|

CHEMICAL KINETICS

Forum Expert
Posted on
26 Feb 2009 10:47:33 IST
Posts: 1104
26 Feb 2009 10:47:33 IST
0 people liked this
24
1420
CHEMICAL KINETICS
Engineering Entrance , JEE Main , JEE Advanced , Chemistry , Physical Chemistry

CHEMICAL KINETICS

A reaction is of zero order when the rate of reaction is independent of the concentration of materials. The rate of reaction is a constant. When the limiting reactant is completely consumed, the reaction stops abruptly.

The zero order rate law for the general reaction

is written as the equation

Zero Order Reaction

which on integration of both sides gives

When t = 0 the concentration of A is [A]0. The constant of integration must be [A]0.

Now the integrated form of zero-order kinetics can be written as follows

Plotting [A] versus t will give a straight line with slope -k.

A general unimolecular reaction

where A is a reactant and P is a product is called a first-order reaction.

The rate is proportional to the concentration of a single reactant raised to the first power.

The decrease in the concentration of A over time can be written as:

First Order Reaction

Equation (2) represents the differential form of the rate law. Integration of this equation and determination of the integration constant C produces the corresponding integrated law.

Integrating equation (2) yields:

The constant of integration C can be evaluated by using boundary conditions. When t = 0, [A] = [A]0. [A]0 is the initial concentration of A.

Substituting into equation (3) gives:

Therefore the value of the constant of integration is:

Substituting (5) into (4) leads to:

Plotting   ln[A]   or   ln[A] / [A]0   against time creates a straight line with slope   -k. The plot should be linear up to a conversion of about 90%.

Equation (6) can also be written as:

This means that the concentration of A decreases exponentially as a function of time.

The rate constant k can also be determined from the half-life t1/2. Half-life is the time it takes for the concentration to fall from [A]0 to [A]0 / 2 ==> [A] = 1/2 [A]0 .

According to equation (6) is obtained:

A and B react to produce P:

If the initial concentration of the reactant A is much larger than the concentration of B, the concentration of A will not change appreciably during the course of the reaction The concentration of the reactant in excess will remain almost constant. Thus the rate's dependence on B can be isolated and the rate law can be written

Pseudo First Order Reaction

Equation (1) represents the differential form of the rate law. Integration of this equation and evaluation of the integration constant C produces the corresponding integrated law.

Substituting [B] = c into equation (1) yields:

Integrating equation (2) gives:

The constant of integration C can be evaluated by using boundary conditions. At t = 0 the concentration of B is c0.

Therefore

Accordingly is obtained:

If the decrease in concentration of B is followed by photometric measurement the Beer' Law must be taken into account.

Combining equation (4) and Beer' Law

A = absorbance, e = molar absorbtivity with units of L · mol -1 cm -1
c = concentration of the compound in solution, expressed in mol · L -1

gives the relationship between k' and lnA:

According to equation (7), a plot of lnA versus time should lead to a straight line whose slope is the pseudo-first order rate constant k'. The value of k' can then be divided by the known, constant concentration of the excess compound to obtain the true constant second order k:

The pseudo-first order rate constant k' can be also determined from the half-life t1/2.

The rate of a second order reaction is proportional to either the concentration of a reactant squared, or the product of concentrations of two reactants.

For the general case of a reaction between A and B, such that

the rate of reaction will be given by

Second Order Reaction

1. Initial concentrations of the two reactants are equal:

Equation (1) can be written as:

Separating the variables and integrating gives:

Provided that [A] = [A]0 at t = 0 the constant of integration C becomes equal to 1 / [A]0.

Thus the second order integrated rate equation is

A plot of  1 / [A]   vs   t   produces a straight line with slope   k and intercept   1 / [A]0  . The plot should be linear up to a conversion of about 50%.

2. Starting concentrations of the two reactants are different:

If [A]0 and [B]0 are different the variable x is used.

Equation (1) becomes

where [A]0 - x = [A], [B]0 - x = [B] and x is the decrease in the concentration of A and B.

Equation (5) can be integrated after separation of the variables and partial fraction expansion. The result is:

where C is the constant of integration.

Using the condition that x = 0, when t = 0, the value of C can be found

and equation (6) becomes

If [A]0 > [B]0, then a plot of

The Arrhenius equation is often written in the logarithmic form:

If the experimental method yields reactant concentrations rather than x, the equivalent form of equation (8) is

against t will have a positive slope, equal ([A]0 - [B]0) k.

 It is a well-known fact that raising the temperature increases the reaction rate. Quantitatively this relationship between the rate a reaction proceeds and its temperature is determined by the Arrhenius Equation: Ea = activation energy R = 8.314 J/mol·K T = absolute temperature in KelvinsA = pre-exponential or frequency factorA = p · Z, where Z is the collision rate and p is a steric factor.Z turns out to be only weakly dependant on temperature. Thus the frequency factor is a constant, specific for each reaction.

Because equivalent amounts of A and B are reacting, [A] can be expressed in terms of [B].

If [B] = x , [A] = [A]0 - (x0 - x)

Provided that the initial concentration of A is twice the initial concentration of B (see Kinetic equations - Reaction Second Order - Download PDF file) equation (10) becomes

Summary

 Reaction Order Differential Rate Law Integrated Rate Law Linear Plot Slope of Linear Plot Units of Rate Constant 0 - d[A] / dt = k [A] = [A]0 - kt [A]  vs  t - k mol · L-1 · s-1 1st - d[A] / dt = k [A] [A] = [A]0 e - kt ln[A]  vs  t - k s-1 2nd - d[A] / dt = k [A]2 1 / [A] = 1 / [A]0 + kt 1 / [A]  vs  t k L · mol-1 · s-1

Arrhenius Equation

 Effective collisions The Arrhenius equation is based on the collision theory which supposes that particles must collide with both the correct orientation and with sufficient kinetic energy if the reactants are to be converted into products. The two animations are taken from Effects of temperature, concentration, catalysts, inhibitors on reaction rates Ineffektive collisions

 Bestimmung von Ea

"Two-Point Form" of the Arrhenius Equation

The activation energy can also be found algebraically by substituting two rate constants (k1, k2) and the two corresponding reaction temperatures (T1, T2) into the Arrhenius Equation (2).

Substracting equation (4) from equation (3) results in

Rerrangement of equation (5) and solving for E a yields

A plot of lnk versus 1/T produces a straight line with the familiar form y = -mx + b, where

x = 1/T
y = lnk
m = - Ea/ R
b = lnA

The activation energy Ea can be determined from the slope m of this line: Ea = -m · R

The value of the activation energy Ea is rounded to one decimal place. The value of lnA shall be expressed with an accuracy of two decimal places.

An accurate determination of the activation energy requires at least three runs completed at different reaction temperatures. The temperature intervals should be at least 5°C.

• 1
• 2
• GO
• Go to Page...

Blazing goIITian

Joined: 17 Mar 2008 20:56:33 IST
Posts: 1243
26 Feb 2009 10:50:50 IST
0 people liked this

wow! nice article

Forum Expert
Joined: 7 Aug 2008 15:43:23 IST
Posts: 1104
26 Feb 2009 10:51:00 IST
0 people liked this

actually !i have cd of chemistry,from where it has taken.....i hope it will be beneficial for u..........do comments!

Hot goIITian

Joined: 17 Dec 2008 10:41:08 IST
Posts: 141
26 Feb 2009 10:58:30 IST
0 people liked this

gr8 yaar

Forum Expert
Joined: 7 Aug 2008 15:43:23 IST
Posts: 1104
26 Feb 2009 11:02:27 IST
0 people liked this

i have just copy figures and animation only........matter and summary is my efforts.......dont think ,that i just copied and pasted here....so do comments!cheers!!!!!!!!!!!!!

Blazing goIITian

Joined: 25 Dec 2007 15:48:14 IST
Posts: 1104
26 Feb 2009 11:08:49 IST
0 people liked this

nice one...but too long..can be summarized more..but still great effort..

Hot goIITian

Joined: 24 Feb 2009 14:46:58 IST
Posts: 119
26 Feb 2009 11:22:53 IST
0 people liked this

good!

New kid on the Block

Joined: 16 Feb 2009 15:17:27 IST
Posts: 21
26 Feb 2009 11:34:46 IST
0 people liked this

hodi baba!!!!!!!!!nice.........

Forum Expert
Joined: 7 Aug 2008 15:43:23 IST
Posts: 1104
26 Feb 2009 11:39:16 IST
0 people liked this

i have more articles ,beneficial for JEE....if u want check my profile............the most rated one im giving.....check that.......

Forum Expert
Joined: 7 Aug 2008 15:43:23 IST
Posts: 1104
26 Feb 2009 11:39:52 IST
0 people liked this

Cool goIITian

Joined: 15 Jul 2008 15:41:58 IST
Posts: 37
26 Feb 2009 12:10:23 IST
0 people liked this

good!!!!!

Blazing goIITian

Joined: 1 Jun 2008 20:37:23 IST
Posts: 1625
26 Feb 2009 12:11:59 IST
0 people liked this

gr8 job !!!

Cool goIITian

Joined: 26 Feb 2009 14:58:32 IST
Posts: 31
26 Feb 2009 15:04:50 IST
0 people liked this

marvellous!

Hot goIITian

Joined: 4 Feb 2008 01:00:29 IST
Posts: 167
27 Feb 2009 09:14:54 IST
0 people liked this

nice

Blazing goIITian

Joined: 11 Sep 2008 12:38:20 IST
Posts: 1159
27 Feb 2009 13:33:08 IST
0 people liked this

it looks stupid when you donot mention the source------http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/kinetics.htm

Blazing goIITian

Joined: 11 Sep 2008 12:38:20 IST
Posts: 1159
27 Feb 2009 15:01:16 IST
0 people liked this

seems to be good copy paste from "cd"!!!!!!!!!!!! is the internet a form of cd????????

Forum Expert
Joined: 7 Aug 2008 15:43:23 IST
Posts: 1104
27 Feb 2009 16:27:03 IST
0 people liked this

man!go and see the cd of chem master.....i think this guy is not intrested in the article.....

Blazing goIITian

Joined: 11 Sep 2008 12:38:20 IST
Posts: 1159
27 Feb 2009 20:44:37 IST
0 people liked this

no need to show "cd".............you can refer the source i have mentioned!!!!!!! donot try to cover your mischief!!!!!!!

Blazing goIITian

Joined: 28 Oct 2008 11:51:55 IST
Posts: 552
27 Feb 2009 21:09:12 IST
0 people liked this

debotosh what if ....if he had copied 4m a cd...or 4m ny other source.... t main thing is he has provided t help through t article 2 all those who dont have such a material n were in need of it... so v shuld atleast give credit 4 the article written....... saharsha the article is good enough......

Blazing goIITian

Joined: 11 Sep 2008 12:38:20 IST
Posts: 1159
28 Feb 2009 10:31:36 IST
0 people liked this

but putting up such reasons looks awkaward as if trying to cover up some mischief!!!!

 Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a avatar to have your pictures show up by your comment If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

## For Quick Info

Name

Mobile

E-mail

City

Class

Vertical Limit

Top Contributors
All Time This Month Last Week
1. Bipin Dubey
 Altitude - 16545 m Post - 7958
2. Himanshu
 Altitude - 10925 m Post - 3836
3. Hari Shankar
 Altitude - 9960 m Post - 2185
4. edison
 Altitude - 10815 m Post - 7797
5. Sagar Saxena
 Altitude - 8625 m Post - 8064