Question

In the Check Your Learning section how did they get that value of K?

Calculating an Equilibrium Constant using Standard Free Energy Change Given that the standard free energies of formation of Ag'(aq), CI (aq), and AgCl(s) are 77.1 kJ/mol, -131.2 kJ/mol, and -109.8 kJ/mol, respectively, calculate the solubility product, Ksp, for AgCl. Solution The reaction of interest is the following: Ag (aq) +CI (aq) Ksp [Ag ICI] AgCl(s) The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products: OpenStax book is available for free at http://cnx.org/content/col26069/1.5 pter 16 | Thermodynamics 8E AG [AGi (Ag (aq)) + AGi (CI (aq) - [AG; (AgCI(s) 177.1 kJ/mol - 131.2 kJ/mol] - [-109.8 kJ/mol] 55.7 kJ/mol The equilibrium constant for the reaction may then be derived from its standard free energy change: AG RT K'sp e exp(-4)- exp- 10 mol 8.314 J/mol-Kx 298.15 K = e-22470-1,74x 10-O exp(-22.470) This result is in reasonable agreement with the value provided in Appendix J. Check Your Learning Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C. 2NO2(8) N2O4(8) Answer: K6.9

Answer 1

The reaction in discussion is

2N02(g) + N204(9)

Step 1: Look at the table attached and find the Gibbs free energy of formation data for reactant NO₂ (g) and the product N₂O₄ (g). You would get for NO₂, it is 51.30 kJ/mol and for N₂O₄, it is 99.8 kJ /mol

Step 2: Find the overall Gibbs free energy change of the whole reaction. This is given by,

AG reaction = AG.products – AG f,reactants AG reaction = (99.8kJ/mol) - (2 x 51.30kJ/mol) AG action = -2.8kJ/mol

We had used the data for NO₂ in reactant side and N₂O₄ on product side. Note that because in the balanced reaction, there are 2 molecules of NO₂, we multiplied 51.30 kJ/mol by 2.

Further, we convert Gibbs free energy change to J/mol from kJ/mol by multiplying by 1000

-2.8 kJ /mol = -2.8 kJ/mol * (1000 J/kJ) = -2800 J/mol

Step 3: Use the formula to calculate K from Gibbs free energy. This formula is given just before check your learning section.

Thus, we have,

-Greaction K = exp RT

We will put R =8.314 J/mol-K and T = 298.15 K. Thus, we will have,

-(-2800J/mol K = exp (8.314J/mol K) (298.15K) K = exp(1.12957) = 3.095 3.1 =

Thus, the answer based on given data should be 3.1 and not 6.9.

Additional Note-

Note that the data given in the table corresponding to N₂O₄ is different in other sources. The free energy of formation for N₂O₄ in many sources is given as 97.8 kJ.mol^-1 instead of 99.8 kJ.mol^-1 which is given in your table.There appears to be a misprint. If we will take the value 99.8 kJ.mol^-1 for N₂O₄, we will actually get the value 4.8 kJ/mol as free energy instead of 2.8 kJ/mol which will give the value of K as 6.9. Thus, even though in the Appendix G, they gave the value as 99.8 kJ/mol, in calculation of K, they must have taken the value as 97.8 kJ/mol. But I suggest, you must rely on the table given and hence, the answer will be K=3.1.