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Topic: Kinetics Question  (Read 8240 times)

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pkothari13

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Kinetics Question
« 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  ???
« Last Edit: April 03, 2005, 07:44:53 PM by pkothari13 »

objectivist

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Re:Kinetics Question
« Reply #1 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

objectivist

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Re:Kinetics Question
« Reply #2 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

objectivist

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Re:Kinetics Question
« Reply #3 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.

Offline Donaldson Tan

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Question (1)
« Reply #4 on: April 04, 2005, 05:58:24 AM »
isn't half-life a PC game? just kidding.. LOL

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])
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline Donaldson Tan

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Re:Kinetics Question
« Reply #5 on: April 04, 2005, 06:03:05 AM »
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.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline Donaldson Tan

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Question (2)
« Reply #6 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


"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

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