I am confused about why delta G (DG) = 0 when a system is at equilibrium. I know DG=0 because of the equation DG = -RTlnK and K=1. However, I thought K is a constant that is specific to any reaction, and when something is in equilibrium Q=K, and K doesn't have to be 1 (in fact, K usually is never 1, but instead, something very small or big, like 6x10^-5 or 7x10^2). So, if a reaction is at equilibrium (a reaction with an eq. constant K that is NOT equal to 1), then DG will not equal to 0. If this is true, then I don't understand why the textbook tells me DG = 0 when the system is at equilibrium, with K = 1. (I don't understand why K has to be 1 at equilibrium, since K is a constant that can be any number) Please explain. Thanks.
Answer:
You are confusing yourself between ΔG˚ and ΔG. The standard Gibbs free energy change or ΔG˚ is equal to -RT ln K, where k is the equilibrium constant for the reaction. ΔG is the Gibbs free energy for the reaction at that particular point of time. We have equiplibrium when ΔG is zero, not ΔG˚. K does not have to be 0 for this.
The relation between these 2 quantities is given as
ΔG = ΔG˚ + RT ln Q
Substituting the value for ΔG˚, we have
ΔG = -RT ln K + RT ln Q
ΔG = RT ln Q/K
For ΔG greater than 0, K<Q, which means that the reaction will be in backward direction.
For ΔG less than 0, K>Q, so the reaction will be in the forward direction.
For ΔG = 0, K=Q so we have the equilibrium state achieved.
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