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 0.269 moles of PCl3 and 0.269 moles of Cl2 are introduced into a 1.00 L vessel at 500 K. [PCl3] = M [Cl2] = M [PCl5] = M

Answer 1

molarity of PCl3 = no of moles/volume in L

= 0.269/1 = 0.269M

molarity of Cl2 = no of moles/volume in L

= 0.269/1 = 0.269M

--------------PCl3(g) + Cl2(g) ------------> PCl5(g)

I------------ 0.269---- 0.269 ------------------ 0

C-------- -x ---------- -x ------------------------- +x

E--------- 0.269-x ----- 0.269-x ------------------+x

Kc = [PCl5]/[PCl3][Cl2]

83.3 = x/(0.269-x)(0.269-x)

83.3(0.269-x)^2 = x

x = 0.218

[PCl3] = 0.269-x = 0.269-0.218 = 0.051M

[PCl5] =x = 0.218M