Question

Furamate + H2O <--> malate .

At 25 degrees celsius the equilibrium constant K= (activity Malate/activity furamatete)=4.0. The activity of malate and furamate are defined on the molatiry concentration scale (a=c in a dilute solution)

a) What is the standard Gibbs free energy change for the reaction at 25 degrees celsius.

b) What is the Gibbs free energy change for the reaction at equilibrium?

c) What is the Gibbs free energy change when 1 mol of 0.1M furamate is converted to 1 mol of 0.1M malate?

d) What is the Gibbs free energy change when 2 mol of 0.1M furamate is converted to 2 mol of 0.1 malate?

e) If K= 0.8 at 35 degrees celsius, Calculate the standard enthalphy change for the reaction ; assume that the enthalphy is independent of temperature.

f) Calculate the standard entropy change for the reaction; assume tha delta S is independent of temperature.

Answer 1

The balanced chemical reaction in equilibrium is

Furamate + H2O <--> malate, K = 4.0 at 25 DegC(298 K)

(a): The standard Gibb's free energy(G⁰) for a system in equilibrium can be calculated from the following formulae

G⁰ = - RTxlnK = - 8.314 J.mol^-1K^-1x 298Kx ln(4.0) = 3435  J.mol^-1 = - 3.435 KJ.mol^-1

(b) Since the reaction is at equilibrium at 25 DegC, the Gibbs free energy change for the reaction at equilibrium

= G⁰ = - 3.435 KJ.mol^-1

(c) Given K1 = 4.0, T1 = 298 K

K2 = 0.8, T2 = 35 DegC = (273+35) = 308K

The value of K varies with temperature that can be defined by the following relation

ln(K2/K1) = (H⁰ / R)x[1/T1 - 1/T2]

=> ln(0.8/4.0) = (H⁰ / 8.314J.mol^-1K^-1)x[1/298 - 1/308]

=> H⁰ = ln(0.8/4.0)x8.314J.mol^-1K^-1x298x308 / 10 = - 122.8 KJ / mol (answer)

(f) G⁰ =H⁰ - TS⁰  

=> S⁰= (H⁰ - G⁰) / T = [- 122.8 KJ / mol - ( - 3.435 KJ.mol^-1)] / 298K = - 119365 J.mol^-1 / 298K

=>  S⁰  = - 400.5J.mol^-1K^-1