Question

A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by the equation 2NO2 ⇌ N2O4. The partial pressure of NO2 is 61.2 MPa and the partial pressure of N2O4 is 38.8 MPa. Several weights are added to the lid of the container, increasing the total pressure on the container to 200 MPa. What are the most likely partial pressures when equilibrium is established again?3.

Answer 1

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = $\frac{38.8}{61.2^{2} }$

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

0.0104 = $\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}$

0.0104$[P(NO_{2} )]^{2}$ + $P(NO_{2} )$ - 200 = 0

Resolving the second degree equation:

$P(NO_{2} )$ = $\frac{-1+\sqrt{9.32} }{0.0208}$

$P(NO_{2} )$ = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are $P(NO_{2} )$ = 98.7 MPa and P(N₂O₄) = 101.3 MPa