Question

Nitric oxide is formed in automobile exhaust when nitrogen and oxygen in air react at high temperatures. N2(g) + O2(g) 2NO(g) The equilibrium constant Kp for the reaction is 0.0025 at 2127°C. If a container is charged with 8.00 atm of nitrogen and 5.00 atm of oxygen and the mixture is allowed to reach equilibrium, what will be the equilibrium partial pressure of nitrogen?

Answer 1

Answer:

The partial pressure of nitrogen, at the equilibrium is 7.846 atm

Explanation:

Step 1: The balanced equation

N2(g) + O2(g) 2NO(g)

Step 2: Data given

Kp = 0.0025 at 2127 °C

pressure of nitrogen = 8.00 atm

Pressure of oxygen = 5.00 atm

Step 3: ICE chart

Kp = (P(NO))² / (P(N2)*P(O2))   = 0.0025

The initial pressure N2 is 8 atm, there will react X atm. The final pressure will be 8-X atm

The initial pressure O2 is 5 atm, there will react X atm. The final pressure will be 5-X atm

The initial pressure NO is 0, There will be created 2X atm. The final pressure is 2X atm

So Kp = 0.0025 = (2X)² / ((8-X)(5-x))

0.0025( 40-13X + X²) = 4X2

3.9975X² + 0.0325 -0.1 = 0

X = 0.154

Step 4: Calculate partial pressure of nitrogen

P(N2) = 8-X = 8 - 0.154 atm = 7.846 atm

The partial pressure of oxygen is 5- 0.154 atm = 4.846 atm

The partial pressure of NO is 2*0.154 atm = 0.308 atm

To control this we can calculate the Kp ( =0.0025)

Kp = 0.308² /(7.846 *4.846) = 0.0025

The partial pressure of nitrogen, at the equilibrium is 7.846 atm