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

Problem 14.27

Determine Kc at 298 K for the reaction 2 CH4(g) C2H2(g) + 3 H2(g), given the following data at 298 K.

CH4(g) + H2O(g) CO(g) + 3 H2(g)    Kp = 1.2×10–25

2 C2H2(g) + 3 O2(g)4 CO(g) + 2 H2O(g)    Kp = 1.1×102

H2(g) + ½ O2(g) H2O(g)    Kp = 1.1×1040

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

First, find Kp for the target reaction using the Law of Multiple Equilibria. Then, use the relationship Kp = Kc(RT)n to find the value of Kc.

The target reaction has 2 CH4(g) as one of the reactants and 1 C2H2(g) as a product. This tells us that the first reaction must have all of the stoichiometric coefficients doubled and that the Kp for the given reaction must be squared. The stoichiometric coefficients of the second reaction must all be halved and then the reaction reversed so this changes the Kp to the inverse of the square root of the given reaction.

2 CH4(g) + 2 H2O(g)2 CO(g) + 6 H2(g)    Kp = (1.2×10–25)2 = 1.4×10–50

2 CO(g) + H2O(g) C2H2(g) + O2(g)     Kp = 1/(1.1×102)½ = 9.5×10–2

The sum of these two reactions is given below, which conveniently eliminates CO(g). Kp for the summed reaction is found by multiplying the two equilibrium constants.

2 CH4(g) + 3 H2O(g) C2H2(g) + 6 H2(g) + O2(g)     Kp = (1.4×10–50)( 9.5×10–2) = 1.3×10–51

This is not the target reaction. To reach the target, the third reaction given above must be used with all of its stoichiometric coefficents multiplied by three. This means that the equilbrium constant must be cubed.

3 H2(g) + O2(g) 3 H2O(g)    Kp = (1.1×1040)3 = 1.3×10120

Now, adding the last two reactions together gives the target reaction. Kp for the target reaction is found by multiplying equilibrium constants.

2 CH4(g)C2H2(g) + 3 H2(g)    Kp = (1.3×10–51)( 1.3×10120) = 1.7×1069

Finally, to find Kc for the final reaction, n = [1 + 3] – [2] = 2 so Kc = Kp/(RT)n = (1.7×1069)/(0.0821×298)2 = 2.8×1066