Question

Phosphorous pentachloride decomposes according to the reaction


PCl5(g)=PCl3(g)+Cl2(g)


A 12.3 g sample of PCl5 is added to a sealed 1.50 L flask and the reaction is allowed to come to equilibrium at a constant temperature. At equilibrium, 31.8% of the PCl5 remains. What is the equilibrium constant, Kc, for the reaction?


Question is asking for Kc

Answer 1

Answer: The equilibrium constant, $K_c$, for the reaction is 0.061.

Explanation:

Initial concentration of $PCl_5$  = $\frac{\text {given mass}}{\text {Molar mass}\times Volume in L}}=\frac{12.3g}{208.2g/mol\times 1.50L}=0.039M$  

Equilibrium concentration of $PCl_5$ = $\frac{31.8}{100}\times 0.039=0.012M$  

The given balanced equilibrium reaction is,

                            $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

Initial conc.              0.039 M                0 M        0 M

At eqm. conc.     (0.039-x) M              (x) M   (x) M

Given : (0.039-x) = 0.012

x = 0.027

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[Cl_2]\times [PCl_3]}{[PCl_5]}$

Now put all the given values in this expression, we get :

$K_c=\frac{0.027\times 0.027}{0.012}=0.061$

The equilibrium constant, $K_c$, for the reaction is 0.061.