A reaction A<-->B starts with equal concentration of A and B. AT equilibrium, there is twice as much B as A. What was the value of change in Gibbs Free energy when the reaction originally started if the temperature was 300k?
To find the value of the change in Gibbs Free Energy (ΔG) when the reaction originally started, we need to use the equilibrium constant (K) and the formula for Gibbs Free Energy:
ΔG = -RT ln(K)
where:
ΔG is the change in Gibbs Free Energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
K is the equilibrium constant
Given that the reaction is A <--> B and that, at equilibrium, there is twice as much B as A, we can write the equilibrium constant expression as:
K = [B]^2 / [A]
Since the initial concentration of A and B is equal, we can assume their concentrations to be x moles each. Therefore, at equilibrium, [B] = 2x and [A] = x.
Substituting these concentrations into the equilibrium constant expression, we get:
K = (2x)^2 / x^2
K = 4x^2 / x^2
K = 4
Now, we can plug in the values into the formula for ΔG:
ΔG = -RT ln(K)
ΔG = -(8.314 J/(mol·K))(300 K) ln(4)
Calculating this expression, we find:
ΔG ≈ -(8.314 J/(mol·K))(300 K) ln(4)
ΔG ≈ -3540.2 J/mol
Therefore, the value of the change in Gibbs Free Energy when the reaction originally started, at a temperature of 300 K, is approximately -3540.2 J/mol.