An important metabolic step is the conversion of furamate to malate. In aqueous

Question

An important metabolic step is the conversion of furamate to malate. In aqueous solution, an enzyme (furamase) allows equilibrium to be attained. Furamate+H_2O rightarrow malate At 25 degree C the equilibrium constant K=(a_M/aF)=4.0. The activity of malate is CIM, and the activity of furamate is aF, defined on the molarity concentration scale (a=c in dilute solution). What is the standard Gibbs free-energy change for the reaction at 25 degree C? What is the Gibbs free-energy change for the reaction at equilibrium? What is the Gibbs free-energy change when 1 mol of 0.1 M furamate is converted to 1 mol of 0.1 M malate? What is the Gibbs free-energy change when 2 mol of 0.1 M furamate is converted to 2 mol of 0.1 M malate? If K=8.0 at 35 degree C, calculate the standard enthalpy change for the reaction; assume that the enthalpy is independent of temperature.

Explanation / Answer

The balanced chemical reaction is

Fumarate + H2O -- > Malate , K = am / af = 4.0

(a) T = 25 DegC = 25+273 = 298 K

Now the standard Gibbs free energy change an be calculated from the following formulae

DeltaG0 = - RTxlnK = - 8.314 JK-1mol-1x 298K x ln4 = - 3435 J = - 3.435 KJ(answer)

(b): For a chemical reaction in equilibrium, standard Gibbs free energy is always 0.

(c) Given [malate] = 0.1 M, [fumarate] = 0.1 M

Reaction quotient, Q = [malate] / [fumarate] = 0.1M / 0.1M = 1

DeltaG = DeltaG0 + RTxlnQ = - 3.435 KJ + RTxln(1) = - 3.435 KJ (answer)

(d):Here also Q = 1 and hence ln(1) = 0,

Hence DeltaG = DeltaG0 =  - 3.435 KJ (answer)

(e): Given K = 8.0, T = 35 DegC = 35 + 273 = 308 K

DeltaG = - RTxlnK = - 8.314 JK-1mol-1x 308K x ln8 = - 5325 J (answer)

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