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Physical Chemistry

Blazing goIITian

Joined:	15 Jul 2008
Post:	315

15 Nov 2009 16:30:03 IST

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Show that for a first order reaction, time required 99.9% of reaction to complete is 10 times the t

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Show that for a first order reaction, time required 99.9% of reaction to complete is 10 times the time required for completion of half reaction.

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jee-do or die

Blazing goIITian

Joined: 15 Aug 2008

Posts: 585

15 Nov 2009 16:43:03 IST

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k = 2.303 /t . log [ a-x ] / [ a] ----A

where k is rate of reaction

t ----time taken

a ----initial concentration

x -- reactants those converted into products

we know that

t1/2 = half life

k = 0.693/t50%------1

u can derive ---1 by putting x = a/2 in ---A

for 99% reaction to be completed we have a = 100

x =99.9

from ---A

k = 2.303/t . log [ 100] / [ 100-99.9]

k = 2.303 /t 99% . log [ 100]/[ 0.1]

k = 2.303/t99% . log [ 10^3]

k = [2.303 . 3] / t99%----2

equate 1 and 2

we have k = 0.693/ t50% = 2.303 * 3 / t99%

t99% / t50% = 2.303 * 3 / 0.693 = 9.97 = 10[ approx]

STONE COLD RATTLESNAKE

Blazing goIITian

Joined: 15 Jul 2008

Posts: 315

15 Nov 2009 16:50:05 IST

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Oh nice but is there any other easy method

jee-do or die

Blazing goIITian

Joined: 15 Aug 2008

Posts: 585

15 Nov 2009 16:53:03 IST

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I THINK THIS IS SHORTEST POSSIBLE ONE

BUT I WILL SEE WHETHER EXPERTS HAVE ANY SHORTER VERSION OF THIS

STONE COLD RATTLESNAKE

Blazing goIITian

Joined: 15 Jul 2008

Posts: 315

15 Nov 2009 16:58:23 IST

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For a first order reaction, the time required for 99% completion is

jee-do or die

Blazing goIITian

Joined: 15 Aug 2008

Posts: 585

15 Nov 2009 17:00:55 IST

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t99%= 2.303 * 3 / K

STONE COLD RATTLESNAKE

Blazing goIITian

Joined: 15 Jul 2008

Posts: 315

15 Nov 2009 17:02:36 IST

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The time required for 10% completion of a first order reaction at 298 K isequal to that required for its 25% completion at 308 K. If the value of A is4 × 10^10 s−1. Calculate k at 318 K and Ea.

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