Chem enthalpy change

Mg + CuSO4 -> MgSO4 + Cu
1.0g Mg and 50ml(=50g) 0.500mol/dm^3 CuSO4
CuSO4 is limiting reactant
use q=mcT/mol of limiting reactant
and enthalpy change= - q/mol to find the enthalpy change of this reaction with uncertainties
change in T = 32C
uncertainty measuring cylinder: +- 1ml
thermometer: +-0.5C
CuSO4: 0.50 +- 0.01mol/dm^3

I can calculate the values but I'm not sure about the uncertainty...

Also I can't seem to find the literature value for this reaction; does anyone know where to find it?

Reply 1

UtterlyUseless69

Original post by SedraS

First and foremost, let’s identify the limiting reagent:

n(CuSO4): 0.50 M x 50/1000 dm^3 = 0.025 mol
n(Mg): (1 g)/(24.3 g/mol) = 0.04115… mol

So CuSO4 is the limiting reagent and the uncertainty in the mass of magnesium is probably irrelevant for the purposes of this question, since it’s in a pretty big excess.

If you want the fractional uncertainties in each measurement (these can be converted to percentage uncertainties by multiplying by 100 later)

Two readings were taken using the thermometer to record the temperature change, so the fractional uncertainty would be

UC in ΔT = 2 x (0.5°C)/(32°C) = 0.03125

Similarly, the fractional uncertainties in the volume and concentration of the solution would be:

UC in [CuSO4] = (0.01 M)/(0.50 M) = 0.02
UC in V(CuSO4) = (1 ml)/(50 ml) = 0.02

But by assuming that the value of the density of the solution is exact, we can say that the percentage in the volume is the same as the percentage uncertainty in the mass of the solution. At least in this case, you are assuming the density of the solution is 1 g cm^-3.

The A level way of finding an overall percentage uncertainty when you’re multiplying or dividing data with some uncertainty is to add the percentage uncertainties together.

Remember, we assume we know c exactly and so we assume we know the density exactly.

UC in n(CuSO4) = UC in V(CuSO4) + UC in [CuSO4] = 0.02 + 0.02 = 0.04

UC in ΔH = UC in V(CuSO4) + UC in c + UC in ΔT + UC in n(CuSO4) = 0.02 + 0 + 0.03125 + 0.04 = 0.09125

This equates to a 9.125% uncertainty.

As for the literature value for CuSO4 (aq) + Mg (s) —> MgSO4 (aq) + Cu (s), I’d probably calculate it from literature values of enthalpies of solution and enthalpies of formation with an over-convoluted Hess cycle. I’ll see what I can find and get back to you later if I find anything.

(edited 1 year ago)

Reply 2

SedraS

Thank you so much!

Reply 3

UtterlyUseless69

So I worked out the theoretical value to be -200.5 kJ/mol. The method is dubious, as it involves the Kapustinskii equation, which usually gives a 10% error in estimated lattice energies and also the enthalpies of formation of the solid sulphates have not been corrected for a temperature of 298 K.

The data used and where I found it can be accessed in the spoiler below:

Spoiler

EDIT: More simply, I suppose you could also use the fact that the ionic compounds are dissociated and so you can rewrite the equation as follows:

Cu(2+) (aq) + SO4(2-) (aq) + Mg (s) —> Mg(2+) (aq) + SO4(2-) (aq) + Cu (s)

Which simplifies to

Cu(2+) (aq) + Mg (s) —> Mg(2+) (aq) + Cu (s)

As in my above post, the formation enthalpies of Mg (s) and Cu (s) are both 0. A table online can be used to find the formation enthalpies of the relevant aqueous cations: https://www.thoughtco.com/heats-of-formation-table-603969

This gives an enthalpy change of -526.4 kJ/mol, which is VERY different to the estimated value above. It goes to show how poor some approximations are and why they shouldn’t be taken as gospel.

An alternative method based on a Hess cycle with solution enthalpies (these values were taken from exam questions I found on the internet) and formation enthalpies gives an estimate of -509.8 kJ/mol, indicating that the actual value is most probably something in the -500 kJ/mol sort of region.