Question

1. The following reactions can be coupled to give alanine and oxaloacetate:

?????????+???????? ⇋?????????????+??????? Δ??303°′=−1000 ?/???

?????????+???????????? ⇋?????????????+????????? Δ??303°′=−4810 ?/???

Write the equation for Keq and solve for the below combined reaction at 303.15 K:

b)In the cytoplasm of a certain cell, reaction components were found to be at the following concentrations: pyruvate = 10-3 M, aspartate = 10-3 M, alanine = 10-4 M, oxaloacetate = 10-6 M.Calculate the Gibbs free energy change for the reaction in part (a) under these conditions.

c) What conclusion can be made about the direction of this reaction under cytoplasmic conditions?

Answer 1

?????????+???????? ⇋?????????????+??????? Δ??303°′=−1000 ?/??? ...equation(1)

?????????+???????????? ⇋?????????????+????????? Δ??303°′=−4810 ?/??? ...equation(2)

Reversing equation(2) we get,

?????????????+????????? ⇋ ?????????+????????????, Δ??303°′= +4810 ?/??? ...equation(3)

Adding equation(1) and equation(3) we get the equation for K_eq which is,

???????? + ????????? ⇋ ???????????? + ???????, Δ??303°′= (-1000 + 4810) J/mol = +3810 J/mol ...equation(4)

We have Δ?? = -2.303 * RT * logK_eq ...equation(5)

where, R is the universal gas constant whose value is 8.314 J mol^-1 K^-1

T is the reaction temperature, Δ?? is the free energy change of the reaction and K_eq is the equilibrium constant for the reaction. Here T = 303.15 K

By putting these values in equation(5) we get,

-2.303 * 8.314 * 303.15 * logK_eq = 3810

logK_eq = -3810 / (2.303 * 8.314 * 303.15) = -0.65

So K_eq = 10^-0.65 = 0.22
Therefore the value of K_eq is 0.22

b) From equation(4) equilibrium constant K_eq for the reaction can be written as

K_eq = ([????????????] * [???????]) / ([????????] * [?????????]); ( [] signifies concentration of the corresponding species)

Therefore under cytoplasmic condition, K_eq = (10^-6 * 10^-4) / (10^-3 * 10^-3) = 10^-4

Therefore from equation(5) we get

Δ?? = -2.303 * 8.314 * 303.15 * log(10^-4) = 23217.82

So, under cytoplasmic condition the value of Δ?? is 23217.82 J/mol

c) As the value of Δ?? is positive for the reaction under cytoplasmic condition, the reaction will be non-spontaneous i.e. it will go to the backward direction favoring the reactants.