Question

For the equilibrium: 2 NO (g) <----> N2(g) + O2 (g), Kp=2400. If initially, only NO is present at a partial pressure of 37.30 atm, what will the partial pressures of N2 and O2 be at equilibrium?

1827 atm
38.08 atm
1.725 atm
36.55 atm

Answer 1

The reaction is 2 NO (g) <----> N2(g) + O2

partial pressures

Initial 37.30 0 0

Change -2p +p +p

Equilibrium 37.30-p p p

Kp = pN2 X pO2 / (pNO)^2

2400 = p^2 / (37.30-p)^2

3339096 - 179040p + 2400p^2 = p^2

2399p^2 + 3339096 -179040 p = 0

On solving

p = 36.55atm

Thus partial pressure of N2 and O2 = 36. 55 atm

Answer 2

`p_(N_2)^(eq)=p_(O_2)^(eq)}=18.27atm`

Step-by-step explanation:

Hello,

In this case, for the given chemical reaction, the law of mass action turns out:

`Kp=(p_(N_2)^(eq)p_(O_2)^(eq))/((p_(NO)^(eq))^2)`

Whereas the equilibrium pressures, based on the stoichiometry and the change
`x`, changes to:

`Kp=((x)(x))/((37.30-2x)^2)=2400`

Solving for
`x` via quadratic equation, one obtains:

`x_1=18.27atm\\x_2=18.84atm`

In such a way, the feasible solution is 18.27 atm since the other pressure lead to a negative pressure of NO at the equilibrium, therefore, the equilibrium pressures of nitrogen and oxygen that are equal, result in:

`p_(N_2)^(eq)=p_(O_2)^(eq)}=x_1=18.27atm`

Best regards.