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The heat of combustion of 2-propanol at 298.15 K, determined in a bomb calorimeter, is -33.41 kJ/g. For the combustion of one mole of 2-propanol, determine (a) $\Delta U$ and (b) $\Delta H$

So I managed to determine $\Delta U$ by first finding the amount grams in one of 2-propanol and then using that number with 33.41 kJ/g...

I'm not able to determine Delta H as I'm not sure what to do... Please help and guide me.

Thanks!

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Are you supposed to determine $\Delta_rH$ for the reaction (which is $\Delta_cH$ for the combustion of 2-propanol) or are you supposed to determine $\Delta_fH$ for the formation of 2-propanol?

If the first case, it might be hard.

The change in internal energy $U$ is equal to the sum of the heat transferred $q$ and the work done $w$. If we assume that only pressure volume work is done, then:

$$\Delta U=q+w = q-p\Delta V $$

The negative sign in front of $p\Delta V$ is to ensure that a decrease in volume is work being done on the system (system gains energy).

As a thermodynamic potential, internal energy is only easy to use if the pressure is constant (and $\Delta V$ is measurable), which is not true in a bomb calorimeter, or the volume is constant (so $\Delta V =0$), which is true in a bomb calorimeter.

At constant pressure: $\Delta U=q$.

Enthalpy is defined such that $$\Delta H = \Delta U + \Delta(pV) = q - p\Delta V + p\Delta V + V\Delta p $$ $$\Delta H = q + V\Delta p$$

Thus, enthalpy change is only easy to use if the volume is constant (true in a bomb calorimeter) and the pressure change is knowable (you were not given that data) or if the pressure does not change (so that $\Delta p=0$).

The second case is simpler.

If you were asked to determine $\Delta_f H$ for the formation of 2-propanol, you probably are encouraged to assume that for the combustion reaction $\Delta_rH\approx \Delta_cU$ and go from there. Then, you will need a balanced equation (2-propanol + oxygen forms what?) and some standard enthalpies of formation for the products.

$$\Delta_rH \approx \sum_{products}{\Delta_fH}-\sum_{reactants}{\Delta_fH}$$

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