ Not quite. You can either multiply those reactions by integers, or use the opposite reaction. Essentially you are trying to derive the equation that you have above by canceling out common elements on both sides of the products and reactants. In addition, you are keeping track of the enthalpies and find the sum. However, I do not see any reaction with carbon dioxide as a product or reactant. $\endgroup$
    – user3735
    Commented Mar 13, 2014 at 3:37
  • $\begingroup$ @Jun-GooKwak You're right; on the problems sheet I got the question from, it's actually a $CO _{2}$ in the first one, not an $O _{2}$. $\endgroup$
    – evamvid
    Commented Mar 13, 2014 at 3:41
  • $\begingroup$ Ah, excellent. It's solvable now. I'll do this example for you and I think it will be quite clear soon. $\endgroup$
    – user3735
    Commented Mar 13, 2014 at 3:43

1 Answer 1


Find the $\Delta H^\circ_f$ of the following reaction:

$$\ce{C + O2 -> CO2}$$

We have our list of reactions with known enthalpies:

$\ce{SrO + CO2 -> SrCO3} \quad \Delta H^\circ_f = \pu{−234 kJ}$

$\ce{2SrO -> 2Sr + O2} \quad \Delta H^\circ_f = \pu{1184 kJ} $

$\ce{2SrCO3 -> 2Sr + 2C + 3O2} \quad \Delta H^\circ_f = \pu{2440 kJ} $

Hess's Law takes advantage of the fact that enthalpy is a state function, looking at our reactions, we need to find a way to derive what we want:

Carbon dioxide is on the reactants side, however, we want it on the products, so we can run that reaction reverse; however we need to change the sign of enthalpy if we do that:

$\ce{SrCO3 -> SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$

We need to get rid of $\ce{SrO}$, thus, we use our second equation:

$\ce{2SrO -> 2Sr + O2 } \quad \Delta H^\circ_f = \pu{1184 kJ} $

Since, there's only one strontium oxide in the first reaction, we divide by the integer 2 across the chemical equation including the enthalpy of formation:

$\ce{SrO -> Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $

Now, with equation 3, we need carbon and oxygen on the reactants side, and need to remove $\ce{SrCO3}$ and also divide again by 2, since we only have one strontium.

$\ce{Sr + C + 3/2O2 -> SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $

Now we add up all our reactions and enthalpies, and cancel out common elements on both sides:

$\ce{SrCO3 -> SrO + CO2 } \quad \Delta H^\circ_f = \pu{234 kJ}$

$\ce{SrO -> Sr + 1/2O2 } \quad \Delta H^\circ_f = \pu{592 kJ} $

$\ce{Sr + C + 3/2O2 -> SrCO3 } \quad \Delta H^\circ_f = \pu{-1220 kJ} $

$\ce{C + O2 -> CO2} \quad \Delta H^\circ_f= (\pu{234 kJ}) + (\pu{592 kJ}) + (\pu{-1220 kJ}) = \pu{-394 kJ}$


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