Determine the heat of reaction for the incomplete combustion of carbon. 2C solid plus O₂ gas react to form 2CO gas. Use the following thermochemical equations. I) 2C solid plus 2O₂ gas react to form 2CO₂ gas. ΔH equal to negative 787 kilojoules per mole. II) 2CO gas plus O₂ gas react to form 2CO₂ gas. ΔH equal to negative 566 kilojoules per mole. A) Negative 221 kilojoules per mole. B) Negative 345 kilojoules per mole. C) Negative 1353 kilojoules per mole. D) Negative 1919 kilojoules per mole. E) Negative 2140 kilojoules per mole.
The heat of a reaction is equivalent to the enthalpy change of the system. Our system, which we consider to be the reactants transforming into the products, is going to lose energy or gain energy because of the reaction. So the enthalpy of the system is going to increase if the energy in is greater than the energy out. This would be the case if we were dealing with an endothermic reaction. However, in the opposite scenario, where the energy out is greater than the energy in, the enthalpy of the system will decrease. This would be the case for an exothermic reaction. But we don’t know the heat of reaction or the enthalpy change for this particular case. But we do have the enthalpy changes for two similar processes.
We can see that we have carbons in common with one of the given equations. And we have carbon monoxide in common with the other. And we have oxygen in common with both. So there’s a good chance we can construct a Hess cycle to figure out the enthalpy change for the target reaction. And it’s most simple. Hess’s law tells us that the change in enthalpy for a process is independent of how you get there. So if we go directly from carbon and oxygen to carbon monoxide, the change in enthalpy would be the same if we went from carbon and oxygen to X, to Y, to Z, and then to carbon monoxide.
So our job is to take the two thermochemical equations given and work out a way of constructing them into a Hess cycle, which gives us an alternate route where we know all the enthalpy changes. We have carbon dioxide in both equations. So it’s a good bet. That’s a stop of point on the way from the reactants to the products. If we react together carbon and oxygen, we can produce carbon dioxide. But we need to add one extra oxygen in order to achieve it with two carbon atoms. Now, we can do this if we end up with that many oxygens being lost along the way. We can also get to CO₂ from carbon monoxide by adding more oxygen. Because we’re adding oxygen on both sides and coming together to form the same product, they cancel out.
The enthalpy change for the first step is negative 787 kilojoules per mole. The enthalpy for the second part, where we’re going from our products to carbon dioxide, is negative 566 kilojoules per mole. But we want to get from the reactants to the products, which means we need to flip the second arrow. This means we’re going in the other direction, which means that change in enthalpy sign is reversed. So we go from negative 566 kilojoules per mole to positive 566 kilojoules per mole. You could also see how the oxygens cancel out. Because we’re adding an oxygen along one leg and then taking it away across the next.
Now, we can calculate our heat of reaction, which is equal to the change in enthalpy of the system, which is equal to negative 787 kilojoules per mole plus 566 kilojoules per mole. This is equal to negative 221 kilojoules per mole. This means that even though we’re dealing with an incomplete combustion, the reaction is still exothermic. Just not quite as exothermic as going all the way to carbon dioxide. So our final answer for the heat of reaction for the incomplete combustion of carbon is negative 221 kilojoules per mole. This means when we react one mole of carbon atoms with half a mole of oxygen molecules to produce one mole of carbon monoxide, we release 110.5 kilojoules of energy.