Question

Initially, 0.65 mol of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.15 mol of PCl3 in the flask. What is the equilibrium concentration of PCl5? Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer : The equilibrium concentration of $PCl_5$ is, 0.50 M

Explanation : Given,

Initial moles of $PCl_5$ = 0.65 mole

Volume of solution = 1.0 L

Moles of $PCl_3$ at equilibrium = 0.15 mole

The balanced equilibrium reaction will be,

                          $PCl_5\rightleftharpoons PCl_3+Cl_2$

Initial moles     0.65        0         0

At eqm.           (0.65-x)     x         x

Moles of $PCl_3$ at equilibrium = x = 0.15 mole

Moles of $Cl_2$ at equilibrium = x = 0.15 mole

Moles of $PCl_5$ at equilibrium = (0.65-x) = (0.65-0.15) = 0.50 mole

Now we have to calculate the concentration of $PCl_3,Cl_2\text{ and }PCl_5$ at equilibrium.

Formula used : $Concentration=\frac{Moles}{Volume}$

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}=\frac{0.15mole}{1.0L}=0.15M$

$\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.50mole}{1.0L}=0.50M$

Therefore, the equilibrium concentration of $PCl_5$ is, 0.50 M