Question

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

Homework no.3. Bloenergete READ THESE IMPORTANT INSTRUCTIONS! These instructions are in force, wherever applicable, for all homework assignments and exams throughout the semester. When working problems, follow the guidelines laid out in the sheet I passed out to you about following the rules of good math "grammar Especially, be sure to set up the initial equation, show all important work steps, and report the answer fully as a math "sentence" with variable, equal sign, and answer, complete with correct dimensional units. I wil deduct one point for each instance of not following these steps of good math grammar, even if the answer is numerically correct. Also, as in all science, neatness and organization are important. And now, the actual homework assignment: I. Consider the following hypothetical chemical pathway (A steps (A → B, B → C, C → D): D) consisting of three individual reaction A. Given the Koo, values of each individual step, what is the equilibrium constant, K of the entire pathway (A → D)? (Hint: Recall from General Chemistry that individual Koi values are multiplicative to obtain the total Ka) B. Calculate Δ0°, of each individual reaction step, ie, Δ0rA B), etc. State whether each individual reaction step is or is not spontancous. How were you able to decide whether or not each reaction step is spontaneous? C. Determine ΔGo, of the overall pathway (A-D) by summing the individual AGo values from part B. According to this procedure, is the overall pathway (A → D) spontaneous? How can you tell? D. Now, calculate ar, of the overall pathway (A → D) directly, using the overall Koo, calculated from part A. According to this procedure, is the overall pathway (A D) spontaneous? How can you tell? E. Is it necessary for each individual reaction step in a pathway to be spontaneous for the entire pathway to be spontancous? Explain why or why not. F. Compare the value of AG for the overall pathway that you calculated in part C, by summing the individual Δ0°, values, to the value of ΔGo, that you calculated in part D directly for the overall pathway. How do these two values of ΔG。, compare? The fact that they are equal to each other hint: if they are not equal, then you worked the problems incorrectly-illustrates what important characteristic of free energy?

Please use this set of rules/guidlines

Answer 1

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.

Answer 2

A. Keq for the entire pathway(A->D) :-

Keq^0'(A->D) = Keq^0'(A->B) X Keq^0'(B->C) X Keq^0'(C->D)

Thus substituting the reapective values as given in the problem we have

Keq^0'(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 G^0' = - RT^0'lnKeq^0'

for the step A->B the delta G^0' is

delta G^0' = -(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 G^0' should be negative . In the reactive delta G^0' is negative thus the reaction A-> B is spontaneous.

Next for the step B->C the delta G ^0' is

delta G^0' = -( 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 G^0' for the step C->D is

delta G^0' = -(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 G^0' of the overall pathaway is

delta G^0' (A->D) = delta G^0'(A->B) + delta G^0'(B->C) + delta G^0'(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 G^0' using Keq^0' calculated in answer A.

delta G^0' = - RT^0'ln(Keq^0')

=-(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 G^0' 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.