Nitrogen dioxide is one of the many oxides of nitrogen (often form another form of NOx, dinitrogen tetroxide A chemical engineer studying this reaction fils a 500. mL flask at 7.9 °C with 4.9 atm of nitrogen dioxide gas. He thèn raises the temperature considerably and when the mixture has come to equilibrium determines that it contains 2.7 atm of nitrogen dioxide gas The engineer then adds another 1.2 atm of nitrogen dioxide, and allows the mixture to come to equilibrium again. Calculate the pressure of dinitrogen tetroxide after equilibrium is reached the second time. Round your answer to 2 significant digits

Kp = 0.15

1.48 atm

Explanation:

The equilibrium in question is

2 NO₂ (g)   ⇄   N₂O₄ (g)

and we need to determine Kp which equals pN₂O₄ / pNO₂².

We know that at equilibrium we have 2.7 atm of NO₂ and originally we had 4.9 atm of NO₂. Thus the amount of NO₂ gas  reacted was (4.9 - 2.7) atm. Thus, at equilibrium we have

pNO₂ = 2.7 atm

pNO₂ reacted = (4.9 - 2.7) atm = 2.2 atm

Calculating the pressure N₂O₄ produced:

(1 atm N₂O₄ / 2 atm NO₂) x 2.2 atm NO₂ = 1.1 atm N₂O₄

We can now calculate Kp:

Kp = pN₂O₄ / pNO₂² = 1.1 / 2.7² = 0.15

For the second part we can tell some of the added NO₂ will be consumed according to LeChatelier´s principle to restore equilibrium, therefore we can  formulate the following relation:

Kp = (1.1 + x ) / ( 3.90 - 2x)² = 0.15

After doing some algebra we are going to have the quadratic equation:

1.1 + x = .15 ( 3.9 - 2x)²

1.1 + x = .15 ( 15.2 - 15.6 + 4x²)

1.1 + x = 2.28 - 2.34 x  +0.60 x²

0 = 1.18 -3.34 x +0 .60 x²

After solving this quadratic equation we get two roots 5.18  and 0.38. The first one is a physically impossible, the other is 0.38 .

Therefore the pressure of dinitrogen tetraoxide after equilibrium is reached the second time is 0.38 atm  + 1.1 atm = 1.48 atm

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