查看完整版本 : chemical equilibrium and Beer law

peter555 2012-12-17 09:12 PM

chemical equilibrium and Beer law

唔該樓下的解答了我不懂的題目:):loveliness:
剛剛又遇上兩題不懂既題目 希望大家都可以幫幫我>_<
THX:loveliness:

Beer Law (我唔知點解計多左個位-_-)
[IMG]http://i248.photobucket.com/albums/gg189/peter2981/DSC_0480_zps303fa526.jpg[/IMG]

chemical equilibrium
[IMG]http://i248.photobucket.com/albums/gg189/peter2981/DSC_0481_zps884a2173.jpg[/IMG]

唔該>_<!!

[[i] 本帖最後由 peter555 於 2012-12-18 10:59 AM 編輯 [/i]]

jmlo 2012-12-17 11:37 PM

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7(b)
Note that each mole of Ba(OH)2 gives two moles of OH-
So [OH-] = 2*0.02 = 0.04M
pH = 14 - pOH = 14 - (-log (4 x 10^-2)) = 14 - 1.40 = 12.60

jmlo 2012-12-18 03:38 AM

回覆 1# 的帖子

3(a)
Amount of NaOH used in neutralization = 5*0.0666 = 0.333 mol
Hence, the amount of unreacted acid = 0.333 mol
Amount of unreacted ethanol = 0.333 mol
Therefore, the amount of ester present in the solution at eqm = 0.667 mol
Similarly, the water formed from esterification = 0.667 mol

The eqm constant is thus equal to
Kc = [ester][water]/[acid][ethanol]
= (0.667/V)(0.667/V)/(0.333/V)(0.333/V)
= 4.01

FYI: The actual volume V of the solution is not important as they will be cancelled out during calculation.

3(b)
We know that
            CH3COOH  +  C2H5OH   <----->   CH3COOC2H5 + H2O
Initial        1.00             0.05                           0                 0
Change       -x                -x                            +x               +x
Eqm          1-x              0.05-x                         x                 x

Therefore,
Kc = 4.01 = (x/V)(x/V)/((1.00-x)/V)((0.05-x)/V)
3x^2 - 6x + 2 = 0
x = 0.4226 or 1.5774 (rejected as x must be smaller than 0.5)

The molar mass of CH3COOC2H5 = 88 g/mol
Amount of CH3COOC2H5 at eqm = 0.4226 mol
The mass of CH3COOC2H5 at eqm = 0.4226*88 = 37.2 g

[[i] 本帖最後由 jmlo 於 2012-12-18 06:42 AM 編輯 [/i]]

peter555 2012-12-18 10:56 AM

:smile_44: 很厲害呀 樓上的
很強 唔該哂>_<

jmlo 2012-12-18 12:58 PM

回覆 1# 的帖子

Q4. I also got an answer which is 10 times larger than the given answer. :smile_27:

Q. Chemical equilibrium
The same approach can be used to solve for the practice question. The only thing you need to know is the atmospheric pressure at 1525m. This can be done by using the barometric formula which relates atmospheric pressure and altitude. There exists such online calculators. An example is:
[url]http://planetcalc.com/938/[/url]

When you plug in P (sea level) = 760 mmHg, T = 25C, H = 1525m, you obtain P(at 1525m) = 638 mmHg, which is 0.8394 atm. Using this you can find the partial pressure of O2 and calculate [O2].

peter555 2012-12-18 01:06 PM

回覆 5# 的帖子

:smile_44: 可能錯左
阿Sir好似講過好似少左個位-_-
至於另一題
我試下計先
THX :)!!!
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查看完整版本: chemical equilibrium and Beer law