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
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?
Well this is the rxn given:
2 POCl3(g) → 2 PCl3(g) + O2(g)
Well, it sounds like you're on the right track, but maybe you just took a wrong turn. Let's see if we can steer you in the right direction with a little humor!
So, you calculated the temperature of 300.768 K and it didn't give you the right answer? Well, that's a bit chilly! Sometimes, numbers can be colder than you think. But don't worry, we'll try to warm things up.
Just a thought - did you remember to convert your units correctly? You see, in your calculations, you used dG in kJ, dS in J/K, and T in K. Maybe that's what's causing your answer to be a bit frosty. Make sure all your units are playing on the same temperature field!
Oh, and don't forget about the sign convention for dG. If your reaction is spontaneous at standard conditions, then dG should be negative. Maybe give that a little check too!
But hey, even if your calculations feel like walking on thin ice, don't give up! Take a deep breath, take another crack at it, and I'm sure you'll find the temperature where this reaction is spontaneously laughable!
To determine the temperatures at which the reaction is spontaneous at standard conditions, you need to use the relationship between Gibbs free energy change (ΔG°) and temperature (T) given by ΔG° = ΔH° - TΔS°.
Given the values you provided:
ΔG° = 490 kJ (convert to J, so it becomes 490,000 J)
T = 298 K
ΔS° = 177 J/K
Plug these values into the equation to solve for ΔH°:
490,000 J = ΔH° - (298 K)(177 J/K)
490,000 J = ΔH° - 52,746 J
Rearrange the equation to solve for ΔH°:
ΔH° = 490,000 J + 52,746 J
ΔH° = 542,746 J
Now, you have the value of ΔH°, and ΔS° is given. Since ΔH° and ΔS° do not depend on temperature, you can calculate the temperature at which the reaction becomes spontaneous using the formula:
T = ΔH° / ΔS°
T = 542,746 J / 177 J/K
T ≈ 3068 K
Therefore, the correct temperature at which the reaction is spontaneous at standard conditions is approximately 3068 K, not 300.768 K as you initially calculated.
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