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### Topic: Rate Law Problems  (Read 20094 times)

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#### Joules23

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##### Rate Law Problems
« on: February 01, 2008, 04:47:37 PM »
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7) A compound decomposes by a first-order reaction. The concentration of compound decreases from 0.1124 M to 0.0905 M in 5.2 min. What fraction of the compound remains after 7.6 min?

Using the first order integration law i got K=6.946e-4 (used 312sec for t)
Then again i used integrated law to find [A] and i used 456 sec for t..
I got [A]=.08189
When it asks "What fraction of the compound remains .."  do i just
[A]/.1124 = remaining fraction?
.08189/.1124 =.7285

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10) Experimental data for the reaction below have been plotted in the following three different ways (with concentration units in mol/L).
A  2 B + C

(a) What is the concentration of A after 7 s?
.0125 M

(b) What are the first three half-lives for this experiment?
first half-life
s
second half life
s
third half-life
s

I need help with part (b).. This is what im doing

t1/2=1/([A]ok)
k=slope=10
[A]o for first reaction = (1/20)= .05
1/(.05x10) = 2 .. but thats incorrect?
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« Last Edit: February 01, 2008, 05:33:20 PM by Joules23 »

#### Yggdrasil

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##### Re: Rate Law Problems
« Reply #1 on: February 01, 2008, 09:08:48 PM »
7 is correct.  10a is correct.

For 10b, your calculation for [A]o is wrong.  You calculated the concentration of A when t=1.  You want [A] when t=0.

#### Joules23

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##### Re: Rate Law Problems
« Reply #2 on: February 01, 2008, 11:26:19 PM »
Okkkk.. so first half life.. [A]o = (1/10)=.1
second half life.. [A]o = (1/20)=.05
third half life..[A]o = (1/30)=.0333..

plug those values into the second order half life formula?

#### Yggdrasil

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##### Re: Rate Law Problems
« Reply #3 on: February 02, 2008, 08:13:03 AM »
Third half life [A]o is wrong.  After one half life, you have 1/2 left.  After two half lives, you have (1/2)*(1/2) = 1/4 left.