At 200 °C, the equilibrium constant (Kp) for the reaction belowis 2.40 × 103.
2NO (g) N2 (g) + O2 (g)
A closed vessel is charged with 36.1 atm of NO. At equilibrium, thepartial pressure of O2 is atm.

A. 35.7
B. 6.00
C. 294
D. 1.50 × 10-2
E. 18.1

The partial pressure of O2 is 17.9 ( closest to 18.1 atm) option E

Explanation:

Step 1: Data given

Temperature = 200 °C

the equilibrium constant (Kp) = 2.40 * 10³

Pressure NO = 36.1 atm

Step 2: The balanced equation

2 NO ⇔ N2 + O2

Step 3: The initial pressure

NO = 36.1  atm

N2 = 0 atm

O2 = 0 atm

Step 4: Calculate pressure at the equilibrium

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

NO = (36.1  - 2x)

N2 = x atm

O2 = x atm

Step 5: Calculate the partial pressure of O2

Kp = (pN2)*(pO2) / (pNO)²

2400 = (x²) / (36.1  - 2x)²

√(2400) = x / (36.1 - 2x)

x = 48.99 *(36.1 - 2x)

x = 1768.5 - 97.98x

98.98x = 1768.5

x = 17.9

The partial pressure of O2 is 17.9 ( closest to 18.1 atm) option E