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Problem 14.27

Determine K_{c} at 298 K for the reaction 2 CH_{4}(*g*) C_{2}H_{2}(*g*) + 3 H_{2}(*g*), given the following data at 298 K.

CH

_{4}(g) + H_{2}O(g) CO(g) + 3 H_{2}(g) K_{p}= 1.2×10^{–25}2 C

_{2}H_{2}(g) + 3 O_{2}(g)4 CO(g) + 2 H_{2}O(g) K_{p}= 1.1×10^{2}H

_{2}(g) + ½ O_{2}(g) H_{2}O(g) K_{p}= 1.1×10^{40}

(*Hint:* How are K_{c} and K_{p} for the reaction related?)

First, find K_{p} for the target reaction using the Law of Multiple Equilibria.
Then, use the relationship K_{p} = K_{c}(RT)^{n} to find the value of K_{c}.

The target reaction has 2 CH_{4}(*g*) as one of the reactants and
1 C_{2}H_{2}(*g*) as a product. This tells us that the first reaction must have all of the stoichiometric coefficients doubled and that the K_{p} 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 K_{p} to the inverse of the square root of the given reaction.

2 CH

_{4}(g) + 2 H_{2}O(g)2 CO(g) + 6 H_{2}(g) K_{p}= (1.2×10^{–25})^{2}= 1.4×10^{–50}2 CO(

g) + H_{2}O(g) C_{2}H_{2}(g) + O_{2}(g) K_{p}= 1/(1.1×10^{2})^{½}= 9.5×10^{–2}

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

2 CH

_{4}(g) + 3 H_{2}O(g) C_{2}H_{2}(g) + 6 H_{2}(g) + O_{2}(g) K_{p}= (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 H

_{2}(g) + O_{2}(g) 3 H_{2}O(g) K_{p}= (1.1×10^{40})^{3}= 1.3×10^{120}

Now, adding the last two reactions together gives the target reaction. K_{p} for the target reaction is found by multiplying equilibrium constants.

2 CH

_{4}(g)C_{2}H_{2}(g) + 3 H_{2}(g) K_{p}= (1.3×10^{–51})( 1.3×10^{120}) = 1.7×10^{69}

Finally, to find K_{c} for the final reaction, n =
[1 + 3] – [2] = 2 so K_{c} = K_{p}/(RT)^{n}
= (1.7×10^{69})/(0.0821×298)^{2} = 2.8×10^{66}