Chemical equilibrium ... Help

2010-11-28 8:58 pm

1)Hydrogen sulphide (H2S) decomposes according to the following reaction, for which Kc= 9.3 x 10^-8 at 700oC:
2H2S(g) ⇌ 2H2(g) + S2(g)

if 0.45 mole of H2s is placed in a 3.0L container, what is the equilibrium concentration at 700oC?

2)Compound A decomposes according to the equation
A(g) ⇌ 2B(g) + C(g)

A sealed 1.00-L reaction vessel initially contains 1.75 x 10^-3 mole of A,1.25 x 10^-3 mole of B and 6.50 x 10^-4 mole of C at 100oC. WHen equilibrium is reached, the concentraion of A is 2.15 x 10^-3 M. What are the equilibrium concentraions of B and C?

回答 (1)

用戶26240

2010-11-29 1:59 am

✔ 最佳答案

1)
2H2S(g) ⇌ 2H2(g) + S2(g)

Let y mol dm⁻³ be the equilibrium concentration of S2.
Since Kc is very small, thus assume that 0.45 >> y.

At equilibrium:
[H2S] = 0.45 - 2y ≈ 0.45 mol dm⁻³
[H2] = 2y mol dm⁻³
[S2] = y mol dm⁻³

Kc = [H2]²[S2]/[H2S]²
(2y)²(y)/(0.45)² = 9.3 x 10⁻⁸
y = 1.68 x 10⁻³
2y = 3.36 x 10⁻³

At equilibrium:
[H2S] = 0.45 mol dm⁻³
[H2] = 3.36 x 10⁻³ mol dm⁻³
[S2] = 1.68 x 10⁻³ mol dm⁻³

2)
A(g) ⇌ 2B(g) + C(g)

Increase in concentration of A = (2.15 x 10⁻³) - (1.75 x 10⁻³) = 4 x 10⁻⁴ mol dm⁻³

Equilibrium concentration of B = (1.25 x 10⁻³) - 2(4 x 10⁻⁴) = 4.5 x 10⁻⁴ mol dm⁻³
Equilibrium concentration of C = (6.50 x 10⁻⁴) - (4 x 10⁻⁴) = 2.5 x 10⁻⁴ mol dm⁻³

2010-11-30 00:24:41 補充：
If the initial concentration of C is 6.5 x 10⁻³ mol dm⁻³ (but NOT 6.5 x 10⁻⁴ as you have typed), the equilibrium concentration
[C] = (6.5 x 10⁻³) - (4 x 10⁻⁴) = 6.1 x 10⁻³ mol dm⁻³

Check your question.

參考: 土扁, 土扁

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