A 250 ml flask containing NO ( g ) at 0.46 atm is connected to a 100 ml flask containing oxygen gas at 0.86 atm by means of a stop cock at 350 K. the gases are mixed by opening the stop cock where the following equilibrium is established. undersetg2 NO + O 2 longrightarrow undersetg2 NO 2 leftharpoons undersetgN 2 O 4 The first reaction is complete while the second is at equilibrium. If the total pressure is 0.37 atm , K p is

Question

A 250 ml flask containing NO ( g ) at 0.46 atm is connected to a 100 ml flask containing oxygen gas at 0.86 atm by means of a stop cock at 350 K. the gases are mixed by opening the stop cock where the following equilibrium is established. undersetg2 NO + O 2 longrightarrow undersetg2 NO 2 leftharpoons undersetgN 2 O 4 The first reaction is complete while the second is at equilibrium. If the total pressure is 0.37 atm , K p is (A) 0.645 atm -1 (B) 1.290 atm -1 (C) 20 atm -1 (D) 5.46 atm -1

Tardigrade Admin · Accepted Answer

Partial pressure of gases before reaction

NO = 0.46 \times \frac{250}{350} = 0.3285 a t m

O_{2} = 0.86 \times \frac{100}{350} = 0.2457 a t m

Here, NO is limiting reagent:
Partial pressure after reaction
Ist step:

NO = 0, O_{2} = 0.0814,

N O_{2} = 0.3285 a t m

Now, let

P

be the decrease in partial pressure of

N O_{2}

due to dimerization

0.37 = 0.0814 + 0.3285 - P + P /2

P = 0.0798

K_{P} = \frac{P _{N_{2} O_{4}}}{P _{N O_{2}}^{2}} = (\frac{P}{2}) \frac{1}{( 0.3285 - P ) ^{2}}

= \frac{0.0399}{( 0.2487 ) ^{2}} = 0.645 a t m^{- 1}

$0.645 a t m^{- 1}$

$1.290 a t m^{- 1}$

$20 a t m^{- 1}$

$5.46 a t m^{- 1}$

0.645atm−1

1.290atm−1

20atm−1

5.46atm−1

$0.645 a t m^{- 1}$

$1.290 a t m^{- 1}$

$20 a t m^{- 1}$

$5.46 a t m^{- 1}$