Question

Finding `DeltaH` given heat released and mass in bomb calorimetry?

When 8.23 g of the organic compound cinnamaldehyde, C9H8O, was burned in a bomb calorimeter at 298K, the heat released was 302.9 kJ. Calculate the enthalpy of combustion of cinnamaldehyde at this temperature.

Answer 1

`DeltaH_C = -"4870 kJ/mol"` to 3 sig figs.

---

Note that you are using a bomb calorimeter. That is a constant-volume system, not a constant-pressure system.

That adds a complication, in that for a thermodynamically-closed system,

`color(green)(DeltaH_C = DeltaE_C + Delta(PV)_C)` `" "bb((1))`

where `DeltaE_C` is the change in internal energy of combustion and `Delta(PV)_C` is the change in enthalpy due to pressure-volume work.

At constant volume, `q = DeltaE`, and at constant pressure, `q = DeltaH`. And so, you were given `q = bb(DeltaE_C)` in `"kJ"` so far.

The balanced reaction is:

`"C"_9"H"_8"O"(s) + 21/2"O"_2(g) -> 9"CO"_2(g) + 4"H"_2"O"(l)`

where we assume the water to be a liquid in a closed rigid container, having condensed down from initially forming a gas.

For your scale of reaction, you have `"8.23 g"` of cinnamaldehyde. That means you have:

`8.23 cancel"g" xx "1 mol"/(132.16 cancel"g cinnamaldehyde") = "0.0623 mols"`

For ideal gases, `PV = nRT`, and so,

`DeltaH_C = DeltaE_C + Delta(nRT)`.

The temperature of the surroundings (water + room) is typically constant in a bomb calorimetry experiment, as it is done in a huge, probably `"2-L"` water reservoir.

Thus,

`DeltaH_C ~~ DeltaE_C + Deltan cdot RT_"room"`

For the reaction as-written, the change in mols of gas dominates the pressure-volume work due to the combustion. So we take:

`DeltaH_C ~~ DeltaE_C + Deltan_"gas" cdot RT_"room"` `" "bb((2))`

and get (`21/2"O"_2(g) -> 9"CO"_2(g)`):

`Deltan_"gas" = 9 - 21/2 = -"1.5 mols"`

On the scale of the actual reaction, we have only `6.23%` of `"1 mol"` of cinnamaldehyde, i.e. we expect the change in mols of gas to actually be:

`Deltan_"gas" = -"1.5 mols" xx "0.0623 mols cinnamaldehyde used"/"1 mol cinnamaldehyde as-written"`

`= -"0.0935 mols gas"`

We now have enough info. Using `(1)`,

`DeltaH_C = -"302.9 kJ" + (-0.0935 cancel"mols gas" cdot "0.008314472 kJ/"cancel"mol"cdotcancel"K" cdot 298.15 cancel"K")`

`= -"303.13 kJ"`

And in terms of `"kJ/mol"` as by convention,

`color(blue)(DeltaH_C) = -"303.13 kJ"/(8.23 cancel"g") xx (132.16 cancel"g")/"1 mol"`

`= color(blue)(-"4870 kJ/mol")`

to 3 sig figs.