Practice Problem 9

Use the value of deltaGo obtained in Practice Problem 7 to calculate the equilibrium constant for the following reaction at 25C:

N2(g) + 3 H2(g) equilibr.gif (63 bytes) 2 NH3(g)


Solution

Practice Problem 7 gave the following value for deltaGo for this reaction at 25C:

deltaGo = -32.96 kJ

Now we turn to the relationship between deltaGo for a reaction and the equilibrium constant for the reaction:

deltaGo = - RT ln K

Solving for the natural log of the equilibrium constant gives the following equation:

Ex21_9a.gif (1097 bytes)

Substituting the known value of deltaGo, R, and T into this equation gives the following result:

Ex21_9b.gif (1644 bytes)

The equilibrium constant for this reaction at 25C is therefore 6.0 x 105:

K = e13.3 = 6.0 x105

It is easy to make mistakes when handling the sign of the relationship between deltaGo and K.  It is therefore a good idea to check the final answer to see whether it makes sense.   deltaGo for the reaction in this exercise is negative.  The reaction is therefore spontaneous, and the equilibrium should lie on the side of the products.  The equilibrium constant should therefore be much larger than 1, which it is.

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