Question

Consider the reaction below. At 500 K, the reaction is at equilibrium with the following concentrations. [PCI5]= 0.0095 M [PCI3] = 0.020 [CI2] = 0.020 M What is the equilibrium constant for the given reaction?

Answer 1

Answer: The equilibrium constant for the given reaction is 0.0421.

Explanation:

$PCl_5\rightleftharpoons PCl_3+Cl_2$

Concentration of $[PCl_5]$ =  0.0095 M

Concentration of $[PCl_3]$ =  0.020 M

Concentration of $[Cl_2]$ =  0.020 M

The expression of the equilibrium constant is given as:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.020 M\times 0.020 M}{0.0095 M}$

$K_c=0.0421$ (An equilibrium constant is an unit less constant)

The equilibrium constant for the given reaction is 0.0421.

Answer 2

Answer: 0.042

Explanation:- Equilibrium constant is the ratio of concentration of products to the concentration of reactants each term raised to their stochiometric coefficients.

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$

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

Given :$[PCl_5]$= 0.0095 M

$[PCl_3]$ = 0.020 M

$[Cl_2]$ = 0.020 M

Putting the given values n the equation:

$K_c=\frac{[0.020][0.020]}{[0.0095]}=0.042$