Question

Hydrogen chloride decomposes to form hydrogen and chlorine, like this:

2HCl(g) + H2(g) â Cl2(g)

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen chloride, hydrogen, and chlorine has the following composition:

compound pressure at equilibrium
HCl 84.4 atm
H2 77.9 atm
Cl2 54.4

Required:
Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answer 1

Solution :

Given :

Partial pressure of HCl, $P_{HCl}$ = 84.4 atm

Partial pressure of $H_2$, $P_{H_2}$ = 77.9 atm

Partial pressure of $Cl_2$, $P_{Cl_2}$ = 54.4 atm

Reaction :

$2HCl (g) \leftrightharpoons H_2(g) + Cl_2(g)$

Using equilibrium concept,

$k_p=\frac{(P_{H_2})(P_{Cl_{2}})}{(P_{HCl})^2}$

$k_p=\frac{77.9 \times 54.4}{(84.4)^2}$

$k_p=0.594$

$k_p=0.59$  (in 2 significant figures)

or $k_p=5.9 \times 10^{-1}$