Answer:
(a) -20 kJ/mol
(b) ΔG = 0
(c) A negative value
Explanation:
(a) Calculate the change in standard free energy (ΔG°) for this equilibrium reaction.
We can calculate the standard Gibbs free energy using the following expression:
ΔG° = R.T. lnKc
where,
R is the ideal gas constant
T is the absolute temperature
Kc is the equilibrium constant (in this case Kc = Ka because the reaction is an acid dissociation)
If we replace this expression with the given values, then:
ΔG° = R.T. lnKc
ΔG° = [tex]8.314 \times10^{-3} \frac{kJ}{mol.K} \times 298.15K \times ln3.3\times10^{-4} =-20kJ/mol[/tex]
(b) What is the value of ΔG at equilibrium?
Let us consider a more general expression for equilibrium and non-equilibrium situations:
ΔG = ΔG° + R.T. lnQ
where,
ΔG is Gibbs free energy in that situation
Q is the reaction quotient
At equilibrium, ΔG = 0, Q = Kc, and we get the first expression.
(c) If the reaction were spontaneous in the forward direction, what type of value would you expect for ΔG?
A negative ΔG implies that the forward reaction is spontaneous.
