# What if soot (unburned carbon) accumulated on the bottom on the small can? would this produce a greater or a lower heat value than expected?

May 10, 2017

Incomplete combustion will result in significant heat loss..............

#### Explanation:

On our first introduction to thermochemistry we learn that formation of bonds is exothermic, and the breaking of bonds is endothermic. To put this more simply, energy is RELEASED on bond formation, and energy is REQUIRED to BREAK BONDS.

We can take a hydrocarbon, say hexane (and I choose this for certain reasons which I will relate later), and we can represent its complete combustion by the following stoichiometric reaction:

${C}_{6} {H}_{14} \left(l\right) + \frac{19}{2} {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + 7 {H}_{2} O \left(l\right) + {\Delta}_{1}$

Of course, this is an exothermic reaction, however strong $C - H$, $C - C$, and $O = O$ bonds have been broken. But conversely, even stronger than $C = O$, and $H - O$ bonds have been FORMED. And of course the difference in bond enthalpies of bonds made versus bonds broken can be quantitatively measured.

Now, certainly in internal combustion or diesel engines, combustion is INCOMPLETE. Carbon monoxide, and carbon (as particulate soot) can clearly be detected in engine exhausts. Longer chain hydrocarbons, especially diesel fuels, are particularly prone to this. If we rewrite the combustion reaction to represent this, we gets...........

${C}_{6} {H}_{14} \left(l\right) + 8 {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + C O \left(g\right) + C \left(s\right) + 7 {H}_{2} O \left(l\right) + {\Delta}_{2}$

Now in the second reaction, we still have to break $O = O$, and $C - H$, and $C - C$ bonds, but NECESSARILY we make FEWER $C = O$ and $H - O$ because of the incomplete combustion. And given that this is the case we can confidently infer that ${\Delta}_{1} \text{>>"Delta_2}$

And given some bond enthalpy data (which will be available in your text), we could calculate the difference ${\Sigma}_{\text{bonds formed"-Sigma_"bonds broken}}$, for each reaction; and where combustion is incomplete, $\Delta {H}_{\text{rxn}}$ will be reduced in magnitude.

Sorry to belabour the point; if there is a further query or point of clarification, fire away.