Question

The equilibrium constant, Kp, for the following reaction is 0.497 at 500K.PCl5(g) PCl3(g) + Cl2(g)If an equilibrium mixture of the three gases in a 18.4 L container at 500K contains PCl5 at a pressure of 0.471 atm and PCl3 at a pressure of 0.651 atm, the equilibrium partial pressure of Cl2 is atm.

Answer 1

Answer: The equilibrium partial pressure of chlorine gas is 0.360 atm

Explanation:

For the given chemical equation:

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl2(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{Cl_2}\times p_{PCl_3}}{p_{PCl_5}}$

We are given:

$K_p=0.497\\p_{PCl_3}=0.651atm\\p_{PCl_5}=0.471atm$

Putting values in above equation, we get:

$0.497=\frac{p_{Cl_2}\times 0.651}{0.471}\\\\p_{Cl_2}=0.360atm$

Hence, the equilibrium partial pressure of chlorine gas is 0.360 atm