Question

At 1000 K, a sample of pure NO2 gas decomposes. 2 NO2(g) equilibrium reaction arrow 2 NO(g) + O2(g) The equilibrium constant KP is 158. Analysis shows that the partial pressure of O2 is 0.29 atm at equilibrium. Calculate the pressure of NO and NO2 in the mixture.

Answer 1

Answer: The pressure of NO and
`NO_2` in the mixture is 0.58 atm and 0.024 atm respectively.

Step-by-step explanation:

We are given:

Equilibrium partial pressure of
`O_2` = 0.29 atm

For the given chemical equation:

`2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)`

Initial: a

At eqllm: a-2x 2x x

Calculating for the value of 'x'

`\Rightarrow x=0.29`

Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm

Equilibrium partial pressure of
`NO_2` = a - 2x = a - 2(0.29) = a - 0.58

The expression of
`K_p` for above equation follows:

`K_p=(p_(O_2)* (p_(NO))^2)/((p_(NO_2))^2)`

We are given:

`K_p=158`

Putting values in above expression, we get:

`158=(0.29* (0.58)^2)/((a-0.58)^2)\\\\a=0.555,0.604`

Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.

So,
`a=0.604`

Equilibrium partial pressure of
`NO_2` = (a - 0.58) = (0.604 - 0.58) = 0.024 atm

Hence, the pressure of NO and
`NO_2` in the mixture is 0.58 atm and 0.024 atm respectively.