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

Problem 14.28.

Determine Kc at 298 K for the reaction ½ N2(g) + ½ O2(g) + ½ Cl2(g)NOCl(g), given the following data set at 298 K.
½ N2(g) + O2(g)NO2(g) Kp = 1.0×10–9 (1)
NOCl(g) + ½ O2(g)NO2Cl(g) Kp = 1.1×102 (2)
NO2(g) + ½ Cl2(g)NO2Cl(g) Kp = 0.3 (3)

(Hint: How are Kc and Kp for the reaction related?)




Answer:

Since the equilibrium constant data is given in terms of Kp for the known reactions, first find Kp for the unknown reaction and then, lastly, convert it to Kc.

The target reaction is:

½ N2(g) + ½ O2(g) + ½ Cl2(g)NOCl(g)

Reaction (1) has nitrogen and oxygen as reactants but a wrong product. Reaction (3) can be used to eliminate the nitrogen dioxide product from (1) and also introduces chlorine:
½ N2(g) + O2(g)NO2(g) Kp = 1.0×10–9 (1)
NO2(g) + ½ Cl2(g)NO2Cl(g) Kp = 0.3 (3)

Adding the two reactions together means that we multiply equilibrium constants:
½N2(g) + O2(g) + ½ Cl2(g)NO2Cl(g) Kp = (1.0×10–9)(0.3) = 3×10–10 (4)

(The NO2(g) found on both reactants and products sides has been eliminated.)

This has all the correct reactants, but the product is wrong and the stoichiometry is not quite right. If reaction (2) is reversed (meaning Kp is inverted), it can be added to reaction (4) to give the target reaction:
½ N2(g) + O2(g) + ½ Cl2(g)NO2Cl(g) Kp = 3×10–10 (4)
NO2Cl(g)NOCl(g) + ½ O2(g) Kp = 1/1.1×102 = 9.1×10–3 – (2)

Adding (4) and – (2) gives the desired reaction, both eliminating the NO2Cl and correcting the stoichiometry.

½ N2(g) + ½ O2(g) + ½ Cl2(g)NOCl(g)

With Kp = Kp(4)×Kp(– 2) = (3×10–10)( 9.1×10–3) = 3×10–12

To find Kc, use the relationship Kp = Kc(RT)n

n = 1 –(½ + ½ + ½ ) = –½

R = 0.0821 L•atm/mol•K

T = 298 K

so

3×10–12 = Kc[(0.0821)(298)]–½

Kc = 3×10–12[24.5] = 1×10–11