Consider the following hypothetical chemical pathway (A---->D) consisting of three individual reaction steps (A-->B,B-->C,C-->D)
This is the set of questions
Please use this set of rules/guidlines
A. K(net) = K1*K2*K3
= 32680
Free energy change = -2.303 * 8.3*298*log(32680) = -25714.439 joules
B. K1= 10^4
change in free energy = -2.303 * 8.3*298*log(10^4) = -22784.9608 joules
Spontaneous. As change is free energy is negative.
K2= 8.6 *10^-3
change in free energy = -2.303 * 8.3*298*log(8.6 *10^-3) = 11853.431joules
Non-Spontaneous. As change is free energy is positive
K3=3.8 *10^2
change in free energy = -2.303 * 8.3*298*log(3.8 *10^2) = -14695.06703 joules
Spontaneous. As change is free energy is negative
C. Net Change in free energy = -22784.9608 +11853.431 -14695.06703 = -25626.59 joules
D.
K(net) = K1*K2*K3
= 32680
Free energy change = -2.303 * 8.3*298*log(32680) = -25714.439joule
Spontaneous. As change is free energy is negative.
E. No. each individual step does not necessarily need to be spontaneous.
the overall change in free energy should be negative. and not the individual steps. all the individual free energies are added up. is the total sum is negative, then spontaneous otherwise non spontaneous.
F. change in free energy by summing up free energy changes = -25.6 kJ
Change free energy by first calculating net equilibrium constant = -25.7 kJ
which are nearly equal
Both will be same as they correspond to same reaction.
A. Keq for the entire pathway(A->D) :-
Keq0'(A->D) = Keq0'(A->B) X Keq0'(B->C) X Keq0'(C->D)
Thus substituting the reapective values as given in the problem we have
Keq0'(A->D) = 1X10^4 X 8.6X10^-3 X 3.8 X 10^2 = 3.2680 X 10^4
B. ATQ we have to calculate the delta G0 for the individual reaction.
delta G0' = - RT0'lnKeq0'
for the step A->B the delta G0' is
delta G0' = -(1.987 X 10^-3 kcal mol-1 K-1) X (298K) X (ln (1X10^4) = -5.453 kcal mol-1
the condition for spontaineity is that standard gibbs free energy i.e delta G0' should be negative . In the reactive delta G0' is negative thus the reaction A-> B is spontaneous.
Next for the step B->C the delta G 0' is
delta G0' = -( 1.987X10^-3 Kcal mol-1 K-1) X (298K) X (ln (8.6X10^-3)) = 2.816 Kcal mol-1
As the change in standard gibbs free energy is positive the reaction is not spontaneous for the step B->C
lastly the delta G0' for the step C->D is
delta G0' = -(1.987 X 10^-3 Kcal mol-1 K-1) X (298K) X (ln(3.8X10^2)) = -3.51 Kcal mol-1
the change in standard gibbs free energy is negative , therefore the step C->D is spontaneous.
C. delta G0' of the overall pathaway is
delta G0' (A->D) = delta G0'(A->B) + delta G0'(B->C) + delta G0'(C->D)
= -6.147 Kcal mol-1
As the net change in standard free energy is negative the reaction is spontaneous for the pathway(A->D)
D. delta G0' using Keq0' calculated in answer A.
delta G0' = - RT0'ln(Keq0')
=-(1.987X10^-3 Kcal mol-1K-1)X(298K)Xln(3.2680X10^4) = -6.15 Kcal mol-1.
The reaction is spontaneous as change in standard free energy is negative.
E. It is not necessary that individual reaction should be spontaneous in order to have the overall reaction to be spontaneous. The spontaneity of the entire reaction is judged on the basis whether overall change in standard gibbs free energy is negative or not. As in the reaction given in the problem, although the reaction step B->C is not spontaneous but the overall reaction is spontaneous as overall change ib standard gibbs free energy is negative.
F. The delta G0' calculated in part C and part D is same and the overall change in standard gibbs free energy is negative which means the reaction is spontaneous.
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