COBr2(g) ⇌ CO(g) + Br2(g) 73℃ Kc 0.190 2.00 L COBr2 0.0500 mol

COBr2(g) ⇌ CO(g) + Br2(g) 73℃ Kc 0.190 2.00 L COBr2 0.0500 mol

 

 

Carbonyl bromide decomposes to carbon monoxide and bromine.

COBr2(g) ⇌ CO(g) + Br2(g) Kc is 0.190 at 73℃.

If you place 0.0500 mol of COBr2 in a 2.00 L flask and heat it to 73℃,

what are the equilibrium concentrations of COBr2, CO, and Br2?

What percentage of the original COBr2 decomposed at this temperature?

 

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[COBr2] = 0.0500 mol / 2.00 L = 0.0250 M

 

 

 

COBr2(g) ⇌ CO(g) + Br2(g)

K = [CO] [Br2] / [COBr2]

( 참고 https://ywpop.tistory.com/7136 )

 

0.190 = (x) (x) / (0.0250 – x)

 

x^2 + 0.190x - (0.190)(0.0250) = 0

 

 

 

근의 공식으로 x를 계산하면,

( 참고 https://ywpop.tistory.com/3302 )

 

x = 0.0223669 ≒ 0.0224 M

 

 

 

[COBr2] = 0.0250 – 0.0224 = 0.0026 M

[CO] = [Br2] = 0.0224 M

 

 

 

[검산]

(0.0223669)^2 / (0.0250 – 0.0223669)

= 0.1899959 ≒ 190

 

 

 

percentage of COBr2 decomposed

= (분해된 농도 / 처음 농도) × 100

= (x / C) × 100

= (0.0224 / 0.0250) × 100 = 89.6%

 

 

 

 

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