N2O4 1.588 g 298 K 500 cm3 1.0133 bar N2O4 NO2 mol fraction

N2O4 1.588 g 298 K 500 cm3 1.0133 bar N2O4 NO2 mol fraction

 

 

Nitrogen tetroxide is partially dissociated in the gas phase

according to the reaction N2O4(g) ⇌ 2NO2(g)

A mass of 1.588 g of N2O4 is placed in a 500 cm3 glass vessel

at 298 K and dissociates to an equilibrium mixture at 1.0133 bar.

a) What are the mole fractions of N2O4 and NO2?

b) What percentage of the N2O4 has dissociated?

Assume that the gases are ideal.

 

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N2O4의 몰질량 = 92.01 g/mol 이므로,

1.588 g / (92.01 g/mol) = 0.017259 mol N2O4

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

N2O4의 초기압력을 계산하면,

P = nRT / V

= (0.017259 mol) (0.083145 bar•L/mol•K) (298 K) / (0.5 L)

= (0.017259) (0.083145) (298) / (0.5)

= 0.85526 bar

 

 

 

ICE 도표를 작성하면,

 

................N2O4(g) ... ⇌ ... 2NO2(g)

초기(bar): 0.85526 ............ 0

변화(bar): –x ..................... +2x

평형(bar): 0.85526–x ........ 2x

 

 

 

평형에서,

혼합물의 압력 = 1.0133 bar 이므로,

(0.85526–x) + 2x = 1.0133

 

x = 1.0133 – 0.85526 = 0.15804 bar

 

 

 

0.85526–x = 0.85526 – 0.15804 = 0.69722 bar N2O4

 

2x = 2 × 0.15804 = 0.31608 bar NO2

 

 

 

a) What are the mole fractions of N2O4 and NO2?

> X_N2O4 = 0.69722 / 1.0133 = 0.6881

> X_NO2 = 0.31608 / 1.0133 = 0.3119

 

 

 

b) What percentage of the N2O4 has dissociated?

(0.15804 / 0.85526) × 100 = 18.48%

 

 

 

 

[키워드] N2O4(g) ⇌ 2NO2(g) 기준문서, N2O4(g) ⇌ 2NO2(g) ice 기준문서