# What is the molar heat of combustion when "25000 g" of methanol is combusted in a bomb calorimeter to raise the temperature of the surrounding water by 40^@ "C" if the heat capacity of the calorimeter is "10.4 kJ/"^@ "C"?

Jul 3, 2017

There's probably an error in this question... I get $- \text{0.5331 kJ/mol}$, which is about 1000 times smaller than it should be.

Since we are in a bomb calorimeter, a constant-volume calorimeter, we have from the first law of thermodynamics:

${q}_{V} = \Delta U + {\cancel{P \Delta V}}^{0}$,

where:

• ${q}_{V}$ is the heat flow at constant volume.
• $\Delta U$ is the change in internal energy.
• The work done, $- P \Delta V$, is defined here as the system doing work with respect to the surroundings.

Since we know the change in temperature and the combined heat capacity (presumably in $\text{kJ/"^@ "C}$...), we can find the heat involved in the reaction:

$| {q}_{V} | = {C}_{V} \Delta T$

$= \text{10.4 kJ/"^@ "C" xx 40^@ "C}$

$=$ $\text{416 kJ}$

I interpret the molar heat of combustion to be $\Delta {\overline{U}}_{C}$, rather than the change in enthalpy (since the process is at constant volume, and we aren't given the volume of the calorimeter or water).

$\Delta {\overline{U}}_{C} = {q}_{V} / {n}_{L R}$,

where ${n}_{L R}$ is the mols of the limiting reactant, i.e. the methanol in the presence of oxygen.

Assuming complete combustion:

$2 {\text{CH"_3"OH"(g) + 3"O"_2(g) -> 4"H"_2"O"(g) + 2"CO}}_{2} \left(g\right)$

The molar internal energy of combustion is then the heat released out from the reaction towards the surrounding water:

color(blue)(DeltabarU_C) = -("416 kJ")/(25000 cancel("g CH"_3"OH")) xx (32.04 cancel("g CH"_3"OH"))/"1 mol"

$=$ $\textcolor{b l u e}{- \text{0.5331 kJ/mol}}$

This is an absurdly low number, so there's something off with the data... Are you sure it isn't $\text{25.000 g}$? The heat of combustion is off by over 1000 times.