[EngRPI]

My friend and I were solving a problem that asked calculate the heats of reaction ΔH at 298 K and at 750 K for the oxidation of sulfur dioxide at one atmosphere
SO2(g) + O2(g) = SO3(g)

We calculated it out and got the answer as the heat of reaction at 298 K is -98.89 kJ/mol and the heat of reaction at 750 K is -105.508 kJ/mol

We were just wondering after solving the problem, chemically, why does enthalpy increase when temperature increases?
Is it as simple as Kirchoff's law of thermodynamics saying that when temperature increases and the heat capacity stays positive then the enthalpy also increases?

Thanks!

[mjc123]

Be careful here. We talk, for brevity, about the heat (or enthalpy) of reaction, but of course what this means is the enthalpy change of reaction, as the symbol ΔH indicates. Enthalpy increases with temperature, but the enthalpy change of a reaction may increase or decrease. In this case, ΔH decreases (becomes more negative) with increasing temperature. Do you know a formula for how ΔH changes with temperature?

Oh, and balance your equation correctly!

[orthoformate]

ΔH=∫CpdT

or: H_f=H_i+∫CpdT

or if Cp is constant: H_f=H_i+Cp(T_f-T_i)

So this equation is about enthalpy of formation. meaning if we are talking about SO₂ and assuming that heat capacity is constant, then the heat of formation will increase for SO₂ as temperature increases.

We are having a hard time understanding this conceptually. Assuming constant pressure, enthalpy is equal to heat. so if we talk about forming SO₂ from its elements at STP the heat of formation is -296.84 Kj/mol. if we increase the temperature to 750K and the Cp is a constant w.r.t. temperature then the heat of formation for SO₂ will increase, meaning the formation of SO₂ from its elements will not give off as much heat.

why is this? our understanding was that O₂(g) and S(g) contained chemical potential that was not dependent on temperature. Almost like a compressed spring, which when relaxes gives off the potential energy it contained.

Can you help us understand this concept?

[mjc123]

Again, be careful about notation. In your equation ΔH is the change in enthalpy of a substance with change in temperature, without reaction or phase change. If we are talking about the enthalpy change of a reaction (e.g. heat of formation), we need to take account of the enthalpy of both reactants and products. Can you see how to do this? (It may help to draw a Hess's law cycle: Reactants at T1  Reactants at T2    Products at T2  Products at T1)

our understanding was that O2(g) and S(g) contained chemical potential that was not dependent on temperature.

I don't know what this means. It's certainly not the usual definition of "chemical potential", which is temperature-dependent.

[orthoformate]

We phrased the question incorrectly.

SO₂ + (1/2) O₂  SO₃ @298K
-296.84 Kj/mol + 0 KJ/mol  -395.7 Kj/mol

Products-Reactant= ΔH Formation at 298K

-395.7 - (-296.84) = -98.86 Kj/mol

SO₂ + (1/2) O₂  SO₃ @750K

For simplicity lets do this Cp=4.0 for all substances and does not change

ΔH for SO₂= 4.0J/K*452K=1808 J

the difference in enthalpy of formation for SO₂ is 1808 J, meaning now instead of giving off 395.7 Kj/mol of heat upon formation @298, the formation of SO₂ requires 1413 Kj/mol of heat to be put in.

We are thinking that this increase in energy is due to entropy effects that occur at higher temperatures.

dG=-SdT at constant pressure

[mjc123]

(Quite apart from failing to distinguish between J and kJ!) You are missing the point. You must, as I said, consider both reactants and products. Suppose we know the ΔH at T1, and all the Cps (which we assume are constant with temperature). How do we work out ΔH at T2. Consider the Hess's law cycle:
Reactants (T2)  Reactants (T1)  Products (T1)  Products (T2)
ΔH = Cp(reactants)*(T1-T2) + ΔHrxn(T1) + Cp(products)*(T2-T1)
The overall process is Reactants(T2)  Products (T2) and ΔH is ΔHrxn(T2)
ΔHrxn(T2) = ΔHrxn(T1) + {Cp(products) - Cp(reactants)}*(T2-T1)
or dΔH/dT = ΔCp
So for your reaction (for which, incidentally ΔH is not a "heat of formation"), dΔH/dT = Cp(SO₃) - Cp(SO₂) - 0.5Cp(O₂)
I don't know why you assume a value of 4. You can easily work it out as all the reagents are gases. (Assume as a first approximation that no vibrations are activated. This is not quite true, and becomes less true as T increases - Cp varies quite significantly over the temperature range. But for now make the assumption.)
As for the heat of formation. you don't seem to be sure whether you're working it out for SO₂ or SO₃. But again, you must consider all reagents.
dΔH_f(SO₂)/dT = Cp(SO₂) - Cp(S) - Cp(O₂)
You might be misled by the assumption that heat of formation of elements is 0. But this only applies to standard heat of formation at a specified temperature (usually 298K). The enthalpy of the elements will vary with temperature according to Cp like any other chemical, and this must be taken into account.