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Example

Obtain the value of K_c for the following reaction at a temperature 300 K using the following data.  A 1.355-mol sample of PCl₅ was placed into a 5.00 L and allowed to reach equilibrium.  At equilibrium the mixture was found to contain 0.188 mol of Cl₂.  

PCl₅(g) <==> PCl₃(g) + Cl₂(g) 

We will set up a table similar to the ones we used in the previous topic, except this time we want molar concentrations instead of moles.  This is because we need molar concentrations in the equilibrium constant expression.  From the information given in the problem, the initial concentration of PCl₅ is 1.355 mol / 5.00 L = 0.271 M and the equilibrium concentration of  Cl₂ is 0.188 mol / 5.00 L = 0.0376 M.  Now we can build our table.

Concentrations(M)	PCl5(g)	<==>	PCl3(g)	+	Cl2(g)
Starting	0.271		0		0
Change	-x		+x		+x
Equilibrium	0.271 - x		x		x = 0.0.0376

Thus the equilibrium concentrations will be [PCl₅] = (0.271 - x) M = (0.271 - 0.038) M = 0.233 M and [PCl₃] = [Cl₂] = 0.0376.  Next we will set up theequilibrium constant expression and plug in these values.

     [PCl3][Cl2]
Kc = -----------
       [PCl5]

     [0.0376][0.0376]
Kc = ---------------   = 6.07 x 10-3
         [0.233]

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