# Hess's Law - enthalpy of combustion

The enthalpy of combustion of a solid carbon to form carbon dioxide is $-393.7\ \mathrm{kJ/mol}$ carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is $-283.3\ \mathrm{kJ/mol}\ \ce{CO}$. Use these data to calculate $\Delta H$ for the reaction $$\ce{2C(s) + O_2 (g) -> 2CO(g)}$$

Attempt at solution: From the description, I first wrote down my two equations: $$1)\ \ce{C + O_2 -> CO_2} \\ 2)\ \ce{CO + O -> CO_2}$$

I want to have a $\ce{2CO}$ as a product. In the second equation I see $\ce{CO}$ as reactant, so I multiply that equation by two and reverse it $$\ce{2CO_2 -> 2O + 2CO}$$ The enthalpy thereby changes by $-2(-283.3\ \mathrm{kJ/mol}) = 566.6\ \mathrm{kJ/mol}$. Now I somehow need to get rid of that $\ce{2O}$ on the reactant side, but I don't see how to do it. I tried adding the first equation but that doesn't work.

Edit: The two correct equations are $$1)\ \ce{C + O_2 -> CO_2} \\ 2)\ \ce{2CO + O_2 -> 2 CO_2}$$

We reverse the second equation to get $$\ce{2CO_2 -> 2CO + O_2}$$ The enthalpy thereby changes to $283.3\ \mathrm{kJ/mol}$. Adding the first equation to the preceding one gives $$\ce{C + O_2 + 2CO_2 -> CO_2 + 2CO + O_2}$$ or after deleting common species $$\ce{C + CO_2 -> 2CO}$$ We still need an $\ce{O_2}$ at the reactant side, so we add equation 1 again to obtain $$\ce{2C + O_2 -> 2CO}$$ This is what we wanted. But how do I compute the enthalpy now?

The combustion reaction for carbon monoxide is $$\huge{\ce{2CO (g) + O2 (g) \rightarrow 2CO2(g)}}$$

Steps to solve the problem:

We primarily use the formula: $$\Delta H_{p,T} = \Delta H_\text{formation: products} - H_\text{formation: reactants}$$

1. Take each enthalpy in the reactions given as a variable. For example, $\Delta H_\text{f}$ for $\ce{CO}$ is $x$, for $\ce{CO2}$ is $y$, and so on. Then, instead of mixing the reactions, try to reach relations between variables. For example, from the combustion of $\ce{CO}$, we conclude that $2y − 2x=−283.3~\mathrm{kJ}$

• Note that the enthalpy of the formation of the most stable form of the elements is zero.
2. Assuming that the solid carbon in the question is graphite (i.e. the most stable form of the carbon in nature) repeat 1 for the reaction of its combustion. From there, you'll be able to calculate $y$.

3. Insert $y$ in the first equation to get $x$.
4. Repeat step 1 for the last equation. Note that two of the species have an enthalpy of formation of 0.

Use the following formula: $$\text{sum of enthalpies of combustion of reactants}- \text{sum of enthalpies of combustion of products} = \text{enthalpy change of the reaction}$$ So for $\ce{2C + O2 -> 2CO}$:

• Sum of enthalpies for combustion of reactants: $2\times-393.17$, as $-393.17$ is the enthalpy value for the combustion of one mole of solid carbon.
• Sum of enthalpies of combustion of products: $2\times -283.3$, as $-283.3$ is the enthalpy change for the combustion of one mole of carbon monoxide.
• Therefore enthalpy change of the reaction: $$(2\times -393.17)-(2\times -283.3)= -219.7~\mathrm{kJ/mol}$$Take care with the double minus signs when using your calculator.
• You should use units properly and consistently. Note that the k for kilo is not capitalised, the m in mol is neither; that units are separated from their value by a space and that you cannot add (or subtract) numbers from each other to generate units out of thin air. Units must be added to units to remain units. – Jan Feb 4 '17 at 20:05