0.200 mol of H2S is placed in an empty 1.00 L container

0.200 mol of H2S is placed in an empty 1.00 L container

 

 

The equilibrium constant, K, is 4.20×10^(-6)

for the following reaction:

2H2S(g) ⇌ 2H2(g) + S2(g)

Find the equilibrium concentrations of all entities

when 0.200 mol of H2S(g) is placed in a 1.00 L container.

 

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

 

[H2S] = 0.200 mol / 1.00 L = 0.200 M

 

 

 

ICE 도표를 작성하면,

 

............... 2H2S(g) ... ⇌ ... 2H2(g) . + . S2(g)

초기(M): . 0.200 ............... 0 ............... 0

변화(M): . –2x .................. +2x ........... +x

평형(M): . 0.200–2x ......... 2x ............. x

 

 

 

K = [H2]^2 [S2] / [H2S]^2

 

4.20×10^(-6) = [(2x)^2 (x)] / (0.200–2x)^2

---> 이 식을 정리하면, 3차 방정식이 나온다.

 

 

 

Calculate the equilibrium concentrations for the following reaction

when the 0.200 mol of H2S is placed in an empty 1.00 L container.

(This one will be hard to solve by hand:

use an online tool once you set up the expression.)

2H2S(g) ⇌ 2H2(g) + S2(g) K = 4.20×10^(-6) at 1100 K

 

 

 

필자는 3차 방정식을 풀 능력이 없다.

이 때문에, 어쩔 수 없이,

0.200–2x ≒ 0.200 이라 근사처리해서,

근사값이라도 풀어보겠다.

 

4x^3 / (0.200)^2 = 4.20×10^(-6)

 

x = [(4.20×10^(-6)) (0.200)^2 / 4]^(1/3)

= 0.003476 M

 

 

 

근사값:

[H2S] = 0.200 – 2x

= 0.200 – 2(0.003476) = 0.193048 M

 

[H2] = 2x

= 2(0.003476) = 0.006952 M

 

[S2] = x = 0.003476 M

 

 

 

 

[검산]

K = [0.006952]^2 [0.003476] / [0.193048]^2

= 4.51×10^(-6)

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/12929 ] A 1.00 L reaction container initially contains 9.28×10^(-3) moles of H2S

 

 

 

[키워드] 2H2S(g) ⇌ 2H2(g) + S2(g)_ICE table, 3차 방정식 기준문서