본문 바로가기
일반화학/[15장] 화학 평형

2NOBr(g) ⇌ 2NO(g) + Br2(g) density 4.495 g/L 4.086 g/L

by 영원파란 2022. 10. 22.
반응형

2NOBr(g) ⇌ 2NO(g) + Br2(g) density 4.495 g/L 4.086 g/L

 

 

A sample of gaseous nitrosyl bromide, NOBr,

was placed in a container fitted with a frictionless, massless piston,

where it decomposed to form nitrogen monoxide and bromine.

2NOBr(g) ⇌ 2NO(g) + Br2(g)

 

The initial density of the system was recorded to be 4.495 g/L.

After equilibrium was reached, the density was noted to be 4.086 g/L.

Determine the value of the equilibrium constant for the reaction.

 

If Argon gas is added to the system at equilibrium at constant temperature,

what will happen to the equilibrium position?

What happens to the value of K?

 

---------------------------------------------------

 

밀도_1 / 밀도_2

= (m / V_1) / (m / V_2)

( 참고: 질량 보존의 법칙에 따라, 반응물의 질량 = 생성물의 질량 )

 

= V_2 / V_1

= n_2 / n_1

( 참고: 아보가드로의 법칙 https://ywpop.tistory.com/1979 )

 

= (4.495 g/L) / (4.086 g/L)

= 1.100

---> 초기/평형 밀도 비로부터, 평형/초기 몰수 비 계산.

 

 

 

n_NOBr = W / M

= 4.495 g / (109.91 g/mol)

= 0.040897 mol

= 0.04100 mol

 

 

 

ICE 도표를 작성하면,

 

................. 2NOBr(g) . ⇌ . 2NO(g) . + . Br2(g)

초기(mol) . 0.041 .............. 0 ................ 0

변화(mol) . –2x ................. +2x ............. +x

평형(mol) . 0.041–2x ........ 2x ............... x

 

 

 

n_2 / n_1 = (0.041 – 2x + 2x + x) / 0.041 = 1.100

 

0.041 + x = 1.100 × 0.041

 

x = (1.100 × 0.041) – 0.041 = 0.0041 mol

 

 

 

K = (x) (2x)^2 / (0.041 – 2x)^2

= (0.0041) (2×0.0041)^2 / (0.041 – 2×0.0041)^2

= 0.000256

 

 

 

답: K = 0.000256

 

 

그리드형(광고전용)

댓글1