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### +2 UNIT 6 PAGE- 2

MOLECULARITY AND MECHANISM OF REACTION
According to collision theory a chemical reaction takes place due to collision between the particles of the reactants. The number of reacting species(atoms, molecules or ions) which must collide simultaneously in order to bring about the chemical reaction is called the molecularity of the reaction. The molecularity of the reaction can be 1, 2 or 3 . For example, the decomposition of ammonium nitrite is a unimolecular reaction.
NH4NO2 * N2 + 2 H2O
Similarly, the reaction involving simultaneous collision between two species is a bimolecular reaction.
2 H I (g) * H2 (g)+ I2(g)
In the same way, the reaction between NO and O2 is trimolecular reaction.
2 NO + O2 * 2 NO2
In most of the reactions, the molecularity does not exceed three. It is because the chances of simultaneous collisions between three or more particles are rare. In general , for elementary reactions, i.e., single step reactions, the molecularity of the reaction can be obtained from balanced chemical equation. However, in many reactions molecularity of the reaction as obtained from the balanced chemical equation may come out to be more than three. For example, in the reaction between HBr and O2 , the molecularity is five as given by balanced equation.
4 HBr + O2 * 2 H2O + 2 Br2
Since molecularity greater than three is not possible, therefore , the above reaction does not involve the simultaneous collision of all the reacting species in a single step. In fact, such chemical reactions proceed through a sequence of steps. Each step is an elementary step and involves the simultaneous collision of two or three species only. Such chemical reactions proceed through more than one steps are termed as complex reactions.
The detailed description of various steps (the stepwise reactions are not necessarily observable; they are merely proposed pathways from reactants to products which logically fit into all the available experimental data.) of the chemical reaction is called mechanism of reaction. For example, the above reaction occurs by the following steps :
HBr + O2 * HOOBr
HOOBr + HBr * 2 HOBr
[HOBr + HBr * H2O + Br2 ] x 2
------------------------------------------
4 HBr + O2 * 2 H2O + 2 Br2
It is clear that from the above discussions that a complex reaction occurs in two or more steps. The term molecularity of overall reaction has no significance in this case.
RATE LAW OR RATE EQUATION
In complex reactions, the rate expression written on the basis of the overall balanced equation has no significance at all. The true rate expression for such complex reactions can be evaluated on the basis of the experimental data only. For example, in the reaction between NO2 and F2 to yield nitryl fluoride, NO2F,
2 NO2 (g) + F2 (g) * 2 NO2F(g)
the expected rate expression is :
Rate = k [NO2]2 [F2]
But it is experimentally found that the rate of this reaction is proportional to the product of single concentration term of NO2 and and F2. Thus the experimental rate of the reaction is given as :
Rate = k [NO2] [F2]
Such a mathematical expression that gives the true rate of a reaction in terms of concentration of the reactants , which actually influence the rate ,is called rate law.
In general, for any hypothetical reaction:
a A + b B * c C + d D
The rate law expression may be written as :
Rate = k [A]m [B]n
In the above rate law expression, the numerical value of m and n may or may not be same as a and b . The constant k in the rate law expression is called rate constant, velocity constant or specific reaction rate. Now, if the concentration of each of the reactants involved in the reaction is unity, i.e.,
[ A] = [B] = 1 mol L1 , then
rate = k x 1 x 1 = k
Thus, the rate constant of a reaction at given temperature may be defined as the rate of the reaction when the concentration of each of the reacting species is unity.

Characteristics of rate constants
i) The value of k is different for different reactions.
ii) It is a measure of intrinsic rate of reaction. This means larger the value of k, faster will be the reaction. Similarly, small value of k reflects slower reaction.
iii) At a fixed temperature, the value of k is a constant and characteristic of a reaction.
iv) For a particular reaction, k is independent of concentration but depends on temperature.
The important differences between the rate of the reaction and rate constant of the reaction are given in the following TABLE.
TABLE
Rate of reaction Rate constant of reaction
1. It is the speed at which the reactants are converted into the products.
2. It depends upon the concentration of reactant species at that moment.

3. It decreases with the
progress of reaction
generally. 1. It is a constant of proportionality in the rate law expression.

2. It refers to the rate of reaction at specific point when concentration of every reacting species is unity.
3. It is a constant and does
not depend on the
progress of the reaction.

RATE CONTROLLING STEP
A complex reaction proceeds through more than one steps. Out of the various steps of the reaction, the slowest step will decide the rate of overall reaction because the reaction cannot take place faster than the slowest step. The slowest step of the complex reaction is called the rate controlling step or rate determining step.
The rate law expression of a complex reaction gives an indication about the slowest step in the mechanism of a reaction. For example, the decomposition of nitrous oxide:
2 N2O * 2 N2 + O2
the rate law is found to be :
Rate = k [N2O]
Since the rate of the reaction depends only upon the single power of N2O, it indicates that only one molecule of N2O is involved in the slow step. Thus, with this knowledge and chemical intuition a possible mechanism of the reaction may be proposed as :

The above postulated mechanism is consistent with the rate law expression.
Note: It must be noted that there is no way to verify the absolute validity of the proposed mechanism. All we can do is to postulate a mechanism that is consistent with experimental data. If further investigations reveal information that is not consistent with postulated mechanism, it is to be modified accordingly so as to explain all the observations.

Problems
10. For the reaction:
2 NO2 + F2 * 2 NO2F
The experimental rate law :
Rate = k [NO2] [F2]
Propose the mechanism of the reaction.
11. For the reaction :
2 NO + Br2 * 2 NOBr
if proceeds by the following mechanism:

Suggest a rate law consistent with the mechanism.
ORDER OF A REACTION
It is an important parameter for every chemical reaction. It is always determined experimentally and cannot be written from the balanced chemical equation. It refers to the number of reacting particles whose concentration terms determine the reaction rate.
The Order of a reaction may be defined as the sum of the powers to which concentration terms in the rate law are raised to express the observed rate of a reaction.
For example, for a hypothetical reaction :
a A + b B + c C * Products
if the rate law expression for this reaction is :
Rate = k [A]m [B]n [C]p
Then the order of the reaction is equal to ( m + n + p ). These powers or exponents i.e., m, n and p have no relation to the stoichiometric coefficients a, b and c. Order of the reaction with respect to A is m, with respect to B is n and with respect to C is p. If the sum of the powers is equal to one, the reaction is first order reaction. If the sum of the powers is two or three, the reaction is second order or third order reaction respectively. The order of the reaction can also be zero or fractional.
Some examples of reactions of different orders
a) Reactions of first order
i) Decomposition of Nitrogen pentoxide.
N2O5(g) * 2 NO2(g) + (1/2) O2(g)
Rate = k [N2O5] ; Order = 1
ii) Decomposition of Hydrogen peroxide.
H2O2 * H2O + (1/2)O2
Rate = k[H2O2] ; Order = 1
iii) Dehydrohalogenation of Chloroethane.
C2H5Cl * C2H4 + HCl
Rate = k [C2H5Cl] ; Order = 1
iv) Decomposition SO2Cl2.
SO2Cl2 * SO2 + Cl2
Rate = k [SO2Cl2] ; Order = 1
b) Reactions of second order
i) Decomposition of nitrogen peroxide :
2 NO2 * 2 NO + O2
Rate = k [NO2]2 ; Order = 2
ii) Reaction between hydrogen and iodine to give HI.
H2 + I2 * 2 HI : Rate = k [H2] [I2] ; Order = 2
c) Reactions of Third order :
i) The reaction between nitric oxide and oxygen .
2 NO(g) + O2(g) * 2 NO2(g)
Rate = k [NO]2 [O2] ; Order = 3
ii) Reaction between nitric oxide and chlorine :
2NO + Cl2 **2 NOCl Rate = k [NO]2 [Cl2] ; Order = 3
It may be noted that the order of a reaction may or may not be a whole number. It can have zero or fractional values also.
d. Reactions of fractional order
i) Combination of CO and Cl2.
CO + Cl2 * COCl2
Rate = k [CO]2 [Cl2]1/2 ; Order = 2.5
ii) Decomposition of carbonyl chloride.
COCl2 * CO + Cl2
Rate = k [COCl2]3/2 ; Order = 3/2
iii) Decomposition of acetaldehyde.
CH3CHO * CH4 + CO
Rate = k [CH3CHO] 3/2 ; Order = 3/2
e. Zero order reaction : A number of zero order reactions are known in which the rate of reaction is independent of the concentration of the reactants. For example, the decomposition of ammonia at the surface of metals like gold, platinum etc. is a zero order reaction.

It has been observed that the rate of reaction is independent of the concentration of ammonia, i.e.,
Rate = k [NH3]0
 Order of the reaction = zero.
Units of rate constants
The rate is the change in concentration with time. Therefore, the rate of reaction is expressed by concentration units divided by time units. If the concentrations are expressed in mol L1 and time in seconds, then the units for rate of a reaction are
mol L1 s1 as :

The units of rate constants of different orders are different.
For an nth order reaction:

If concentration is expressed in mol L1 and time in seconds,

For First order reaction , the unit of rate constant is s1 .
The unit of second order rate constant is L mol 1 s1
The unit of third order rate constant is L2 mol2 s1
For gaseous reactions of nth order , k has units of ;
(atm) 1 n s1
INTEGRATED RATE EXPRESSION
This is the most common method for studying the kinetics of chemical reactions. The instantaneous rate of a reaction is given by differential rate law equations. For example, for a general reaction :
a A + b B  Products
The differential form of the rate law is transformed to integrated form of rate law by simple mathematics (calculus). The integrated form is more convenient because it helps to understand the variation of concentrations of the reactants with time. The integrated rate equations for different orders can be derived.
Zero order reaction
A reaction is said to be of zero order, if its rate is independent of the concentration of the reactants. Consider a general reaction
A  Product
Let [A] be the concentration of the reactant A and ko is the rate constant for the zero order reaction. For the zero order reaction, the rate of reaction is independent of the concentration of A.

Thus,

This form of the rate law is known as differential rate equation. Rearranging the above equation,
 d [A] = k0 d t
Itegrating the above equation, we get,

 [A] = k0 t + I …………(2)
where I is the constant of integration. The value of I can be calculated from the initial concentration . For example, the initial concentration of A be [A]0 at t = 0 . Then equation becomes
 [A]o = k0 x 0 + I
I =  [A]0
Substituting the value of I in (2) we get,
 [A] = k0 t  [A]0
k0 t = [A]0  [A]

where [A]o is the initial concentration of A and [A] is the concentration at time t.
Alternately, if the initial concentration of A is ‘a’ moles per litre and x moles of reactants gets changed to products in time t. Then concentration of A left after t be (a  x).
A  Product
Initial concentration : a 0
Concentration at time t : (a  x) x
 [A]o = a , [A] = a  x so that
[A]o  [A] = a  (a  x) = x
Thus equation (3) becomes,

The above equation is general equation for zero order reaction. The amount of the substance reacted is proportional to time.
First order reaction
Consider the general first order reaction
A  Products.
The differential rate law equation is  d[A] = k [A]
dt
The differential form of rate law is transformed to integrated form of rate law by simple calculus. This equation is used to calculate the rate constant k.
Let us illustrate this method by applying it to a first order reaction. Consider a first order reaction. For the first order reaction, the rate of this reaction is directly proportional to the concentration of the reactant A. Thus :
Rate =  d[A] = k1 [A] ……….(5)
dt
This form of rate law is known as differential rate equation. Rearranging the above equation :
 d[A] = k1dt ……….(6)
[A]
Integrating the above equation we get:

 ln [A] = k1 t + I ……(7)
where I is the constant of integration. The value of I can be calculated from the initial concentrations. For example, the initial concentration of A can be [A]0 at t = 0. Then Equ.(7) becomes :
 ln [A]0 = k 1 x 0 + I
 I =  ln [A]0
Substituting the value of I in Eq.(7) , we get :
 ln [A] = k1 t  ln [A]0
Rearranging ,
ln [A]0  ln [A] = k 1 t
or ln [A]0 = k 1 t
[A]
or k 1 = 1 ln [A]0 ……..(8)
t [A]
Changing the above expression to log base 10
( ln x = 2.303 log x), we get:

The Equ .(9 ) is integrated rate expression for first order reaction.
The above equation may also be written in an alternate form. Let the initial concentration of A is 'a' moles per litre. Suppose at time t , x moles of reactants get changed into products. Then the concentration of A left after time t is (a  x). Therefore, [A]0= a and [A] = (a  x ).
Thus Equ. (9) becomes :

Significance of integrated Rate equation
The integrated Rate equation can be used in the following ways :
i) Order of reaction : All reactions of first order obey the following equation :

Starting with the known concentration of A, the concentration of the reactant [A], at different intervals of time (t) may be measured. The value of [A] at different time intervals are substituted in the above equation. If the value of k comes out to be a constant, then the reaction is of first order.
(ii) Calculation of Rate constant : The integrated rate equation can be used to calculate the rate constant of a reaction. If the reaction is known to be first order , then by substituting initial concentration of reactant [A]0 and concentrations at different times (t) in the above equation , the value of k can be calculated.
The rate constants can be calculated graphically . The Equation (9) may be written as :

This equation is comparable to the equation of a straight line ( y = m x + c ). When a graph is plotted between log[A] against corresponding time interval we get a straight line. This is shown in Fig.

Plot of t versus log[A] to calculate rate
constant for First Order Reaction.

The slope of this line is equal to :
Slope =  k 1
2.303
From this the rate constant can be calculated.
k 1 =  2.303 (slope)
EXPERIMENTAL DETERMINATION OF ORDER OF A REACTION
The following methods employed to determine the rate law , the rate constant and order of a reaction.
1. GRAPHICAL METHOD : This method is used to determine the rate law of the reaction which involves only one reactant species. The various steps involved are :
i) The concentrations of reacting substances are determined at different intervals by suitable method.
ii) A graph is plotted between concentration and time.
iii) From the plot of concentration vs time , the instantaneous rates corresponding to different concentrations determined by drawing tangents to the curve and subsequently calculating their slopes.
iv) The rate of reaction (calculated above step 3) is plotted versus concentration [A] or (concentration)2 , [A]2 and so on.
(a) If rate of reaction rate remains constant in the rate versus concentration graph, it means that the rate is independent of the concentration of the reactant, i.e.,
Rate = k [A]0 = k
Therefore, the reaction is of zero order.
(b) If a straight line is obtained in rate versus concentration graph, it means that the rate is directly proportional to the concentration of the reactant, i.e.,
Rate = k [A]
Therefore, the reaction is of first order.
(c) If a straight line is obtained in the rate versus (concentration)2 graph, it means that
Rate = k [A]2
Therefore, the order of the reaction is 2.
(d) Similarly , if we get straight line in the rate versus (concentration)3 graph, then
Rate = k [A]3
And the order of the reaction is 3.
In general, , if we get straight line by plotting graph of the rate versus (concentration)n, where n = 1,2,3… so on, then
Rate = k [A]n
And order of the reaction is n.

These graphs are shown below.

Illustration
Consider the decomposition of nitrogen pentoxide
2 N2O5(g)  4 NO2(g) + O2(g)
A convenient method to follow the reaction is to measure at different times , the increase in pressure accompanying the reaction. From the measured values it is possible to first calculate the partial pressure of N2O5 and then the concentration in moles per litre of N2O5. The molar concentrations so obtained for different times are plotted in the form of curve(Fig 5). The values of slopes(or reaction rate) progressively decreases as the concentration of N2O5 decreases. This shows that the reaction depends on its concentration. On plotting rate against concentration, i.e., the rate against [N2O5], a straight line is obtained (Fig 6).

It means that the rate of reaction is proportional to the concentration of N2O5 or :
Rate  [N2O5]
which in turn means that the rate law is :
Rate = k [N2O5]
However, we do not get straight line by plotting rate of a reaction against [N2O5]2. This means that the reaction is not of second order.
Thus the correct rate law for this reaction is :

and the order of the reaction is one.
Calculation of Rate constant
The rate constant can be calculated by substituting the values of rate and concentration of [N2O5] in the rate expression :

Note : From the above discussion, we notice that the coefficient of N2O5 in the balanced equation is 2 while the exponent in the rate law is 1. Thus the order of the reaction is not same as the coefficient in the balanced chemical equation.
Graphical method for integrated rate equation
Graphical method can also be applied for integrated rate equation. In this method, appropriate function of concentration is plotted against time. The resulting curve will be straight line only for the case in which the appropriate integrated equations has been used. The integrated rate equations for zero, first and second order reactions are given below :

As shown below, straight lines are obtained for a plot of [A] versus t for a zero order reaction, of log [A] versus t for a first order reaction and 1/[A] versus t for second order reaction as shown below.

The curves also help to calculate the value of k from the slope of the straight line.
2. INITIAL RATE METHOD
The graphical method cannot be applied for the reactions which involve more than one reactants. The rates of such reactions can be determined by initial rate method. This method involves the following steps:
(a) The initial rate of the reaction, i.e., the rate at the beginning of the reaction is measured. This may be taken as the rate over an initial time interval that is short enough so that concentrations of the reactants do not change appreciably from their initial values. This corresponds to slope of the tangent to the concentration versus time graph at t = 0.
(b) The initial concentration of only one reactant is changed(keeping other concentration constant) and the rate is determined again. From this the order with respect to that particular reactant is calculated.
(c) The procedure is repeated with respect to each reactant until the overall rate law is fully determined.
(d) The sum total of the individual orders with respect to each reactant gives the order of the reaction.
Illustration
The method is illustrated by taking hypothetical reaction :
2 A + 2 B  Products.
The experimental data for this reaction is given below:
Experiment Concentration
[A] [B] Rate of reaction
I 0.01 0.01 0.005
II 0.02 0.01 0.020
III 0.02 0.03 0.060
The general form of the rate law may be written as :
Rate = k [A]p[B]q
Then, the expression for initial rate is :
(Rate)0 = k [A]op[B]oq
where subscript zero denotes initial values. The problem involves the determination of p and q.
Consider the experiments I and II and substituting the values we get,
(Rate)1 = k (0.01)p(0.01)q = 0.005 …….(1)
(Rate)2 = k (0.02)p(0.01)q = 0.020 ……..(2)
Dividing Eq(2) by Eq.(1):

(2)p = 4 or (2)p = 22  p = 2
Similarly, comparing experiments II and III,
(Rate)2 = k ( 0.02)p(0.01)q = 0.020 ………(3)
(Rate)3 = k (0.02)p (0.03)q = 0.060 ………(4)
Dividing equation (4) by (3), we get :

3q = 3 or q = 1
Therefore , the order with respect to A is 2 and the order with respect to B is 1.
Thus, the rate law may be written as :
Rate = k [A]2 [B]
3. Ostwald Isolation method
This method was introduced by Ostwald in 1902 and is used to find the order of a reaction with respect to one reactant at a time. The total order of the reaction is equal to the sum of the orders of reaction for individual reactants. This method is based on the principle that if the concentrations of all but one reactant are taken in excess, then during the course of the reaction, the concentration of those reactants taken in excess will remain almost constant and hence variation in rate will correspond to the concentration of that reactant whose concentration is small. This process is repeated with other reactants and order with respect to each reactant is determined. The overall order will be the sum of all these orders. For example, consider the general reaction :
a A + b B + c C  Products
Suppose we isolate A by taking B and C in large excess and get the order of the reaction with respect to A ( say p) . Similarly, we isolate B by taking A and C in excess and isolate C by taking A and B in excess and get order with respect to B and (say q) and C (say r) respectively.
Overall order of the reaction n = p + q + r
Problems
12. Three experimental runs were carried out for the reaction between Cl2 and NO.
Cl2 + 2 NO(g)  2 NOCl(g).
The following rate law are determined.
Run Initial concentration (mol/L)
[Cl2] [NO] Initial rate
mol / L / s
1 0.020 0.010 2.4 x 104
2 0.020 0.030 2.16 x 103
3 0.040 0.030 4.32 x 103
Determine :
i) The order with respect to Cl2 and NO.
ii) The rate law. (iii) The rate constant.
13. The decomposition of N2O5 in carbon tetrachloride solution has been investigated.
N2O5(solution) 2 NO2(solution) + (1/2)O2(g)
The reaction has been found to be first order with
first order rate constant 6.2 x 104 s1. Calculate the
rate of reaction when :
(a) [N2O5] = 1.25 mol L-1 and (b) [N2O5] = 0.25 mol L1
(c) What concentration of N2O5 would give a rate 2.4 x 103 s1 ?
14. With the help of the following rate expressions of the reaction,
find out the overall order of the reactions and order with
respect to each reaction ?
2 NO(g) + O2(g)  2 NO2(g) ; Rate = k[NO]2[O2]
2 N2O(g)  2 N2 + O2(g) ; Rate = k [N2O]
2 NO(g) + 2 H2(g)  N2(g)+2H2O(g) ; Rate = k[NO]2[H2]
15. Identify the reaction order from the following rate constants ;
i) k = 2.3 x 105 L mol1 s1
ii) k = 2.3 x 101 s1
16. What is the rate of reaction and order of the reaction , if the mechanism is :
2 NO + H2  N2 + H2O2 (slow)
H2O2 + H2  2 H2O (fast)
17. The experimental data for the reaction :
2 A + B2  2 AB is :
Experiment [A] [B2] Rate
(mol L1 s1)
1 0.50 0. 50 1.6 x 104
2 0.50 1.00 3.2 x 104
3 1.00 1.00 3.2 x 104
Write the most probable rate equation for the reaction
18. For the following reactions, state the order with respect to each reactant and overall order :
a) 3 NO(g)  N2O(g) + NO2(g)
Rate = k [NO]2
b) H2O2 + 3 I + 2 H+  2 H2O + I3(aq)
Rate = k [H2O2][ I]
c) CH3CHO(g)  CH4(g) + CO(g)
Rate = k [CH3CHO]3/ 2
d) CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g)
Rate = k [CHCl3][ Cl2]1/2
What are the dimensions of the rate constants in each case.
19. For the following reaction : 2 A + B + C  A2B + C
the rate law has been determined to be :
Rate = k[A] [B]2 : k = 2.0 x 106 mol2Ls1
(a) For the reaction, determine the initial rate of reaction with [A] = 0.1 mol L1;
[B] = 0.2 mol 1; [C] = 0.8 mol L1.
(b) Determine the rate after 0.04 mol L1 of A has reacted.
20. The rates of reaction starting with initial concentrations 2 x 103 M and 1 x 103 M are equal to 2.4 x 104 M s1 and 0.60 x 104 M s1 respectively. Calculate the order of the reaction with respect to the reactant and also the rate constant.
21. The initial rate of the reaction : A + 5 B + 6 C  3 L + 3 M
has been determined by measuring the rate of disapperance of A under the following conditions :
Expt No [A]o M [B]o M [C]o M Initial rate M min1
1 0.02 0.02 0.02 2.08 x 103
2 0.01 0.02 0.02 1.04 x 103
3 0.02 0.04 -.02 4.16 x 103
4 0.02 0.02 0.04 8.32 x 103
Determine the order of the reaction with respect to each reactant and overall order of the reaction. What is the rate constant ? Calculate the initial rate of the reaction when the concentration of all reactants is 0.01 M. Calculate the initial rate of change in concentration of B and L.
22. The pressure of a gas decomposing at the surface of a solid catalyst has been measured at different times and results are given below;
t (s) 0 100 200 300
P (Pa) 4.00 x 103 3.50 x 103 3.00 x 103 2.5 x 103
Determine the order of the reaction, its rate constant and half life period.
23. The rate of decomposition of N2O5 in CCl4 solution has been studied at 318 K and the following results are obtained :
t min 0 135 339 683 1680
C M 2.08 1.91 1.68 1,35 0. 57
Find the order of the reaction and calculate its rate constant. What is the half-life period ?
4. Half Life Period Method
The order of a reaction can also be determined by another method known as half life period method. This is discussed below.
HALF LIFE PERIOD OF A REACTION
Half life of a reaction is defined as the time during which the concentration of the reactants is reduced to half of the initial concentration or it is the time required for the completion of half of the reaction. It is denoted by t1/2 . Let us calculate the half life of a First Order reaction.
For a First Order reaction, we know that :

In general for an nth order reaction :

Thus it is clear from the above expression , that the half-life period or half-change time for first order reaction does not depend upon the initial concentration of the reactants. Similarly, the time required to reduce the concentration of reactant to any fraction of initial concentration for this type of reactions is also independent of the initial concentration.
In addition to calculation of t1/2, the time required to complete different fractions of a first order reaction can also be calculated.
For example, the time required to complete 1/3 rd of the reaction will be given as :

In general the time required to complete any fraction of a first order reaction is given as :

where n = 2 to 9.

Half life period of zero order and second order reactions
For zero order reaction,
k t = [A]o  [A]
For half life period, t½ ,

Similarly for second order reactions.

These are summed up below:

These results give very interesting observation that only the half lives of first order reactions are independent of the concentrations. The half-lives versus concentration [A]o or 1/[A]o plots are shown below.

DEPENDENCE ON REACTION RATE ON TEMPERATURE
The rate of a chemical reaction is significantly affected by a change in temperature. For most of the chemical reactions, the rate increases with rise in temperature. The rate usually becomes doubled to trebled for each 10 rise in temperature.

Temperature coefficient
The increase in the rate of reaction with rise in temperature is usually expressed in terms of a quantity known as temperature coefficient. It is defined as the ratio of rate constant of a reaction at two different temperatures separated by 10C. The two temperatures generally taken are 35C (308 K) and 25C (298 K).

where k toC is the rate constant for the reaction at tC and k t + 10OC is the rate constant for the same reaction at t + 10C. For most homogeneous reactions the value of temperature constant lies between 2 and 3. This means that the rate constant and hence rate of reaction increases two- to- three fold for every 10 rise in temperature. For example, temperature coefficient of the reaction 2 HI  H2 + I2 is 1.7. In some cases the value of temperature coefficient is found even greater than three. For example temperature coefficient for the decomposition of nitrogen pentoxide (N2O5) is found to be equal to 3.8.

## QUESTIONS

Atoms and Molecules
1.


text:
# [Tex/LaTex] How to draw a shaded sphere

Andrew Stacey pointed out that the Rosetta Code entry "Draw a Sphere" doesn't have a TikZ entry yet.

Is there a way to draw a "properly" shaded sphere using TikZ? The ball shading would seem an obvious choice, but it uses a circular highlight, which is not what it would look like in reality. So: is there a way to make a sphere that would resemble a white, matte (i.e. non-shiny) ball that is illuminated by a single white light source?

This would mainly require getting a non-circular, correctly oriented highlight, I believe. Both Yori's and Altermundus' answer already do a great job at using a "softer" highlight, which looks more believable, but Altermundus' highlight is still circular, and while Yoris' is elliptic, it's not oriented correctly, I think.

Compare a circle with the ball shading to the example made with Perl 6:

\documentclass{article}

\usepackage{tikz}

\begin{document}

\begin{minipage}{2.5cm}\centering
\scriptsize That's not a sphere.

\tikz{
\fill [black] (-1,-1) rectangle (1,1);
}
\end{minipage}%
%
\begin{minipage}{2.5cm}\centering
\scriptsize \emph{That's} a sphere.

\includegraphics[width=2cm]{Sphere-perl6.png}
\end{minipage}
\end{document}

I'm following the earlier post Is there a way to tune ball shading in TikZ ?, particularly Stefan Kottwitz's answer. He showed how to use \pgfdeclareradialshading to change the radial shading. Changing the parameters for the radial and adding some clipping, I can produce this:

Is that sphere enough? Perhaps with some more tweaking it is possible to get an even better result. The code to produce this is:

\documentclass{article}
\usepackage{tikz}

\begin{document}
\makeatletter
color(0bp)=(tikz@ball!0!white);
color(17bp)=(tikz@ball!0!white);
color(21bp)=(tikz@ball!70!black);
color(25bp)=(black!70);
color(30bp)=(black!70)}
\makeatother

\begin{tikzpicture}
\fill [black] (-1,-1) rectangle (1,1);
\begin{scope}
\clip (0,0) circle (1);
\shade [ball color=white] (-0.1,0) ellipse (1.2 and 1);
\end{scope}
\end{tikzpicture}

\end{document}

Or, with a more ellipsoidal shading, one can get this:

The for this is:

\documentclass{article}

\usepackage{tikz}

\begin{document}

\makeatletter
color(0bp)=(tikz@ball!0!white);
color(10bp)=(tikz@ball!0!white);
color(15bp)=(tikz@ball!70!black);
color(20bp)=(black!70);
color(30bp)=(black!70)}
\makeatother

\begin{tikzpicture}

\fill [black] (-1,-1) rectangle (1,1);

\begin{scope}
\clip (0,0) circle (1);
\draw [fill=black!70] (0, 0) circle (1);
\begin{scope}[transform canvas={rotate=45}]
\shade [ball color=white] (0,0.5) ellipse (1.8 and 1.6);
\end{scope}
\end{scope}

\end{tikzpicture}

\end{document}

It's a matter of experimenting and fiddling around with the parameters to see what looks best according to you (judging from the discussion above, this seems quite subjective matter). Notice that you need to specify transform canvas={rotate=45} in the inner scope, not rotate=45 because the latter does not rotate the fill.

format: plain


text:
# Math Help - Alternating Group

1. ## Alternating Group

I need the Proof:

The Alternating Group of degree 4 ( i.e., A4) has no subgroup of order 6.

2. Originally Posted by manik
I need the Proof:

The Alternating Group of degree 4 ( i.e., A4) has no subgroup of order 6.
What have you tried? What do you know?


text:
# A pair of mutually “bounding” sequences.

Given $({a_n})_{n=1}^{\infty}$, $({b_n})_{n=1}^{\infty}$ convergent sequences and where $$\{n\in\mathbb{N}\mid a_n\le b_n\}\quad\text{and}\quad\{n\in\mathbb{N}\mid b_n\le a_n\}$$ are both unbounded, prove that $$\lim \limits_{n\to \infty}a_n=\lim \limits_{n\to \infty}b_n$$

I would like to know how I can prove it using simple calculus theorem(I only know the definition of limit, arithmetics of limits and the Squeeze Theorem).

Thank you very much for your time and help.

-
Thank you very much for your input, I'll keep it in mind. –  Anonymous Mar 29 '12 at 5:24
Do you know that if a sequence converges to, say, $L$, then any subsequence also converges to $L$? –  Andres Caicedo Mar 29 '12 at 5:28
@Andres Caicedo No, I didn't know that. It's true for any subsequence which is defined for infinity, right? –  Anonymous Mar 29 '12 at 5:39
What do you mean by "defined for infinity"? If what you meant is that the sequence has infinitely many terms (as opposed to a finite subsequence, such as $a_2,a_{17},a_{372},a_{373},a_{649}$), then yes. –  Andres Caicedo Mar 29 '12 at 5:42

If you want to proceed from basics, meaning the $\epsilon$-$N$ definition of limit, let the limits of our sequences be $a$ and $b$ respectively.

We show that we cannot have $a<b$, and that we cannot have $a>b$. To show that $a<b$ is not possible, let $\epsilon=(b-a)/3$. There is an $N$ such that if $n >N$ then $a_n$ is within $\epsilon$ of $a$, and $b_n$ is within $\epsilon$ of $b$. But then we cannot have $a_n \ge b_n$, contradicting the fact that the set of $n$ such that $a_n \ge b_n$ is unbounded.

-
Hey, I decleared n1, n2 and N such that N=max{n1,n2} for every n greater then N by definition of limits: $|a_n-a|<\epsilon$ and $|b_n-b|<\epsilon$. How do I show that there is contradiction now? –  Anonymous Mar 29 '12 at 16:33
Draw a picture, and it will be clear. But let's proceed formally. Recall we are assuming that $b>a$. We have $b_n>b-\epsilon$ and $a_n<a+\epsilon$. (Why? Because $b-\epsilon<b_n<b+\epsilon$, and we quoted only the left inequality for $b_n$. Similarly, $a-\epsilon<a_n<a+\epsilon$, and we quoted only the right inequality for $a_n$). So $b_n-a_n <(b-\epsilon)-(a+\epsilon)=(b-a)-2\epsilon>\epsilon$, which in particular forces $b_n>a_n$. –  André Nicolas Mar 29 '12 at 17:36
@André Nicolas did you mean to write to following inequality: $b_n-a_n >(b-\epsilon)-(a+\epsilon)=(b-a)-2\epsilon=\epsilon$ (I'm not near my computer so I can't add this to the comments below, please move this message to its place and not as an answer). Thank you very much! –  Anonymous Mar 30 '12 at 9:39
@Anonymous: Yes. Remember, I am assuming something that turns out to be false ($b>a$) and am showing how this is incompatible with $b_n \le a_n$ for an unbounded collection of $n$. Then one would do a separate proof showing $b<a$ is not possible, but there is no need to bother because of the symmetry. –  André Nicolas Mar 30 '12 at 13:25
@André Nicolas Awesome, one last thing I want to make sure of is that you wrote in the inequality (b−a)−2ϵ>ϵ where I re-wrote it as I think you meant (b−a)−2ϵ=ϵ(instead of > I think you meant = and made a typo) assuming of course $\epsilon=(b-a)/3$. am I correct? (Please re-move it to the comments above/below.) Thank you very much again! –  Anonymous Mar 30 '12 at 14:22

Here's a somewhat different take that uses the arithmetic of sequences in a central way.

Since $\lim_{n\to\infty} a_n =a$ and $\lim_{n\to\infty} b_n = b$, we have by arithmetic of sequences that $\lim_{n\to\infty} \{a_n - b_n\} = a-b$. By hypothesis, $\{a_n-b_n\}$ has infinitely many positive terms and infinitely many negative terms. If $a-b$ is positive, then the negative terms can't get arbitrarily close to $a-b$ as they would have to do.

Formally, for every $N$ there is some $n\ge N$ for which $a_n-b_n<0$ and thus $|(a-b) - (a_n-b_n)| \ge a-b$. Thus $\{a_n-b_n\}\not\to a-b$, contradiction.

On the other hand, if $a-b$ is negative, then the positive terms can't get arbitrarily close to $a-b$, again a contradiction. (The formal statement is almost exactly like the one above.) Therefore $a-b=0$.

-


text:
# Bound on $a,b,c,d$ and truth of statement $\forall \, x \in \mathbb N, \exists \, a,b,c,d \le {l(x)} \in \mathbb N:x=a^b+c^d$

To render the statement in the title in relatively simple English, "For every natural number $x$, there exist four natural numbers $a,b,c,d \le l(x)$, where $l(x)$ is a function of $x$ which bounds these numbers, such that $x=a^b+c^d$. I have attempted to do some research on whether this statement is true, with no success. Obviously if $0$ is included as a possible value for $a,b,c,d$ then the statement is always true and an obvious bound on $l(x)$ is $x$, but I am excluding it for the sake of a more interesting problem. I haven't done much numerical testing of this one, so it is possible that there very low numbers that prove this false.

On a different but related question, we look at the similar statement $$\forall \, x \in \mathbb N, \exists \, a,b,c,d \le {L(x)} \in \mathbb N:x=a*3^b+c*2^d$$. Similarly, the statement is always true if $0$ is included as a possible value of $a,b,c,d$, as $b$ and $c$ can be set to $0$ and $a$ to $x$. If we exclude $0$, for all numbers that I have tested greater than $6$ the statement holds true for some value of $L$. I have, however, completely failed at determining any good bounds on $L$, other than that it appears to be in $O(x^c)$ where $\frac 1 4 \lt c \lt \frac 1 2$. I don't have a proof for that, just numerical evidence.

Truth of these statements, expressions for $l$ and $L$?

-
Take $b=c=d=1, a=x-1$. –  vadim123 Jul 14 '14 at 18:52
So what is your question? –  Crostul Jul 14 '14 at 18:55
@Crostul I'm trying to find a) whether these two statements are always true, under the given conditions, and b) expressions for the functions $l$ and $L$. vadim123: Thank you, I should have thought of that. Is it possible to get an expression for $l$? –  Laertes Jul 14 '14 at 18:59
@AdamHughes: He meant that for any $x$, the statement is true because $b,c,d$ can be $1$ and $a$ can be $x-1$. –  Laertes Jul 14 '14 at 19:19
If you allow either $b$ or $d = 1$, then for any $x > 4$, you can pick $a = \lceil \sqrt{x}\rceil - 1$, $b = 2$, $c = x - a^2$ and $d = 1$. this leads to a bound $$l(x) \le \max( a, c ) \le 2\lceil\sqrt{x}\rceil -1$$ –  achille hui Jul 14 '14 at 19:59

For the second problem, if $u,v$ are relatively prime, then every integer can be written as $xu+yv$, if we allow $x,y\in\mathbb Z$. This is because there exists a solution of $xu+yv=1$ with $x,y\in\mathbb Z$ and we can simply multiply this. Next, if $n=xu+yv$, then also $n=(x\pm v)u+(y\mp u)v$ so that a solution with $0<x\le v$ must exist. We conclude that a solution with $x,y>0$ exists for all $n>uv$.

Now for given $n\ge 10$, let $d=\lfloor \log_2\sqrt {n-1}\rfloor$ and let $b=\lfloor \log_3\frac {n-1}{2^d}\rfloor$. Then

• $n\ge 10$ implies $d\ge 1$ as required
• $2^d\le \sqrt{n-1}$ implies $\frac{n-1}{2^d}\ge\sqrt{n-1}\ge3$ and hence $b\ge1$ as required.
• By the preceding paragraph, there exist integers $a,c>0$ such that $a\cdot 3^b+c\cdot 2^d=n$.
• From $2^{d+1}> \sqrt{n-1}$ we conclude $c<\frac n{2^d}< \frac{2n}{\sqrt{n-1}}=O(\sqrt n)$
• From $3^{b+1}>\frac{n-1}{2^d}\ge\sqrt{n-1}$, we conclude $a<\frac{n}{3^b}<\frac{3n}{\sqrt{n-1}}=O(\sqrt n)$.

This confirms your observation that $L=O(\sqrt n)$ (and you can extract an exact expression from the above).

You cannot expect any substantially better bound: Assume $L=O(n^\alpha)$ with $\alpha<\frac12$, say $L\le Cn^\alpha$ for all $n>n_0$. For $N\in\mathbb N$, there are about $CN^{2\alpha}\log_2 N\log_3N$ tuples $(a,b,c,d)$ with $a,c\le CN^\alpha$ and $3^b, 2^d<N$. For large $N$, we have $CN^{2\alpha}\log_2 N\log_3N\ll N$, hence there are not enough tuples to provide solutions for all $n$ with $n_0<n\le N$.

-


text:
# Tag Info

15

Now that you've got $K_{eq}$, you need to take a look at the expression for the equilibrium constant, which is $$K_{eq} = \frac{a_{\text{ice}}}{a_{\text{water}}},$$ where $a$ is the activity of each species. For a pure liquid and a pure solid, the activity is defined as 1, so the expression on the right hand side of that equation (usually referred to as the ...

15

For a chemical reaction, $$\Delta G = -R T \ln K_{eq}$$ $$\implies K_{eq} = e^{\frac{-\Delta G}{RT}}$$ Thus, in order for a chemical reaction to have $K_{eq}=1$*, it is necessary to have $\Delta G = 0$ For some chemical reactions**, we can achieve this by adjusting the temperature and/or pressure. Thus it is possible. However, it's unlikely that the ...

9

Actually you need concept of coupled differential equations but I suppose you can do a 'trick' to avoid it: $$\frac{dA}{dt} =- a A + b B$$ $$- \frac{dB}{dt} = - aA + bB$$ Are the form of the two equations, you may subtract the two equation to find $\frac{dA}{dt} = - \frac{dB}{dt}$, and by integration: $A(t)+B(t)=C$, setting $t=0$ , we find $A(0)+B(0)=2=... 8 Your textbook's text is correct but the reaction image is incomplete. Reaction of chlorobenzene with ammonia in presence of both the cuprous chloride and cuprous oxide forms aniline but via two different reactions. Your textbook only mentions one of the reactions in the image. In the reaction involving the oxide, cuprous oxide is a reactant and is consumed ... 7 I will start by addressing the posted question: the "state of a substance at a specific temperature and pressure" refers to "the most stable phase of the homogeneous substance at the specific p and T". The phase is indicated in the phase diagram by looking up the appropriate T,p point. At$\pu{20 ^\circ C}$and 1 atm pressure that would ... 5 [OP] The equilibrium constant would not include the solid$\ce{I2}$, but why is this? Let me explain this with a different example. If you have a saturated solution (e.g. lemonade with too much sugar in it) it is at equilibrium. If you add more sugar, the lemonade does not get sweeter. That tells you that the amount of solid does not matter (as long as ... 5$\ce{HO2-}$is a weaker base than$\ce{NaOH}$. but$\ce{O2^2-}$is much stronger base than$\ce{HO2-}$and does not occur in water solutions in significant amount. But its salts ( sometimes in the form of hydrates ) can be precipitated at highly alkaline solutions of hydrogen peroxide. Additionally, lack of product presence supports the respective ... 5 The essential condition is the compound thermal stability. If it decomposes below its melting point, it does not have a triple point. If it decomposes before properties of gaseous and liquid phases converge to each other, it has just estimated, extrapolated triple point. Beyond CP we talk about supercritical fluid, it kind of shares many properties of both ... 5 The rate forwards and back are the same at equilibrium and differ before this is attained. Let the reaction be$A+B=C$. If each concentration is suddenly halved then the reaction is not necessarily any longer at equilibrium and in this case the rates forwards and back differ until a new equilibrium is reached. The figure shows this. The red curve is the ... 5 We all are familiar with the standard state as 1 bar pressure and 273 K.The standard state can be any well defined state. You can define it in any way that is convenient for you. There is no universal definition. We may want to choose different references depending on the calculation we are doing, so we can choose different standard states. In thermodynamics ... 5 Chemical reactions may be performed in sequences (equally known as batches), or continuously. The two differ e.g., by residence time of the reagents and products in this container. The former form is like what you are used to, e.g., in the kitchen. You charge your pot with potatoes, close it, cook them, open it; done. In a flow reactor (the later form), ... 4 Note that all salts, used in humidity calibration, dissociate to their respective hydrated cations and anions when dissolved. From the solution crystallizes the sesquihydrate$\ce{K2CO3·\frac 32 H2O}$("potash hydrate"). Heating this solid above 200 °C gives the anhydrous salt (Wikipedia). It seems you have anhydrous carbonate, that hydrates ... 4 Recall that$\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$Where$\ce{[OH-]}$is the concentration of hydroxide ions. Suppose a solution has$10^{-7}$moles of$\ce{[OH-]}$ions. Then this implies$\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$and$\mathrm p\ce{OH} = -\log_{10}\space [10^{-7}]$which is equal to$7$. Suppose a solution has$10^{-4}$... 4 An exothermic reaction has a reduced equilibrium constant at higher$T$because while the contribution of the change in the entropy of the system is a fixed quantity (for a small$T$change), the effect of transferring heat to the surroundings is reduced at higher$T$(because it causes a smaller change in the entropy of the surroundings). Mathematically ... 4 Let's say you have a reaction that has reached equilibrium. The reaction quotient$Q$will be equal to the equilibrium constant$K$, and the Gibbs energy of reaction will be zero. Now, you decide to change the standard state (without touching your experiment, which is still at equilibrium). For example, if you are a biochemist, you could say that the pH at ... 4 Yes. The universal condition for equilibrium is that, given the current constraints on the system, no further spontaneous change can take place. I.e., that the entropy of the universe (system + surroundings) is maximized. At constant temperature and pressure, a maximization of the entropy of the universe corresponds to a minimization of the Gibbs free ... 4 First of all you made a slight mistake one of the equations is wrong First equilibrium reaction, $$\ce{Ag(aq)+ + e- <=> Ag(s)}$$ Writing the Nernst equation for this reaction $$\ce{E_{Ag|Ag^+} = E_{Ag|Ag^+}^0 - \frac{RT}{F}ln\frac{\ce{[Ag]}}{\ce[Ag^+]}}$$ Second equilibrium reaction, $$\ce{AgCl(s) + e- <=> Ag(s) + Cl-(aq)}$$ Writing the Nernst ... 4 The equation$3$is correct because in the reaction, one gaseous molecule gets dissociated and produces a total of$2$gaseous molecules. The number of gaseous molecules increases if the dissociation reaction proceeds. At a given pressure, the volume occupied by$2$gaseous molecules is twice the volume occupied by$1$gaseous molecule 4 It is the question of equilibrium versus nonequilibrium yield and production rate. Maximizing the ammonia output flow is not the same as maximizing the reaction yield. Approaching the equilibrium yield would lead to just minimal production. OTOH, the top production rate would be near zero reaction yield and unlimited input flow, what is not technically ... 3 Nothing would happen. The reaction would not shift to produce more product. Let's suppose you allow this reaction to equilibrate in a sealed container. Equilibrate means you've reached the equilibrium value for$p_{\text{O}_2}$at that temperature. Now further suppose you were able to remove all the product without changing the free volume (say you ... 3 I am using a saturated sodium chloride solution as an example here because it is more common and easier to reproduce at home. This system is at equilibrium, with salt dissolving and crystallizing at the same rate. If you take away the solid NaCl from a saturated solution, both the forward and the reverse reactions stop. Obviously, salt crystals are no longer ... 3 Note that under the given conditions, methanol cannot exist in the liquid state, however since you are given in the problem that methanol is obtained in liquid state, you will only consider the equilibrium (for problem solving purpose) $$\ce{CO(g) + 2 H2(g) <=> CH3OH(l)}$$ And since$K_p$is related to$K_c$by the relation $$K_p = K_c(RT)^{\Delta n_\... 3 I'd like first to answer your question: My question is, when writing the formula for K_\mathrm{D}, is the organic phase always in the numerator and the aqueous phase in the denominator? The answer is yes. The IUPAC Recommendations 1993 (Ref.1) defines Partition Ratio (K_\mathrm{D}) as follows (also see the Goldbook): Partition Ratio (K_\mathrm{D}): ... 3 Esterification \ce{R_1-COOH + HO-R_2 <=> R1-COO-R2 + H2O} has the equilibrium constant, expressed in compound activities:$$K = \frac{a_\mathrm{ester} \cdot a_\mathrm{\ce{H2O}} }{ a_\mathrm{acid} \cdot a_\mathrm{alkohol}}$$The water activity is decreased by dissolved salts by 2 ways: decreasing the molar fraction of water by dissolved salt ... 3 You don't need gas laws, temperature or K_p. You overcomplicated the problem: K_c can be found by following its definition with an RICE table:$$ \begin{array}{lccccc} &\text{R}\ce{&2 HI(g) &<=> &H2(g) &+ &I2(g)} \\ &\text{I} & 2c_0 && 0 && 0 \\ &\text{C} & -2\alpha c_0 && \alpha c_0 &... 3 Option C is the right answer Solubility of AgCN will be equal to sum of concentrations of Ag+ and [Ag(CN)2]- 3$K_p$is defined in terms of free energies of the gaseous substances in their standard states at 1 atm pressure so must be independent of pressure, i.e. is a constant. With concentrations we use$c_i=n_i/V$for species$i$and for example, if$K_c=c_\ce{A}^ac_\ce{B}^b/(c_\ce{C}^cc_\ce{D}^d)$then for an ideal/perfect gas$K_c=K_p(RT)^{-\Delta n}$where$\...

3

Initially in the solution the following equilibria exist $$\ce{HA <=> H+ +A-}$$ and $$\ce{H2O <=> H+ +OH-}$$ Let initial concentration of $\ce{HA}$ be $a_1$ and that of $\ce{A-}$ be $a_2$. Now to a litre of this buffer solution, let's say you have added $\mathrm dn$ moles of strong acid to it. The acid dissociation constant of $\ce{HA}$ be $K_a$ ...

3

I will continue with the data of $[\ce{H2CO3}] = \pu{10^{-4.97} M}.$ Now, as the $K_2$ of $\ce{H2CO3}$ is very small as compared to its $K_1,$ we can assume that all the $\ce{H+}$ will come from the first dissociation of $\ce{H2CO3}.$  \begin{array}{lccc} & \ce{&H2CO3 &<=> &H+(aq) &+ &HCO3-(aq)} \\ &\text{Initial} & ...

3

The $\ce{Na+}$ ion is a spectator ion in this case. It doesn't participate in the chemical reaction. It is found unchanged on both the reagent and product sides (red top-right and green bottom-right reaction formulas), so it can be canceled out to get the net reaction (red left). From there, he did the usual acid-base calculation to get a formula for \$\ce{Kb}...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
1
JEE Main 2021 (Online) 20th July Morning Shift
+4
-1
AC voltage V(t) = 20 sin$$\omega$$t of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2. The amplitude of the oscillating displacement current for the applied AC voltage is _________. [Take $$\varepsilon$$0 = 8.85 $$\times$$ 10$$-$$12 F/m]
A
55.58 $$\mu$$A
B
21.14 $$\mu$$A
C
27.79 $$\mu$$A
D
83.37 $$\mu$$A
2
JEE Main 2021 (Online) 18th March Evening Shift
+4
-1
In a series LCR circuit, the inductive reactance (XL) is 10$$\Omega$$ and the capacitive reactance (XC) is 4$$\Omega$$. The resistance (R) in the circuit is 6$$\Omega$$. The power factor of the circuit is :
A
$${1 \over 2}$$
B
$${{\sqrt 3 } \over 2}$$
C
$${1 \over {\sqrt 2 }}$$
D
$${1 \over {2\sqrt 2 }}$$
3
JEE Main 2021 (Online) 18th March Evening Shift
+4
-1
The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of resistance R, inductance L is connected to a battery, is :
A
infinite
B
$${L \over R}$$ ln10
C
$${L \over R}$$ ln2
D
$${L \over R}$$ ln5
4
JEE Main 2021 (Online) 18th March Morning Shift
+4
-1
In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :
A
The bandwidth of resonance circuit will increase.
B
The resonance frequency will increase.
C
The quality factor will increase.
D
The quality factor and the resonance frequency will remain constant.
JEE Main Subjects
Physics
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Modern Physics
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text:
# Injection iff Monomorphism in Category of Sets

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## Theorem

Let $\mathbf{Set}$ be the category of sets.

Let $f: X \to Y$ be a morphism in $\mathbf{Set}$, i.e. a mapping.

Then $f$ is an injection if and only if it is a monomorphism.

## Proof

### Necessary Condition

Suppose that $f$ is injective.

Suppose further that we have mappings $g, h: Z \to X$ such that $g \ne h$.

Then necessarily there exists some $z \in Z$ such that $g \left({z}\right) \ne h \left({z}\right)$ by Equality of Mappings.

As $f$ is injective, it follows that:

$\map f {\map g z} \ne \map f {\map h z}$

which, again by Equality of Mappings, means that $f \circ g \ne f \circ h$.

Hence $f$ is monic, by the Rule of Transposition.

$\Box$

### Sufficient Condition

Suppose that $f: X \rightarrowtail Y$ is a monomorphism.

By definition of injection, it will suffice to show that:

$x \ne x' \implies \map f x \ne \map f {x'}$

To this end, consider a singleton $\set a$, and define:

$\bar x: \set a \to X, \map {\bar x} a := x$
$\bar x': \set a \to X, \map {\bar x'} a := x'$

In particular, $\bar x \ne \bar x'$, and so, $f$ being monic, we deduce:

$f \circ \bar x \ne f \circ \bar x'$

It follows that it must be that:

 $\ds \map f x$ $=$ $\ds \map f {\map {\bar x} a}$ $\ds$ $\ne$ $\ds \map f {\map {\bar x'} a}$ Equality of Mappings $\ds$ $=$ $\ds \map f {x'}$

Hence $f$ is injective.

$\blacksquare$


text:
# Sequence that converges

Lee33

## Homework Statement

I was given this homework problem:

Show that if ##a_1,a_2, ... ,## is a sequence of real numbers that converges to ##a##, then $$lim_{n\to \infty}\frac{\sum^n_{k=1} a_k}{n}=a.$$

I was provided a solution but my book never went over such examples or the concrete steps to solve such a problem. I am wondering what are the basic first steps to solving these types of problems?

And if possible, where can I find practice problems like these online? I searched but I couldn't find any.

Last edited:

## Answers and Replies

Homework Helper
Gold Member
Dearly Missed
Hint:
Remember that for every n, you may rewrite:
$$na=\sum_{k=1}^{n}a$$

Homework Helper

## Homework Statement

I was given this homework problem:

Show that if ##a_1,a_2, ... ,## is a sequence of real numbers that converges to ##a##, then $$lim_{n\to \infty}\frac{\sum^n_{k=1} a_k}{n}=a.$$

I was provided a solution but my book never went over such examples or the concrete steps to solve such a problem. I am wondering what are the basic first steps to solving these types of problems?

There are various techniques for limit problems, but since this problem asks you to start with an arbitrary convergent sequence $(a_k)$ the only one which will work is to use what you know about $(a_k)$: for all $\epsilon > 0$ there exists $K \in \mathbb{N}$ such that if $k \geq K$ then $a - \epsilon < a_k < a + \epsilon$.

That suggests taking an arbitrary $\epsilon > 0$ and its corresponding $K$ and splitting the sum as follows:
$$\frac1n \sum_{k=1}^n a_k = \frac1n \sum_{k=1}^{K-1} a_k + \frac1n \sum_{k=K}^n a_k$$
(You are interested in the limit $n \to \infty$, so at some stage you will have $n > K$ and you may as well assume that to start with.)

Your plan is to show that
$$a - \epsilon \leq \lim_{n \to \infty} \frac1n \sum_{k=1}^n a_k \leq a + \epsilon$$
and since $\epsilon > 0$ was arbitrary it must follow that
$$\lim_{n \to \infty} \frac1n \sum_{k=1}^n a_k = a$$
as required.

Mentor
I would use ϵ/2 in one step instead of ϵ, that makes the inequalities easier to show.

Lee33
Thank you very much, pasmith! That cleared some issues I had, thanks for clarifying it for me.


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# A study of ferroelectric thin films deposited on a $LaNiO_3$ barrier electrode by nebulized spray pyrolysis

Murugavel, P and Sharma, Rajat and Raju, AR and Rao, CNR (2000) A study of ferroelectric thin films deposited on a $LaNiO_3$ barrier electrode by nebulized spray pyrolysis. In: Journal of Physics D-Applied Physics, 33 (8). pp. 906-911.

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## Abstract

Thin films of ferroelectric $PbTiO_3$ (PT) and $Pb(Zr_{0.5}Ti_{0.5})O_3$ (PZT) as well as antiferroelectric $PbZrO_3$ (PZ) have been prepared on $LaNiO_3/SiO_2/Si$ substrates by nebulized spray pyrolysis (NSP) of metal-organic precursors. The metallic $LaNiO_3$ (LNO) electrode layer was also deposited by NSP. The ferroelectric films obtained show satisfactory morphology and desirable dielectric properties. Typical values of the coercive field, remnant polarization and dielectric constant (300 K) for the $PT/LNO/SiO_2/Si$ film are 170 kV $cm^{-1}$, 22 \mu C $cm^{-2}$ and 210, respectively, with the corresponding values for the $PZT/LNO/SiO_2/Si$ film being 120 kV $cm^{-1}$, 13 \mu C $cm^{-2}$ and 540, respectively. The $PZ/LNO/SiO_2/Si$ film shows typical antiferroelectric characteristics including the electric-field induced reversible antiferroelectric—ferroelectric transition. The various films deposited on $LNO/SiO_2/Si$ by NSP are comparable in all respects to those prepared on $Pt/Ti/SiO_2/Si$ by the same technique.

Item Type: Journal Article Copyright of this article belongs to Institute of Physics. Division of Chemical Sciences > Materials Research Centre 31 Jul 2008 19 Sep 2010 04:48 http://eprints.iisc.ernet.in/id/eprint/15355


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# Human Genome Sequencing Cost in 2026

### Question

A genome consists of all of the DNA contained in a cell's nucleus. DNA is composed of four bases (abbreviated G, A, T, and C), with the biological information encoded within DNA determined by the order of those bases. Sequencing a genome involves using various techniques to ascertain the order in which these bases are present in the DNA of an individual organism.

The first human genome sequencing was completed around 2001. Since then, technological advancements have enabled the cost of the procedure to drop rapidly. According to the National Human Genome Research Institute, sequencing a human genome cost over US$10 million in 2006, and only around US$562 in 2021. The cost data, graphs, and methods can be viewed here.

What will be the cost of sequencing a whole human genome in 2026 (in 2021 USD)?

This question will resolve as the cost to sequence a whole human genome in the year 2026, accrding to the US National Human Genome Research Institute. To even out erratic estimates, this question resolve as the average of all values reported for 2026 (expected to be quarterly). This value will be indexed to 2021 USD using the CPI. To account for data revisions, this will resolve based on the latest report published by NHGRI by the end of 2027.

If NHGRI does not release any updates in 2027, this question may use any successor organization with similar data or methodology. If no such successor is available, the question may resolve ambiguously.

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