How Does Temperature Affect Equilibrium Constants for Conformers?

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The discussion focuses on calculating the equilibrium constant (Keq) for two conformers with a 1 Kcal/mol energy difference at room temperature and -78°C. The user attempted to use the equation Keq = e^(deltaG/(-R*T)) but received feedback that they incorrectly applied deltadeltaG instead of deltaG. The temperatures used were 293.15K for room temperature and 195.15K for -78°C, with R set at 8.313 J/(mol*K). The user's calculated values for Keq were 0.18 at room temperature and 0.08 at -78°C, which were deemed incorrect. Clarification is sought on how to properly define ΔG in the context of the equation.
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Homework Statement


Hi, I have the following task:

We have two conformers and the difference in energy between them is aprox. 1 Kcal/mol. What is the Keq at room temperature vs. at -78°C?

Homework Equations



Keq=e^(deltaG/(-R*T))

The Attempt at a Solution



I tried to solve the problem the following way:

deltaG = R*T*ln(Keq) --> Keq=e^(deltaG/(-R*T))

T(room)=293.15K, T(-78°C)=195.15K
R=8.313 J/(mol*K), DeltaG= 1 Kcal/mol= 4184 J/mol

then I inserted the values and got Keq(room) = 0.18 and Keq(-78°C) = 0.08 which is apparently wrong. They told me that I used deltadeltaG instead of deltaG in the equation. Can somebody help me?
 
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In the equation you quote, how is ΔG defined?
 

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