Question

PCl 5(g) gives PCl 3(g) + Cl 2(g)
1.0 mol of PCl 5(g), 1.0mol of PCl 3(g) and 1.0 mol of Cl 2(g) are placed in a container of volume 1 dm3 at 250 °C and allowed to reach equilibrium.
At this temperature, the equilibrium mixture contains 1.8 moles of PCl 3.
What is the value of Kc at 250 °C?

Answer 1

Kc = (PCl3)(Cl2)/(PCl5)

Kc =(1mol/lit)(1mol/lit)/1mol/lit)

Kc = 1mol/lit

Equilibrium mixture contains 1.8mols of PCl3,then

Kc = (PCl3)(Cl2)/PCl5)

Kc = (1.8mol/lit)(1mol/lit)/(1mol/lit)

Kc = 1.8 mol/lit