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

Answer:

$p_{N_2}^{eq}=p_{O_2}^{eq}}=18.27atm$

Explanation:

Hello,

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

$Kp=\frac{p_{N_2}^{eq}p_{O_2}^{eq}}{(p_{NO}^{eq})^2}$

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

$Kp=\frac{(x)(x)}{(37.30-2x)^2}=2400$

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.

Answer 2

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