**Practice Problem 9**

Use the value of *G*^{o}
obtained in Practice Problem 7 to calculate the equilibrium constant for the following
reaction at 25°C:

N_{2}(*g*) + 3 H_{2}(*g*) 2 NH_{3}(*g*)

**Solution**

Practice Problem 7 gave the following value for *G*^{o }for this reaction at 25°C:

*G*^{o
}= -32.96 kJ

Now we turn to the relationship between *G*^{o }for a
reaction and the equilibrium constant for the reaction:

*G*^{o}
= - *RT* ln *K*

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

Substituting the known value of *G*^{o}, *R*, and *T* into this
equation gives the following result:

The equilibrium constant for this reaction at 25°C is therefore 6.0
x 10^{5}:

*K* = *e*^{13.3} = 6.0 x10^{5}

It is easy to make mistakes when handling the sign of the relationship between *G*^{o} and *K*.
It is therefore a good idea to check the final answer to see whether it makes
sense. *G*^{o}
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.