Question

a sample of pure no2 gas heated to 1000k decomposes. the equilibrium constant kp= 100 atm analysis shows that the partial pressure of o2 is 0.25atm at equilibrium when the partial pressure of no2 at equilibrium will be

Answer 1

2NO₂(g) <=> 2NO(g) + O₂(g)

Initial P   0   0

Press -2x 2x x

Equi P - 2x 2x x

K base p = Po₂x P² NO/P² NO₂ = (x)(2x)²/ (P-2x)²

The pressure of the oxygen at this point is 0.25 so the above equation will reduce to

1/100 = (0.25) (2x0.25)²/(P-(2x0.25)²

(P - 0.5)² = 25

P = 5.5

Total pressure on summing for NO₂ and NO

P base sum = P NO₂ + P base NO

= 5.5 + 0.5 = 6 atm

Answer 2

Answer:- 0.025 atm

Solution:- The given balanced equilibrium equation is:

2NO_2(g)\leftrightarrow 2NO(g)+O_2(g)2NO2​(g)↔2NO(g)+O2​(g)

From the equation, there is 2:1 mole ratio between NO and O_2O2​ .

Their equilibrium partial pressure would be in the ratio. Equilibrium partial pressure of oxygen is given as 0.25 atm. So, the equilibrium partial pressure of NO = 2(0.25 atm) = 0.50 atm

Equilibrium expression for the given equation is written as:



Let's plug in the values in it and solve it for equilibrium partial pressure of NO_2NO2​ . Let's say the equilibrium partial pressure of nitrogen dioxide is p.

100=\frac{(0.50)^2(0.25))}{p^2}100=p2(0.50)2(0.25))​

100=\frac{0.0625}{p^2}100=p20.0625​

On rearranging the above equation:

p^2=\frac{0.0625}{100}p2=1000.0625​

On taking square root to both sides:

p=\frac{0.25}{10}p=100.25​

p = 0.025 atm

So, the equilibrium partial pressure of NO_2NO2​ is 0.025 atm.