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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: pkothari13 on April 03, 2005, 06:15:43 PM

Title: Kinetics Question
Post by: pkothari13 on April 03, 2005, 06:15:43 PM
I have two questions on Kinetics...

1) The reaction A-----> B + C is second-order in A. When [A]0 is .250 M, the reaction is 30.0% complete in 50.0 minutes.
         a) Calculate the value of the rate constant.
         b) Calculate the half-life for the reaction.

2) A first order reaction is 40.% complete at the end of 50. minutes. What is the value of the rate constant (in min-1)?


I thought of using half-life, but that's when it's 50% done.... so i dont know  ???
Title: Re:Kinetics Question
Post by: objectivist on April 03, 2005, 09:31:25 PM
I think I can help you with the second question.

ln(.40) = -kt

ln(.40) = -k(50)

solve for K. you shoud get 0.0183258
Title: Re:Kinetics Question
Post by: objectivist on April 03, 2005, 09:35:53 PM
Also, For 1 b)


ln(.30) = -kt

ln(.30) = -k(50)

k = 0.024079

half life = 0.693 divided by 0.024079

half life equals 28.78 minutes
Title: Re:Kinetics Question
Post by: objectivist on April 03, 2005, 09:41:09 PM
To sum up for ya..

the equation
 ln([A]1/[A0]) = -kt


is the natural log of the concentration ratio is equal to the negative constant times the time.

your questions gave you a percent. so you can put the percent in for the natural log thing. if they give you concentrations, you take the natural log of the concentrations divided by each other.

when you get "K' you can plug it into the equation half life = 0.693/k



FYI: just in case u dont know about ln (natural log), it is a button on a graphing calculator.
Title: Question (1)
Post by: Donaldson Tan on April 04, 2005, 05:58:24 AM
isn't half-life a PC game? just kidding.. LOL

Quote from: objectivist on April 03, 2005, 09:41:09 PM
the equation
 ln([A]1/[A0]) = -kt

Objectivist had given the reaction profile for a first-order reaction, not a 2nd order one.

the rate law is given by:
- dA/dt = k[A]2
-[A]-2 dA/dt = k
integrating both sides with respect to t,
[A]-1 = kt + C (integration constant)
using initial conditions when t = 0, [A] = [Ao] => C = [Ao]-1
hence, the reaction profile is given by:
[A]-1 - [Ao]-1 = kt

Quote
When [A]0 is .250 M, the reaction is 30.0% complete in 50.0 minutes.
when t = 50min, [Ao] = .250M, [A] = 70.0% x [Ao] = 0.175
0.175-1 - 0.250-1 = k(50)
k = 0.0343 dm3mol-1min-1

to find the half-life of the reaction, ie. when the reactants are 50% are used up, just plug in all the necessary values into the reaction profile equation, given you know k, [Ao], [A] ( = 50% x [Ao])
Title: Re:Kinetics Question
Post by: Donaldson Tan on April 04, 2005, 06:03:05 AM
Quote from: objectivist on April 03, 2005, 09:41:09 PM
when you get "K' you can plug it into the equation half life = 0.693/k

This equation is only valid for first-order reactions.
Title: Question (2)
Post by: Donaldson Tan on April 04, 2005, 06:08:56 AM
integrating the rate law -dA/dt = k[A], you will arrive at the reaction profile:
ln ([A]/[Ao]) = -kt

Quote
A first order reaction is 40.% complete at the end of 50. minutes
t = 50min
[A]/[Ao] = 100% - 40% = 60%
thus you can solve for k