Question

33. Use the following standard-state free energy of formation data to calculate the acid-dissociation equilibrium constant (K_a) at 25^oC:

Compound	Delta Gf (kJ/mol)
HCO2 (aq)	-372.3
H+ (aq)	0.00
HCO2- (aq)	-351.0

a.   6.3 x 10^-5                                                   d.   8.0 x 10^-3

b.   3.4 x 10^-5                                                   e.   4.0 x 10^-5

c.   1.8 x 10^-4

34. Use the K_a Table to identify the acid in question #33 at 25^oC:

	Ka	Name
a	7.2 x 10-4	Nitrous Acid
b	6.6 x 10-4	Hydroflouric Acid
c	6.3 x 10-5	Benzoic Acid
d	1.8 x 10-4	Formic Acid
e	1.8 x 10-5	Ethanoic Acid

Answer 1

33)

we have:

Gof(HCO2(aq)) = -372.3 KJ/mol

Gof(H+(aq)) = 0.0 KJ/mol

Gof(CO2-(aq)) = -351.0 KJ/mol

we have the Balanced chemical equation as:

HCO2(aq) ---> H+(aq) + CO2-(aq)

deltaGo rxn = 1*Gof(H+(aq)) + 1*Gof(CO2-(aq)) - 1*Gof( HCO2(aq))

deltaGo rxn = 1*(0.0) + 1*(-351.0) - 1*(-372.3)

deltaGo rxn = 21.3 KJ

T= 25.0 oC

= (25.0+273) K

= 298 K

deltaG = 21.3 KJ/mol

deltaG = 21300 J/mol

we have below equation to be used:

deltaG = -R*T*ln Kc

21300 = - 8.314*298.0* ln(Kc)

ln Kc = -8.5971

Kc = 1.8*10^-4

Answer: c

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