0.46 mol CH4, 0.64 mol C2H2, 0.92 mol H2

0.46 mol CH4, 0.64 mol C2H2, 0.92 mol H2

 

 

Methane, ethyne, and hydrogen

form the following equilibrium system:

2CH4(g) ⇌ C2H2(g) + 3H2(g)

At 1000℃, the equilibrium mixture was found to contain

0.46 mol methane, 0.64 mol ethyne,

and 0.92 mol hydrogen in a 4.0 L container.

Calculate the value of the equilibrium constant.

 

 

Methane, ethyne, and hydrogen

form the following equilibrium mixture.

2CH4(g) ⇌ C2H2(g) + 3H2(g)

While studying this reaction mixture,

a chemist analysed a 4.0 L sealed flask at 1700℃.

The chemist found 0.46 mol of CH4(g), 0.64 mol of C2H2(g),

and 0.92 mol of H2(g).

What is the value of Kc for the reaction at 1700℃?

 

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▶ 참고: 평형 상수식 쓰기

[ https://ywpop.tistory.com/7136 ]

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평형에서의 농도가 주어진 셈이므로,

평형 상수 계산 문제 중에서,

난이도 최하 문제.

 

 

 

> [CH4] = 0.46 mol / 4.0 L = 0.115 M

> [C2H2] = 0.64 mol / 4.0 L = 0.16 M

> [H2] = 0.92 mol / 4.0 L = 0.23 M

 

 

 

 

 

 

2CH4(g) ⇌ C2H2(g) + 3H2(g)

 

Kc = [C2H2] [H2]^3 / [CH4]^2

= [0.16] [0.23]^3 / [0.115]^2

= 0.1472

 

 

 

답: Kc = 0.15