# What is the thermochemical equation for the combustion of benzene?

May 26, 2015

A thermochemical equation is simply a balanced chemical equation that includes the change in enthalpy that accompanies that respective reaction.

As is the case with all hydrocarbons, which are compounds that only contain carbon and hydrogen, benzene's combustion will lead to the formation of only two products, carbon dioxide, $C {O}_{2}$, and water, ${H}_{2} O$.

The balanced chemical equation for the combustion of benzene, ${C}_{6} {H}_{6}$, is

$2 {C}_{6} {H}_{6 \left(l\right)} + 15 {O}_{2 \left(g\right)} \to 12 C {O}_{2 \left(g\right)} + 6 {H}_{2} {O}_{\left(l\right)}$

Now, in order to have the thermochemical equation, you need to add the change in enthalpy associated with this reaction, which is listed as being equal to -6546 kJ.

$2 {C}_{6} {H}_{6 \left(l\right)} + 15 {O}_{2 \left(g\right)} \to 12 C {O}_{2 \left(g\right)} + 6 {H}_{2} {O}_{\left(l\right)}$, $\Delta {H}_{\text{rxn" = "-6546 kJ}}$

You need to be a little careful here because this is the change in enthalpy that accompanies the combustion of 2 moles of benzene. This means that you can also write

${C}_{6} {H}_{6 \left(l\right)} + \frac{15}{2} {O}_{2 \left(g\right)} \to 6 C {O}_{2 \left(g\right)} + 3 {H}_{2} {O}_{\left(l\right)}$

$\Delta {H}_{\text{rxn" = "-6546 kJ"/2 = "-3273 kJ}}$

This is what the change in enthalpy is when 1 mole of benzene undergoes combustion.

Here's a really cool video of benzene's combustion reaction