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.

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:

$K_p=\frac{p_{O_2}\times (p_{NO})^2}{(p_{NO_2})^2}$

We are given:

$K_p=158$

Putting values in above expression, we get:

$158=\frac{0.29\times (0.58)^2}{(a-0.58)^2}\\\\a=0.555,0.604$

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.