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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 17.54

Use data from Appendix C to determine Kp at 298 K for these reactions.

(a) 2 N2O(g) + O2(g)4 NO(g)

(b) 2 NH3(g) + 2 O2(g)N2O(g) + 3 H2O(g)




Answer:

Find Go for the reaction from the free energy of formations, then use Go = –RTlnKp to find the equilibrium constant.

(a) 2 N2O(g) + O2(g)4 NO(g)

Gof(NO(g)) = 86.57 kJ/mol
Gof(O2(g)) = 0.00 kJ/mol
Gof(N2O(g)) = 104.2 kJ/mol

Go = [4(86.57)] – [2(104.2) + 0.0] = 137.9 kJ

137900 = –(8.314)(298)lnKp
lnKp = –55.7
Kp = e–55.7 = 6×10–25

(b) 2 NH3(g) + 2 O2(g)N2O(g) + 3 H2O(g)

Gof(N2O(g)) = 104.2 kJ/mol
Gof(H2O(g)) = –228.6 kJ/mol
Gof(NH3(g)) = –16.48 kJ/mol
Gof(O2(g)) = 0.00 kJ/mol

Go = [104.2 + 3(–228.6)] – [2(–16.48) + 2(0.00)] = –548.6 kJ

–548600 = –(8.314)(298)lnKp
lnKp = 221.4
Kp = e221.4 = 1×1096