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Question

For the reaction X + Y ⇄ Z with a Keq of 0.822

A) What is the standard free energy change for this reaction at 37 degrees Celsius?

B) If [X] = 10mM, [Y] = 0.21M, [Z] = 125 μM, which direction does the reaction proceed at

37 degrees Celsius?

C) (2 points) If ∆S^o of this reaction is -0.05 kcal/mol, what is the standard change in enthalpy for the

reaction?

Answer 1

X + Y <==> Z .. Keq = 0.822

A). ∆Gº = -RT ln K

∆Gº = -(8.314 J/Kmol)(310K) ln 0.822

∆Gº = 505 J/mol

B). Q = [Z] / [X][Y]

[Z] = 125 uM = 1.25x10^-4 M

[X] = 10 mM = 0.01 M

[Y] = 0.21 M

Q = 1.25x10^-4 / (0.01)(0.21)

Q = 0.0595

Since Q << Keq, the reaction will proceed to the right, toward the products in order to reach equilibrium

C). ∆Sº = -0.05 kcal/mol

convert to kJ/mol: -0.05 kcal/mol x 4.184 kJ/kcal = -0.209 kJ/mol

∆Gº = ∆Hº = T∆Sº

∆Gº = 505 J/mol = 0.505 kJ/mol

∆Hº = ?

∆Sº = -0.209 kJ/mol

T = 310K

Solving for ∆Hº...

0.505 = ∆H - (310)(-0.209)

0.505 = ∆Hº + 64.8

∆Hº = -64.3 kJ/mol

Be sure to check all of the math.