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Topic: Order of reaction  (Read 7177 times)

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Offline xpressmusic

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Order of reaction
« on: May 05, 2011, 12:37:56 PM »
HI could anyone help me in solving this...

In the late 19th century the two pioneers of the study of reaction kinetics, Vernon Harcourt
and William Esson, studied the rate of the reaction between hydrogen peroxide and iodide
ions in acidic solution.
H2O2 + 2I– + 2H+ 2H2O + I2
This reaction is considered to go by the following steps.
step 1 H2O2 + I– --->IO– + H2O
step 2 IO– + H---> HOI
step 3 HOI + H+ + I– --->I2 + H2O
The general form of the rate equation is as follows.

rate = k[H2O2]^a [I–]^b [H+]^c

2(b)Suggest values for the orders a, b and c in the rate equation for each of the following
cases.
numerical value a b c
step 1 is the slowest overall
step 2 is the slowest overall
step 3 is the slowest overall
Explain also plzzz

Offline Schrödinger

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Re: Order of reaction
« Reply #1 on: May 09, 2011, 04:14:28 PM »
Let me give you the following hints :

1 .
The slowest overall step determines the rate of the reaction. Hence, if the slowest step (also rightly called the RDS, Rate Determining Step) is of the form :

A + P ---> C + Q
then rate = k[A][P]

2.
If step 1 is the RDS :

Rate = k[H2O2]1[I-]1

3.
Those steps that are not the RDS are considered to be so fast that they are in equilibrium. Hence, you can use the equilibrium expression to express terms such as [IO-], [HOI] in terms of the reactants present in the original question. Just some algebra.
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Offline xpressmusic

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Re: Order of reaction
« Reply #2 on: May 21, 2011, 02:37:25 AM »
Thanks for replying :)
I tried to do it in the same way but ain't getting my answers.. would you please help me further in doing them and stating your answers...
Thanks in advance

Offline Borek

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Re: Order of reaction
« Reply #3 on: May 21, 2011, 05:21:58 AM »
I tried to do it in the same way but ain't getting my answers

Show you work then, nobody is going to solve the question for you.
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sambru

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Re: Order of reaction
« Reply #4 on: May 23, 2011, 12:31:36 PM »
I tried to do it in the same way but ain't getting solutions.

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