Question

Calculate the pressures of NO, Cl2, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 7.0 atm NO and 3.5 atm Cl2. (Hint: Kp is relatively large; assume the reaction goes to completion then comes back to equilibrium.)

2 NO(g) + Cl2(g) --> 2 NOCl(g)Kp = 2.9 ✕ 103 at 149°C

Answer 1

Answer: The pressures of
`NO`,
`Cl_2`, and
`NOCl` in an equilibrium mixture are 0.4 atm , 0.2 atm and 6.6 atm respectively.

Step-by-step explanation:

Initial pressure of
`NO` = 7.0 atm

Initial pressure of
`Cl_2` = 3.5 atm

The given balanced equilibrium reaction is,

`2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)`

Initial pressure 7.0 atm 3.5 atm 0 atm

At eqm. conc. (7-2x)atm (3.5-x)atm (2x) Matm

The expression for equilibrium constant for this reaction will be,

`K_p=([NOCl]^2)/([NO]^2[Cl_2])`

`K_p=((2x)^2)/((7-2x)^2* (3.5-x))`

we are given :
`K_p=2.9* 10^3`

Now put all the given values in this expression, we get :

`2.9* 10^3=((2x)^2)/((7-2x)^2* (3.5-x))`

`x=3.3atm`

Pressure of
`NO` at equilibrium = (7-2x) =
`(7-2* 3.3)=0.4 atm`

pressure of
`Cl_2` at equilibrium = (3.5-x) =
`(3.5-3.3)=0.2 atm`

pressure of
`NOCl` at equilibrium = (2x) =
`(2* 3.3)=6.6 atm`