Question

Consider the reaction below in a closed flask. At 200 o C, the equilibrium constant (Kp) is 2.40 × 103 . 2 NO (g)  N2 (g) + O2 (g) If 36.1 atm of NO (g) is added to the closed flask at 200 o C, what is the approximate partial pressure of O2 at equilibrium?

Answer 1

Explanation:

Since, the given reaction is as follows.

       $2NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g)$

Initial:    36.1 atm                 0          0

Change:    2x                      x           x

Equilibrium: (36.1 - 2x)       x            x

Now, expression for $K_{p}$ of this reaction is as follows.

            $K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}$

As the initial pressure of NO is 36.1 atm. Hence, partial pressure of $O_{2}$ at equilibrium will be calculated as follows.

              $K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}$

        $2.40 \times 10^{3} = \frac{x \times x}{(36.1 - 2x)^{2}}$

                 x = 18.1 atm

Thus, we can conclude that partial pressure of $O_{2}$ at equilibrium is 18.1 atm.

Answer 2

Answer:

partial pressure O2 = 17.867 atm

Explanation:

Step 1: Data given

Temperature = 200 °C

Kp = 2.40 *10^3

Pressure NO = 36.1 atm

Step 2: The balanced equation

2 NO ⇔ N2 + O2

Step 3: The initial pressure

pNO = 36.1 atm

pN2 = 0 atm

pO2 = 0 atm

Step 4: the pressure at the equilibrium

For 2 moles NO we'll have 1 mol N2 and 1 mol O2

pNO = 36.1 - 2X atm

pN2 = X atm

pO2 = X atm

Step 5: Calculate partial pressures

Kp = pN2 * pO2 / (pNO)²

2.40*10³ = x²/(36.1 - 2x)²

48.99 = x/(36.1-2x)

x = 1768.5 -97.98x

x = 17.867

x = partial pressure O2 = 17.867 atm