Question

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"`?

Answer 1

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

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

`q_V = DeltaU + cancel(PDeltaV)^(0)`,

where:

• `q_V` is the heat flow at constant volume.
• `DeltaU` is the change in internal energy.
• The work done, `-PDeltaV`, 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 `"kJ/"^@ "C"`...), we can find the heat involved in the reaction:

`|q_V| = C_VDeltaT`

`= "10.4 kJ/"^@ "C" xx 40^@ "C"`

`=` `"416 kJ"`

I interpret the molar heat of combustion to be `DeltabarU_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).

`DeltabarU_C = q_V/n_(LR)`,

where `n_(LR)` is the mols of the limiting reactant, i.e. the methanol in the presence of oxygen.

Assuming complete combustion:

`2"CH"_3"OH"(g) + 3"O"_2(g) -> 4"H"_2"O"(g) + 2"CO"_2(g)`

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"`

`=` `color(blue)(-"0.5331 kJ/mol")`

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