Question

Problem A)

The standard reduction potentials for the following reactions are given below.

Pyruvate + 2H+ + 2e- → Lactate E∘ = -0.190 V

NAD+ + 2H+ + 2e- → NADH/H+ E∘ = -0.320V.

Calculate △G∘ for the overall spontaneous reaction making use of the relationship between △E∘ and △G∘.

Problem B)

Calculate the △G∘ʹ for the reaction given the equilibrium constant is 1.97 and the physiological relevant temperature is 37 ∘C.

fructose-6-phosphate → glucose-6-phosphate

Problem C)

Calculate the ratio of [A- ]/[HA] of a 0.1 M phosphate buffer (pKa = 6.86) at pH 7.2..

Answer 1

1)

Add 1st reaction with 2nd reverse reaction ===>

Gives

Pyruvate + NADH ------> lactate + NAD+

Erxn = 0.32 - 0.19 = 0.13 V

delta G = -nFE

n = no of lelectrons = 2 [since 2 electrons are involved ]

F = 96500 C

E = 0.13 V

delta G = -25.090 KJ

2)

DG^0’ = - RT ln K_eq           - equation

Given, K_eq = 1.97            , T = 37⁰C = 310.15 K

R= universal gas constant = 0.00831 kJmol^-1K^-1

Putting the values in equation 1

DG^0’ = - (0.00831 kJmol^-1K^-1) (310.15 K) ln (1.97)

        = - (2.577 kJmol^-1) ln (1.97) = -(2.577 kJmol^-1) (0.678) = - 1.75 kJmol^-1

Hence, DG^0’ = - 1.75 kJmol^-1