The chemical equilibrum

[匿名]

2013-10-22 3:30 am

2.Assume that tartaric acid (molar mass = 150.0 g mol-1) is the only acid present in a bottle of wine.The acid is a weak acid and its dissociation can be represented
by the following equation:
RCOOH(aq)to form RCOO-(aq) + H+(aq) it is a reversible reaction.
Calculate the pH of the wine if the bottle of wine has a total acidity content of 7.00 g dm-3.

3. N2O4(g) undergoes decomposition to give NO2(g) according to the following equation:
N2O4(g)to form 2NO2(g) it is a reversible reaction.

The equilibrium constant,Kc,for the decomposition is 1.51 mol dm-3 at 127C.
0.500mole of N2O4(g) was placed in a 4.00dm3 reaction vessel and heated to 127C. What was the percentage decomposition of N2O4(g) at equilibrium?

更新1:

sorry啊,忘記了第二題本身是提供了Kc Kc= 1.00 *10^-3 mol dm-3

回答 (1)

用戶4366

2013-10-22 7:21 am

✔ 最佳答案

2. RCOOH(aq) <===> RCOO-(aq) + H+(aq)

[RCOOH] = 7/150 = 0.0467 M

Without Ka of the acid, no further calculation is possible.

It is found from the web that Ka1 = 9.2 X 10-4

Let :
[RCOOH] = 0.0467 - x
[RCOO-] = x
[H+] = x

X²/( 0.0467 - x) = 9.2 X 10⁻⁴
X² + 9.2 x 10⁻⁴x - 4.2964X10⁻⁵ = 0
x = 0.0122 M
pH = 1.9

I don't know what's wrong with your question.
A weak acid solution with pH = 1.9 is just unbelievable.

Sorry!

2013-10-21 23:28:33 補充：
一個飲品的 pH 應該在 5 至 7 之間。

2013-10-22 00:00:09 補充：
那樽並不是酒，是「強水」，完全 (無尼頭) 的題目，唔想計啦！

2013-10-24 00:56:28 補充：
9.2 X 10⁻⁴ 和 1.00 X 10⁻³ 差別不大，結果會差不多一樣，解題方法是相同的。

2013-10-24 01:07:20 補充：
N₂O₄ === 2NO₂
0.5(1-α) .. (2)(0.5)α ....... no. of mole

[N₂O₄] = (0.5)(1-α)/4
[NO₂] = (2)(0.5)α/4 = α/4

Kc = [NO₂]² / [N₂O₄] = 1.51

(α/4)² / (0.5)(1-α)/4 = 1.51

α² + 3.02α - 3.02 = 0

α = 0.79 or α = -3.812 (reject <0)

79% dissociated

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