# Reaction enthalpy at non-standard conditions

1. Feb 27, 2014

### giokara

Hi

I am tutoring a pupil in high-school out of necessity (no other teachers avail), while I am not so acquainted anymore with chemistry. The pupil showed me a question concerning the calculation of the reaction enthalpy which looked somewhat as follows:

1. Problem statement, all variables and given/known data

Calculate the reaction-enthalpy for the reaction:

$C_3H_8 (g) + 5O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (g)$

at 30$^\mathrm{o}$C and 1.08 atm.
The formation enthalpies at standard conditions and the heat capacities at constant pressure were known.

2. Relevant equations

3. The attempt at a solution

NOTE: I am not sure anymore about the exact reaction or the exact temperature and pressure, the above is meant as a generic example of the exercise that the pupil showed me.

As I remember, first the formation enthalpies at the non-standard conditions should be computed. This should be done in two steps:

1) Take into account the increase in temperature assuming constant pressure: use $\Delta H = C_p\Delta T$.
2) Take into account the increase in pressure assuming constant temperature.

I cannot remember how the second step should be computed, neither did I find some appropriate formula during a quick search.
Is the current approach the correct one? Could it be possible to give me some more information about how to proceed?

Giorgos

2. Feb 27, 2014

### Staff: Mentor

Yes. Your methodology is correct. As far as the pressure effect is concerned, it is hard to imagine how the pressure correction could be significant at that pressure and with that pressure change.

If you still want to get the pressure correction, you calculate the molar enthalpy change for each of the species individually (as a pure species) using:
$$\frac{∂H}{∂P}=(V-T\frac{∂V}{∂T})dP$$
To do this calculation, you need to know the PVT behavior of each pure species slightly beyond the ideal gas range (with an equation of state). Probably, Van Der Waals would be adequate, but you need to know the Van Der Waals parameters for each species). In my judgement, it's not worth the effort.

Chet

3. Feb 27, 2014

Hi Chet,