Question

2NOCl(g) ↔ 2NO(g) + Cl2(g) with K = 1.6 x10–5. In an experiment, 1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in a 1.00-L container. If x moles of NOCl react, what is the equilibrium concentration of NO?

Answer 1

Answer: 2x

Step-by-step explanation:

Answer 2

Answer:
`3.8* 10^(-3)M`

Step-by-step explanation:

Moles of
`NOCl` = 1 mole

Moles of
`Cl_2` = 1 mole

Volume of solution = 1 L

Initial concentration of
`NOCl` = 1 M

Initial concentration of
`Cl_2` = 1 M

The given balanced equilibrium reaction is,

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

Initial conc. 1 M 0M 1 M

At eqm. conc. (1-2x) M (2x) M (1+x) M

The expression for equilibrium constant for this reaction will be,

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

The
`K_c`=
`1.6* 10^(-5)}`

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

`{1.6* 10^(-5)}=((2x)^2* (1+x))/((1-2x)^2)`

By solving the term 'x', we get :

`x=0.0019`

Concentration of
`NO` at equilibrium= (2x) M =
`2* 0.0019=3.8* 10^(-3)M`