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

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,

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

`$k_p=(77.9 * 54.4)/((84.4)^2)$`

`$k_p=0.594$`

`k_p=0.59` (in 2 significant figures)

or
`k_p=5.9 * 10^(-1)`