[azupol]

The ceiling concentration for the radical polymerization of styrene in solution is 1x10^-6 M at 25°C.  For the radical polymerization of the pure monomer (no solvent), the ceiling temperature is 310°C.  Calculate ΔH° (in kJ/mo) and ΔS° (in J/Kmol).

k_p (rate of propagation)=134 M^-1 s^-1 (at 298 K).
boiling point=145°C
density=.909g/mL
MW=104.15g/mol
R=8.314J/Kmol


I'm not understanding how to approach this.  Can I assume that ΔG° is = at 298K?  Then use 1/[M]_c, the ceiling concentration, as my rate constant, since at equilibrium for radical polymerizations the equilibrium constant is that?

I tried -(8.314)(298)ln(1/1x10^-6)

but I'm not sure that's right.

Any help is appreciated.

[mjc123]

Can I assume that ΔG° is = at 298K?

No. This statement is meaningless.

Then use 1/[M]c, the ceiling concentration, as my rate constant, since at equilibrium for radical polymerizations the equilibrium constant is that?

No, K_eq = k_p/k_d. But I don't think you need the rate constants anyway - this is a thermodynamic thing.
What are K_eq and ΔG° at 298K?
For the pure monomer, I think we're supposed to assume that the activity of liquid styrene is 1. Then what are K_eq and ΔG° at 583K?
Now you have two simultaneous equations in ΔH° and ΔS°.
(I have seen very different values quoted for the ceiling temperature, from 235°C to 395°C. Presumably it hasn't been (can't be?) measured, it has to be extrapolated from the thermodynamic data - and I have seen somewhat different values for those too.)

[azupol]

Sorry, I meant to write "Can I assume that ΔG°=0 at 298K?"

Yes, k_eq =k_p/k_d but if you go through the derivation, it turns out that at equilibrium, this will be equal to 1/[M]_c, where [M]_c is the monomer concentration at the ceiling temperature.

I guess that's the crux of my question, what is K_eq at 583K?  Since this is the ceiling temperature (at least as quoted in the question) this means that the rates of the forward (polymerization) reaction and reverse (depolymerization) reaction must be equal, which means that K_eq=k_p/k_d, and since the rates are equal, K_eq=1, yes?  If this is true, I can take it from there and know how to solve it.  Could you confirm this?

Thanks for your reply.

[Plontaj]

https://en.wikipedia.org/wiki/Ceiling_temperature

Ceiling temperature = Tc = ΔH(Tc) / ΔS(Tc)

If the ceiling temperature is that temperature on which polymerization is reversible then ΔG(Tc) = 0

0 = ΔH(Tc) - T * ΔS(Tc) , Tc = ΔH(Tc) / ΔS(Tc)

If the heat capacity can be assumed as constant, then:
ΔH(Tc) = ΔH(T°) + Cp_liquid*(Tb-T°) + ΔH_boiling + Cp_gas*(Tc-Tb)
ΔS(Tc) = ΔS(T°) + Cp_liquid*ln(Tb/T°) + ΔS_boiling + Cp_gas*ln(Tc/Tb) - R*ln(P/P°)
if not, it's need to integrate Cp(T) and Cp(T)/T

if K = 1/[M] then:

ΔG(T°) = ΔG°(T°) + R*T°*ln(1/[M]) = 0

I guess that's the crux of my question, what is Keq at 583K?  Since this is the ceiling temperature (at least as quoted in the question) this means that the rates of the forward (polymerization) reaction and reverse (depolymerization) reaction must be equal, which means that Keq=kp/kd, and since the rates are equal, Keq=1, yes?

Why Keq = 1 if Keq=kp/kd?

[azupol]

Because at equilibrium, which is reached at the ceiling temperature and corresponding monomer concentrations, the forward rate and reverse rate are equal.  Thus k_p=k_dp so it follows then that K_eq=1
Also this is a more general fact that K_eq=1 at equilibrium.

[azupol]

If anyone is interested, I've solved it.

[M]_c= n/V = n/(n FW/d) = d/FW

ΔH-TΔS=-RT ln([M]_c).  From above, we have [M]_c for both temperatures now, and 2 equations in 2 unknowns.

It's just elementary algebra from here.

NOTE: This will not give the exact enthalpy and entropy, because the initial given values are not exact, these are just taken from the problem set I was assigned.

[Plontaj]

Because at equilibrium, which is reached at the ceiling temperature and corresponding monomer concentrations, the forward rate and reverse rate are equal.  Thus kp=kdp so it follows then that Keq=1
Also this is a more general fact that Keq=1 at equilibrium.

This is not true. If the equilibrium is reached the rate back and forth reaction are equal rp = rd, so then kp*[reagent conc.] = kd*[reagent conc.], you've got constant according with mass-action law, which not necessarily has to be equals 1.

ΔH-TΔS=-RT ln([M]c).  From above, we have [M]c for both temperatures now, and 2 equations in 2 unknowns.

Yes, but in that approach are needed huge simplification, like constant enthalpy and entropy for instance what cannot be fulfilled.