Question

The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K. PCl3(g) Cl2(g) PCl5(g) Calculate the equilibrium concentrations of reactant and products when 0.453 moles of PCl3 and 0.453 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.

Answer 1

Answer:

$[PCl_3]=[Cl_2]=0.068M$

$[PCl_5]=0.385M$

Explanation:

Hello!

In this case, since the equilibrium expression for the considered equation is:

$Kc=\frac{[PCl_5]}{[Cl_2][PCl_3]}$

Which can be written in terms of the reaction extent and the ICE chart and the initial concentrations of 0.453 M as shown below:

$83.3=\frac{x}{(0.453M-x)(0.453M-x)}$

We can solve for x as follows:

$x=0.385M$

In such a way, we obtain the following concentrations at equilibrium:

$[PCl_3]=[Cl_2]=0.453M-0.385M=0.068M$

$[PCl_5]=0.385M$

Best regards!