Question

Q: The value of ΔS° for this reaction is 177 J/K. At what temperatures is this reaction spontaneous at standard conditions? Assume that ΔH° and ΔS° do not depend on temperature.

I got 300.768 K through dG = dH-TdS and T=dH/dS but it's still incorrect.
The values I used:
dG = 490kJ
T = 298K
dS = 177J/K

Answer 1

Firstly, I don't believe your question makes sense because you are asking to find temperatures for spontaneity but are given std. cond.
You would have to find when Delta G is = neg. Do you know Delta H?

Answer 2

Well this is the rxn given:
2 POCl3(g) → 2 PCl3(g) + O2(g)

Answer 3

Standard conditions refer to the ‘Thermodynamic Properties of Substances’ table which is at 25ᵒC & 1 Atm. That is, calculation of ΔGᵒ, ΔHᵒ & ΔSᵒ are based on the values in that table. Then apply the results to the Gibbs Free Energy equation and solve for Temperature.
…………..…..2POCl₃(g) ….. => …. 2PCl₃(g) …….+……. O₂(g)
ΔH(f)ᵒ: … 2(-559.8)Kj …………… 2(-288.7)Kj …...…….. 0Kj
Sᵒ: ……...2(325.3)J/K………..……2(311.6)J/K …….…..1(205)J/K
ΔGᵒ(f) …..2(-514.3)Kj……………. 2(-269.6)Kj……….……0Kj

ΔH(Rxn) = Σn∙ΔHᵒ(f)-products - Σn∙ΔHᵒ(f)-reactants
= [2(-288.7) – 2(-559.8)]Kj = +542.2Kj
ΔS(Rxn) = Σn∙Sᵒ(f)-products - Σn∙Sᵒ(f)-reactants
= (2(311.6)J/K + 1(205)J/K) – (2(325.3)J/K) = +177.5J/K = +0.1776Kj/K
ΔG(Rxn) = Σn∙ΔGᵒ(f)-products - Σn∙ΔGᵒ(f)-reactants
= 2(-269.6)Kj – 2(-514.3)Kj = +489.4Kj

ΔG = ΔH – TΔS => TΔS = ΔH – ΔG => T = (ΔH – ΔG)/ΔS
T = 542.2Kj – 489.4Kj / 0.1776Kj/K = (542.2 – 489.4)Kj/0.1776Kj/K = 297.3K = 24.3ᵒC