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Question

Consider the simple reaction below:

A ⇄ B + C ΔG^o = -3.7 kcal/mol

1. What is the K eq for this reaction under physiological conditions?

2. If a reaction is set up with 0.1μM of A, 0.1M of B and 1mM of C, what is the free energy

change at physiological temperature?

3. The standard entropy change of this reaction is 24 cal/mol*K. What is the standard

enthalpy change at physiological temperature?

Answer 1

A <==> B + C .. ∆Gº = -3.7 kcal/mol x 4.184 kJ/kcal = -15.5 kJ/mol

1). ∆Gº = -RT ln Keq

-15.5 kJ/mol = -(0.008314 kJ/Kmol)(310K) ln Keq

ln Keq = 6.01

Keq = 409

2). [A] = 0.1 uM = 1x10^-7 M

[B] = 0.1 M

[C] = 1 mM = 1x10^-3 M

Q = [B][C] / [A] = (0.1)(1x10^-3) / 1x10^-7 = 1000

∆G = ∆Gº - RT ln Q

∆G = -15.5 kJ/mol - (0.008314 kJ/Kmol)(310K) ln 1000

∆G = -15.5 - (2.58)(6.91) = -15.5 - 17.8

∆G = -33.3 kJ/mol

3). ∆Gº = ∆Hº - T∆Sº

∆Gº = -15.5 kJ/mol

∆Hº = ?

T = 310K

∆Sº = 24 cal/Kmol x 4.184 J/cal = 100 J/mol = 0.100 kJ/mol

Solve for ∆Hº...

-15.5 = ∆Hº - (310)(0.100)

-15.5 = ∆Hº - 31

∆Hº = 15.5 kJ/mol

Be sure to check all of the math.