Chem 301 Problem Day #7
Fall 2008
1. The relationships between K and ΔGrxn are also valid for substances in solution (assuming
ideal solutions). Use this fact to solve the following problem:
The equilibrium constant, Kc, is 8.9x10-5 M at 298 K for the reaction:
fructose-1,6-diphosphate → glyceraldehyde-3-phosphate + dihydroxyacetone phosphate
a. Calculate ΔGorxn for this reaction assuming the behavior to be ideal.
ΔGorxn = -RT lnKc = -8.3145(298) ln(8.9x10-5) = 23.1 kJ/mole
b. Suppose that we have a mixture that is initially 0.010 M in fructose-1,6-diphosphate and
10-5 M in both products. What is the value of ΔGrxn? In which direction will reaction proceed to
reach equilibrium ?
Qc = [G][D]/[F] = (10-5)(10-5)/(0.010) = 1.0x10-8
ΔGrxn = ΔGorxn + RTlnQc
= 23.11x103 + (8.3145)(298) ln(1.0x10-8)
= 22.11x103 – 45.64x103
= - 22.53 kJ/mole
the reaction will proceed towards the products.
c. When the system reaches equilibrium, what will be the concentrations of reactants and
products?
F G + D
I 0.01 10-5 10-5 Kc = (10-5+x) (10-5+x) /(0.10-x)
C -x +x +x 8.9x10-5 = (10-5+x)2 / (0.10 –x)
E 0.01–x 10-5+x 10-5+x either solve with calculator or assume that
Efinal 9x10-3 9x10-4 9x10-4 x is much less than 0.10 (but not much less
than 10-5)
8.9x10-5 = (10-5+x)2 / 0.10
x = 9.3 x 10-4
2. Calculate the molar entropy of mixing and the molar Gibb’s energy of mixing when
synthetic air is produced at room temperature by mixing N2, O2 and Ar so that the final
percentages are 79%, 20%, and 1%, respectively.
mole fraction is directly related to percentage (χ = %/100)
ΔGmix = RT [Σ (χ lnχ)] = 8.3145(298) [0.79 ln(0.79) + 0.20 ln(0.20) + 0.10 ln(0.10)]
= -1.37 kJ/mole
ΔSmix = ΔGmix/T = -1.37/298 = 4.61 J/Kmole
