CHM112 Home Page Reference Materials Homework Assignments Lectures Exams & Quizzes Grades Study Aids URI Home Page


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

Problem 17.42

Use data from Appendix C to determine Ho and So, at 298 K, for the following reaction. Then determine Go in two ways and compare the results.

C(graphite) + H2O(g) CO(g) + H2(g)






Answer:

Find Ho usingand So using.

Then Go can be found from Go = Ho – TSo and

C(graphite)
Hfo = 0.00 kJ/mol
So = 5.74 J/mol•K
Gfo = 0.00 kJ/mol

H2O(g)
Hfo = –241.8 kJ/mol
So = 188.7 J/mol•K
Gfo = –228.6 kJ/mol

CO(g)
Hfo = –110.5 kJ/mol
So = 197.6 J/mol•K
Gfo = –137.2 kJ/mol

H2(g)
Hfo = 0.00 kJ/mol
So = 130.6 J/mol•K
Gfo = 0.00 kJ/mol

Ho = [–110.5 + 0.0] – [0.0 + –241.8] = 131.3 kJ
So = [197.6 + 130.6] – [5.74 + 188.7] = 133.8 J/K = 0.1338 kJ/K
Go = [–137.2 + 0.0] – [0.0 + –228.6] = 91.4 kJ
Go = 131.3 – (298)(0.1338) = 91.4 kJ

Both methods give the same Go, as indeed they should.