How is equilibrium related to Gibbs' free energy?
2 Answers
Every reaction in a closed system is in some sort of equilibrium. Depending on the equilibrium constant
A common type of equilibrium is found in acid/base reactions. For acids and bases, there is a specific equation you can use to determine the extent of this equilibrium called the Henderson-Hasselbalch equation:
#pH = pKa + log (([A^(-)])/([HA]))#
#pH = pKb + log (([BH^(+)])/([B]))#
where
For example, consider the dissociation of acetic acid in water:
#CH_3COOH + H_2O rightleftharpoons CH_3COO^(-) + H_3O^(+)#
The
The amount of acetic acid that dissociates is the amount of dissociated acetic acid that forms. Therefore:
#pH = pKa + log (([CH_3COO^(-)])/([CH_3COOH]))#
#4 = -log(1.8xx10^(-5)) + log (x/(1 - x))#
#4 = 4.74 + log (x/(1-x)) => -0.74 = log (x/(1-x))#
#10^-0.74 = 0.182 = x/(1-x)#
#0.182 - 0.182x = x => 0.182 = 1.182x#
Dissociated acetic acid:
Undissociated acetic acid:
Since the ratio in the Henderson-Hasselbalch equation is:
#[[CH_3COO^(-)]]/[[CH_3COOH]] = 0.154/0.846#
The equilibrium is skewed
Even a reaction like this:
#H_2SO_4(aq) rightleftharpoons HSO_4^(-)(aq) + H^(+)(aq)#
is simply extremely heavily skewed towards the products side.
The
Every reaction in a closed system is in some sort of equilibrium. Depending on the equilibrium constant
For a typical reaction with equilibrium constant
#DeltaG = DeltaG^o + RTlnQ#
where
At equilibrium:
#0 = DeltaG^o + RTlnK#
#DeltaG^o = -RTlnK#
The value
Let's say for some reaction, we had
#DeltaG = DeltaG^o + RTlnQ#
#= -50 + (8.314472*300)lnQ#
Now let's say we already calculated the quantity for
#-30 = (8.314472*300)lnQ#
#color(blue)(Q) = e^("-30/(8.314472"*"300)") = color(blue)(0.988)#
For exact equilibrium,