Ans. #I. Using, dG = dG 0’ + RT (ln Keq) Where, dG = free energy change of the r

Question

Ans. #I. Using, dG = dG0’ + RT (ln Keq)

Where, dG = free energy change of the reaction under given conditions

dG0’ = free energy change of the reaction at standard conditions

R = universal gas constant = 0.001987 kcal mol-1K-1 = (0.0083146 kJ mol-1 K-1)

T = temperature in kelvin

Keq = equilibrium constant

# Given,

[ATP] = 0.0079 M

[ADP] = 0.00104 M

[Pi] = 0.0079 M

pH = 7.00

Temperature, T = 37.00C = 310.15 K

Using [H3O+] = antilog (-pH) = antilog (-7.0) = 1.0 x 10-7 M.

Since [H3O+]= 1.0 x 10-7 M on the reactant side, the [H3O+] remains the same on product side, too.

Now,

Equilibrium constant, Keq = [ADP] [Pi] [H3O+]product / ([ATP] [H3O+]reactant )

Or, Keq = (0.00104 x 0.0079 x 1.0 x 10-7) / (0.0079 x 1.0 x 10-7)

Hence, Keq = 0.00104

# Given, dG0’ = -30.5 kJ/mol

Putting the values in equation 1-

dG= (-30.5 kJ mol-1) + (0.008315 kJ mol-1 K-1) x 310.15 K ln (0.00104)

Or, dG = -30.5 kJ mol-1 + 2.57889725 kJ mol-1 x (-6.8685345658)

Or, d = -30.5 kJ mol-1 – 17.17 kJ mol-1

Hence, dG = -47.67 mol-1

#II. Compare dG and dG0’ :

If dG = dG0’ - the reaction is at equilibrium

If dG > dG0’ – the reaction would go in backward direction

If dG < dG0’ - Spontaneous, the reaction would go in forward direction

Since dG (-47.67 kJ mol-1) < dG0’ (-30.5 kJ mol-1), the reaction is exergonic (i.e. the free energy change of the reaction under given conditions is more negative than that under standard conditions) and spontaneous.

Explanation / Answer

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