# Heat of Combustion

Methanol ($\ce{CH3OH}$) is regarded by many chemists as a possible liquid fuel for the future. The combustion of methanol occurs according to the equation:

$$\ce{2CH3OH(g) + 3O2(g) -> 2CO2(g) + 4H2O(l)}\quad\Delta H=-1506\ \mathrm{kJ\ mol^{-1}}$$

(a) Determine the heat of combustion of methanol, in $\mathrm{kJ\ g^{-1}}$.

(b) If $0.750\ \mathrm{mol}$ of methanol is mixed with $0.750\ \mathrm{mol}$ of oxygen and the mixture is ignited, how much energy will be released?

I have some working for this question, but not sure if it is right, because there is two moles of methanol in the equation.

for 18a)

$$1506\ \mathrm{kJ}\space\text{for}\space1\ \mathrm{mol}$$ $$1=\frac{m}{M}$$ $$1=\frac{m}{32.04}$$ $$m=32.04\ \mathrm g$$ $$\therefore1506\ \mathrm{kJ}\space\text{for}\space32.04\ \mathrm g$$ $$\therefore47\ \mathrm{kJ}\space\text{for}\space1\ \mathrm g$$ $$\therefore\space\text{heat of combustion}=47\ \mathrm{kJ\ g^{-1}}$$

for 18b)

must use $0.75\ \mathrm{mol}$ of oxygen

$$\therefore0.75\times\frac{2}{3}=0.5\ \mathrm{mol}\space\text{of methanol}$$ $$1506\ \mathrm{kJ}\space\text{for}\space1\ \mathrm{mol}$$ $$1506\times0.5\space\text{for}\space0.5\ \mathrm{mol}$$ $$753\ \mathrm{kJ}\space\text{for}\space0.5\ \mathrm{mol}$$ Therefore, $753\ \mathrm{kJ}$ is released for $0.75\ \mathrm{mol}$ of oxygen and $0.75\ \mathrm{mol}$ of methanol. Only $0.5\ \mathrm{mol}$ of methanol can be used.

Not entirely sure if any of this working is correct, any help is appreciated.

• Part (a) is correct. In part (b), you have correctly identified the limiting reactant as oxygen, and hence only 0.5 moles of methanol will be combusted. Since 2 moles of methanol yields 1506 kJ, 0.5 moles would yield 376.5 kJ or 377 kJ (to 3 sig.figs.) – Dr. J. Jun 17 '18 at 11:20
• @Dr.J. Part (a) is not correct. The given enthalpy is for 2 mol of methanol. The molar enthalpy of combustion of liquid (!) methanol at 25 °C actually is 726.1 kJ/mol. – Loong Jun 17 '18 at 16:46
• Correct answer for (a) should be 23.5 kJ/g as pointed out by Loong. Initial comment was made in haste. – Dr. J. Jun 18 '18 at 11:54

# Part A

First, the molar mass (MM) of methanol must be calculated. Sum the atomic masses given on the periodic table when multiplied by the instances of each atom in methanol.

$$\begin{split} \text{MM}_{\ce{CH3OH}} & = 1 \times 12.01~\mathrm{g} + 4 \times 1.01~\mathrm{g} + 1 \times 16.00~\mathrm{g} \\ & = 32.05~\mathrm{g} \end{split}$$

Employ dimensional analysis to change the units of the given quantity.

$$\frac{-1506~\mathrm{kJ}}{1~\mathrm{mol_{rxn}}} \times \frac{1~\mathrm{mol_{rxn}}}{2~\mathrm{mol}_{\ce{CH3OH}}} \times \frac{1~\mathrm{mol}_{\ce{CH3OH}}}{32.05~\mathrm{g}} = -23.49~\frac{\mathrm{kJ}}{\mathrm{g}_{\ce{CH3OH}}}$$

The heat of combustion of methanol in $\mathrm{kJ}~\mathrm{g^{-1}}$ is -23.49.

# Part B

Identify this as a limiting reagent problem. When given equimolar quantities of both reactants, the limiting reagent is the reactant with the greater coefficient in the balanced chemical equation ($\ce{O2 (g)}$ in this instance). Mathematically, determine the number of moles of $\ce{O2 (g)}$ needed to fully combust $0.750~\mathrm{mol}~\ce{CH3OH}$. If this number exceeds the provided moles of $\ce{O2 (g)}$, then $\ce{O2 (g)}$ is the limiting reagent.

$$\frac{0.750~\mathrm{mol}_{\ce{CH3OH}}}{1} \times \frac{3~\mathrm{mol}_{\ce{O2}}}{2~\mathrm{mol}_{\ce{CH3OH}}} = 1.13~\mathrm{mol}_{\ce{O2}}$$

$$1.13~\mathrm{mol}_{\ce{O2}} > 0.750~\mathrm{mol}_{\ce{O2}}$$

Therefore, $\ce{O2 (g)}$ is the limiting reagent.

Knowing $\ce{O2 (g)}$ is the limiting reagent, employ dimensional analysis to find the quantity of energy released.

$$\frac{0.750~\mathrm{mol}_{\ce{O2}}}{1} \times \frac{1~\mathrm{mol_{rxn}}}{3~\mathrm{mol}_{\ce{O2}}} \times \frac{1506~\mathrm{kJ}}{1~\mathrm{mol_{rxn}}} = 377~\mathrm{kJ}$$

377 kJ of energy is released by the given reaction in the listed scenario.