Consider the reaction X + Y Z. Based solely upon this information, do you expect
ID: 521213 • Letter: C
Question
Consider the reaction X + Y Z. Based solely upon this information, do you expect the entropy change in this reaction to be less than, equal to, or greater than 0?
A
less than 0..
B
equal to 0.
C
greater than 0.
Consider the phosphoglucoisomerase reaction: Glucose-6-P = Fructose-6-P G'0 = +1,790 J/mol
At equilibrium, is the concentration of glucose-6-P greater than or equal to or less than the concentration of fructose-6-P?
If the concentration of fructose-6-P in a cell is greater than the concentration of glucose-6-P, is the value of G' greater than or equal to or less than 0 J/mol.
A
less than 0..
B
equal to 0.
C
greater than 0.
Explanation / Answer
Ans. 1. The number of moles of reactants = 1 mol (X) + 1 mol (Y) = 2 mol
The number of moles of product = 1 mol (Z)
Note that the number of moles of product is lesser than the number of moles of reactants. Therefore, the entropy of the system decreases because 1 mol product has lesser degree of freedom than 2 mol reactants.
Therefore, entropy decreases and is less than 0.
Correct option. A. Less than 0.
Ans. 2. Using the equation dG = dG0’ + RT lnK - equation 1
Where, dG = calculated/ experimental free energy change = ?
dG0’ = standard/ theoretical free energy change R = 0.0083146 kJ mol-1K-1
T = temperature in kelvin = (0C + 273.15) K
K = equilibrium constant under given condition
Putting the values in equation 1-
1.790 kJ/ mol = - (0.0083146 kJ mol-1K-1) x 298 K x ln K
Or, 1.790 kJ/mol / (-2.4777508 kJ/mol) = ln K
Or, -0.72243 = 2.303 log K
Or, log K = -0.72243 / 2.303 = -0.31369
Or, K = antilog (-0.31369) = 0.486
Therefore, equilibrium constant, K = 0.486
Now, from the balanced reaction, equilibrium constant, K = [F-6-P] / [G-6-P]
Or, 0.486 = [F-6-P] / [G-6-P]
Or, [F-6-P] = 0.486 x [G-6-P]
Therefore, [F-6-P] < [G-6-P] - at equilibrium.
Or, [G-6-P] is greater than that of [F-6-P].
Ans. 3. If [F-6-P] > [G-6-P], then equilibrium constant K is always greater than 1.
Say, [F-6-P]/ [G-6-P] = 2, that is [F-6-P] is double that of [G-6-P].
So, K = [F-6-P]/ [G-6-P] = 2
Putting the values of K=2 in equation 2-
dG = dG0’ + RT lnK - equation 2
Where, dG = calculated/ experimental free energy change = ?
dG0’ = standard/ theoretical free energy change R = 0.0083146 kJ mol-1K-1
T = temperature in kelvin = (0C + 273.15) K
K = equilibrium constant under given condition.
dG = 1.790 kJ/mol + (0.0083146 kJ mol-1 K-1) x 298 K x (ln 2)
or, dG = 1.790 kJ/mol + 1.72 kJ/mol = 3.51 kJ/mol
Therefore, dG > dG0 – when [F-6-P] > [G-6-P]
And, dG is greater than 0.