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Question

At 298 K and 1 atm pressure, the partial pressures in an equilibrium mixture of N2O4 and NO2 are 0.7 and 0.3 atmosphere respectively. What will be the partial pressure of NO2 when the gases are in equilibrium at 298 K and at a total (equilibrium) pressure of 10 atmospheres?


A

1.078 atm

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B

2.23 atm

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C

0.77 atm

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D

1.60 atm

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Solution

The correct option is A

1.078 atm


The temperature is maintained constant which means that Kp does not change.

When the total pressure is 1 atmosphere,

N2O4 2NO2

Partial pressure at equlibrium 0.7 atm, 0.3 atm.

Kp=(PNO2)2PN2O4=0.3×0.30.7=0.1285

Note: you can see that Kp is small.

Now derive Kp in terms of α and pressure P. We derived the value of Kp from given partial pressures for equilibrium at 1 atm.

Let us assume there were initially n moles of N2O4 (g) and there were no products.

N2O4(g) NO2(g)
t=0 1 0
t=eqm. (1α) 2α

At Equilibrium, total number of moles = (1 + α)

Partial pressure of NO2 (g) = mole fraction of 2NO2 (g) × Total pressure
=2x1+x×10

Partial pressure of N2O4 (g) = mole fraction of N2O4 (g) x Total Pressure P
=1x1+x×10
Assume initially 1 mole of N2O4 is alone present. At equilibrium total pressure P, we have
Kp = 0.1285 = 4Pα21α2×10
(this should be derived)
For α21α2 to be this small, a has to be very small as well. When α is that small, we can approximate 1 - α21

Solving for a, we get α = 0.056

PNO2 = mole fraction of NO2 × Total Pressure P
= 2α1+α 10 atm = 1.078 atm.


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