How Can You Calculate Equilibrium Temperature with Given Thermodynamic Values?

AI Thread Summary
To calculate the equilibrium temperature of a reaction, the user has values for K, ΔS, ΔG, and ΔH, but faces challenges since these values are given at standard conditions (298K). They plan to use the equation ΔG0 = ΔH0 - TΔS0, but are concerned about the accuracy of ΔS and ΔH at different temperatures. The discussion highlights the need to assume constant aggregate states for the substances involved. Additionally, the Van't Hoff equation is suggested to relate K and temperature at standard conditions to the desired equilibrium temperature. The user acknowledges a misunderstanding regarding the necessity of K in their calculations and corrects an earlier mistake in their equation.
Nikitin
Messages
734
Reaction score
27
I'm supposed to find out the temperature of a reaction at equilibrium. I've calculated K=0.187, ΔS0=940J, and I know that ΔG0=1134kJ and ΔH0=1414kJ.

I'm supposed to use the R*T*ln(K)=ΔG0 formula, but my only problem is that ΔG0 is given for standard conditions,, ie where the temperature equals 298K.

But I am supposed to use it anyway in the assignment, ie I need to reform it into ΔG0=ΔH0 + T*ΔS0=T*R*ln(K), and calculate T.

However, ΔS and ΔH for the reaction are given at standard conditions. When the temperature changes they too change, so how is it possible to get a reliable answer?
 
Physics news on Phys.org
help?
 
Can you post the original question? You might be missing some kind of assumption, like the temperature remains constant for the reaction.
 
Here's the reaction. Ca3(PO4)2 (s) + 3 SiO2 (s) + 5 C (s) -> 3 CaSiO3 (s) + 1/2 P4 (g) + 5 CO (g)

I'm supposed to find the equilibrium temperature, with the data from the OP (all of which are at standard conditions). I am also to assume that none of the substances change their aggregate states (ie none of them turn into gases) going from standard temperature to the equilibrium temperature.
 
Can you find out K at standard conditions. Using this, relate K and T at standard conditions with K and T asked in the question by Van't Hoff equation.\mathrm{ln}\frac{K_2}{K_1}=\frac{-\Delta H^{\circ}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )
 
Do you really need K? Isn't it just a matter of finding T such that \Delta G = \Delta H - T \Delta S = 0?
 
AGNuke: I knew K at equil. temperature.. What I was struggling with was assuming that ΔS and ΔH are the same at 298K and 1500K for the reaction... Anyway I feel I understand allot more.

Borek said:
Do you really need K? Isn't it just a matter of finding T such that \Delta G = \Delta H - T \Delta S = 0?
The assignment wanted that the equilibrum temperature be calculated for K=0.187.. ugh I was probably unclear about this.. sorry I'm sleep deprived as heck.

EDIT: When I wrote that ΔG0=R*T*ln(K) in the OP, I meant ΔG0=-R*T*ln(K). I forgot the minus-sign infront of R*T*ln(K).
 
Last edited:
Back
Top