goIIT: IIT JEE AIEEE CET WBJEE VITJEE BITSAT IIIT AICET Engineering Entrance Experts, Student Discussion Community, IIT JEE Preparation Online

Areal velocity dA/dt= L/2m ;L=angular momentum

i.e v*r/2=4.4*10^16

v*r=8.88*16

lemme call da nearest dist r1, farthest as r2, veloc max as v1, min. veloc as v2

now using geometry

a^2=b^2+c^2

a+c=5.56*10^11

now applyin angular momentum conservation

v1*r1=v2*r2

8.8*10^16=v2*5.56*10^11

wich is 1.582*10^5 m/s

culd u please gimme da exact structure of the product

thanx

from what i inferd from ur ques

using energy conservation

3mgh=0.5k(2d1)^2

h=2d1^2/3mg

Re:Solve for k (IIT JEE 2004)

Re:Solve for k (IIT JEE 2004)

Are the answers d1=d2 and h=kd1^2/3mg

Zero Order Reaction

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

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.

First Order Reaction

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:

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 original 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 t_1/2. Half-life is the time it takes for the concentration to fall from [A]₀ to [A]₀ / 2.

According to equation (6) is obtained:

Pseudo First Order Reaction

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

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
P₀= radiant power for radiation entering; P= radiant power for radiation leaving

gives the relationship between k' and lnA:

One needs only monitor the relative concentration of B as a function of time to obtain 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 t_1/2.

Second Order Reaction

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

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

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

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

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 - 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

Svante Arrhenius 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:

E_a = activation energy
R = 8.314 J/mol·K
T = absolute temperature in Kelvins
A = frequency factor
A = 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.

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 Arrhenius equation is often written in the logarithmic form:

Bestimmung von E_a 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 = - E_a/ R
b = lnA

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

The value of the activation energy E_a 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.

"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

Here are some simple arithmetic tecniques

Multiplying vertically and crosswise is a simple technique in vedic math
Suppose you want to multiply 88 by 98.

Not easy,you might think. But with
VERTICALLY AND CROSSWISE you can give
the answer immediately, using the same method
as above.

Both 88 and 98 are close to 100.
88 is 12 below 100 and 98 is 2 below 100.

You can imagine the sum set out like this:

As before the 86 comes from
subtracting crosswise: 88 - 2 = 86
(or 98 - 12 = 86: you can subtract
either way, you will always get
the same answer).
And the 24 in the answer is
just 12 x 2: you multiply vertically.
So 88 x 98 = 8624

now for multiplyin nos>100

write the nos in dis format, say nos are n1 and n2

100 n1-100

100 n2-100

ans is

(n1+n2-100) (n1+n2-200)

eg. 103*104

100 3

100 4

ans is 10712 wich is (100+3+4 ) (3*4)

find X ,give da mechanism for the rean. also

identify the product, give the mechanism

can any one plz tell me da products wid da mechanism , reason etc..

can someone tell me the ans for dis wid the mechanism

canany1 gimme da products wid the reasoning, mechanism ...

i wanna paste pics of org compds 4 ma ques. how do i go abt it

organic plz hlp