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Post time 14-5-2009 02:54 PM | Show all posts |Read mode
Slm all..first tym masuk n post kat board ni. asalnya malas nak bukak thread baru tp aku tgk board soklan kat atas tu mcm xmelibatkan soklan pun so aku terpaksala bukak satu.

Ok aku ada satu problem chemistry in gas. dah try tp xterjwb..berkaitan Dalton's partial pressure:

C2H4 + H2 --> C2H6

Ok problemnya:

Sebelum reaction

Ethylene n H2 tu dlm satu balang, total pressure 52 torr. bilangan mole H2 lagi byk dari ethylene.


Selepas reaction

Sume still dlm balang yg sama, total pressure 34 torr. Volume n temperature tak berubah.

Soklannye berapa mole fraction ethylene dlm original mixture (sebelum reaction)?

Harap ada leh bantu. TQ
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Post time 15-5-2009 12:44 AM | Show all posts
So, ethylene limiting reagent.

Since PV=nRT, n is directly proportional to P. So k*18mol ethylene reacted. Since mule2 ade 52torr gas, so ade k*52mol gas.

So, mole fraction of ethylene initially = 18/52.
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 Author| Post time 15-5-2009 12:01 PM | Show all posts
Thanx try jwb.

Tp answer according to buku 0.48.

Plus aku xbpe paham camne kau dpt 18 tu?
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Post time 15-5-2009 03:11 PM | Show all posts
Originally posted by StealthTuah at 15-5-2009 12:01 PM
Thanx try jwb.

Tp answer according to buku 0.48.

Plus aku xbpe paham camne kau dpt 18 tu?


Alamak, salah pulak.. Hak3... Ok2, aku try buat detailed sket...

Katekan gas tu ideal, so PV=nRT. So n=(V/RT)*P, write n=k*P, where k is a constant. Since mule2 ade 52torr gas dlm balang tu, so ade 52k mol of ethylene + hydrogen.

In the end, tgl 34 torr je, so ade k*34 mol hydrogen+ethane.

Dlm reaction ni, ethylene limiting reagent, so sume ethylene habis bertindak balas. Dlm kes ni, boleh set up simultaneous equation. Katekan mule2 ade n mol ethylene dan m mol hydrogen. So

Bil. mol ethylene + Bil. mol hydrogen = 52k
n + m = 52k

Since sume ethylene habis tindak balas, n mol hydrogen pun bertindak balas. Also, from stoichiometry, n mol ethane dihasilkan. So,

Bil. mol hydrogen + Bil. mol ethane = 34k
(m - n) + n = 34k
m = 34k

So, n = 18k

So, mole fraction of ethylene initially = 18k/52k = 0.35


So, jwpn ni slh. Hrp sape2 leh tlg tgk mne yg aku salah..
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Post time 15-5-2009 03:16 PM | Show all posts
Btw, ethylene, hydrogen, ngan ethane dlm soklan tu sumenye gas ke?

Klu x, gerenti r aku slh. Psl aku kire sumenye gas...
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 Author| Post time 20-5-2009 01:01 PM | Show all posts

Reply #5 aku_EnSeM's post

Ok2 aku paham dah..betul jugak jln kerja kau tu..n sume gas

tak taulak aku kalo2 jwpn buku tu salah tp.. satu lagi soklan payah jwpn yg aku dpt sama lak..

TQ2 anyway


EDIT: sori2 jwpn kau betul..dah cross check ngan org lain..TQ bbyk

[ Last edited by  StealthTuah at 20-5-2009 13:04 ]
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Post time 20-5-2009 08:18 PM | Show all posts
bijaknyer...hoyeayyyyyy
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