Sponsor Area
This reaction follows second order kinetics.
So that, the rate equation for this reaction will
Rate, R = k[X]2 .............(1)
Let initial concentration is x mol L−1,
Plug the value in equation (1)
Rate, R1 = k .(a)2
= ka2
Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
Plug the value in equation (1) we get
Rate, R2 = k (3a)2
= 9ka2
We have already get that R1 = ka2 plus this value we get
R2 = 9 R1
So that, the rate of formation will increase by 9 times.
Rate = k[A]2
If concentration of X is increased to three times,
Rate = k[3A]2
or Rate = 9 k A2
Thus, rate will increase 9 times.
The activation energy for the reaction
2HI (g) → H2 + I2(g)
is 209.5 kJ mol–1 at 581 k. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
2HI (g) → H + I2(g)
Activation energy, Ea = 209.5 kJ mol−1
Multiply by 1000 to convert in j
Ea= 209500 J mol−1
Temperature, T = 581 K
Gas constant, R = 8.314 JK−1 mol−1
According to Arhenious equation
K = A e –Ea/RT
In this formula term e –Ea/RT represent the number of molecules which have energy equal or more than activation energy
Number of molecules = e –Ea/RT
Plug the values we get
Number of molecules
taking antilog of we get
1.47 x 10-19
r= k[A]n =Kan
When concentration is increased three time [ A] =3a
therefore,
Order = 3.
It is first order of rection.
t = 0 0.01 mol 10 mol 0 mol 0 mol
t 0 mol 9.9 mol 0.01 mol 0.01 mol
The concentration of water does not get altered much during the course of the reaction. So, in the rate equation
the term [H2O] can be taken as constant. The equation, thus, becomes
Rate = k [CH3COOC2H5]
where k = k′ [H2O]
and the reaction behaves as first order reaction
Sponsor Area
The units of rate constant for the nth order reaction equal to (mol L–1)1 – n min–1.
The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.
Order of reaction = X + Y.Rate order =
Sponsor Area
Rate = k [A]1/2 [B]3/2 So order = 1/2 + 3/2 = 2, i.e., second order.
For nth order reaction
For first order reaction, k is independent of the concentration of the reactants.
The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t1/2.
(i) Its molecularity is 2.
(ii) Its order is 1.
The unit of a first order rate constant is s–1 therefore
k = 3 × 102 h-1 represents a first order reaction.
(i) Hydrolysis of ester increases with time due to auto catalytic action of CH3 COOH formed.
(ii) Decolorisation of KMnO4 by oxalic acid becomes fast due to auto catalytic action of Mn2+.
Sponsor Area
The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.
Order of a reaction can be 0, 1, 2, 3 and even a fraction.
For example for a reaction
Rate = k [A]x [B]y
order = x + y
The various factors which affect the rate of chemical reactions are:
(i) Concentration of reactants
(ii) Temperature of reaction
(iii) Presence of catalyst
(iv) Nature of reactants
(v) Surface area
(vi) Exposure to radiations.
(c) Rate = k[CH3CHO]3/2
Order w.r.t., CH3CHO = 3/2 = 1.5
Overall order of reaction = 1.5
Rate constant,
The dimensions of rate constant, k are L1/2 mol–1/2 s–1.
(d) Rate =
Order w.r.t.
Order w.r.t.
Overall order of reaction = 1+0.5 = 1.5
Rate constant,
The dimensions of rate constant, k are L1/2 mol–1/2 s–1.
Nitric Oxide. NO reacts with oxygen to produce nitrogen dioxide.
2NO(g) + O2(g) → 2NO2(g)
The rate law for this reaction is
rate = k[NO]2 [O2]
Propose a mechanism for the above reaction.
Rate of the reaction depends upon the slowest step of the elementary processes.
Mechanism is:
NO + O2 → NO3 (fast) ...(i)
NO3 + NO → NO2 + NO2 (slow) ...(ii)
Rate law is: Rate k[NO3] [NO] but NO3 is essentially NO + O2
So, rate = k[NO]2 [O2].
Activated complex is the intermediate compound formed by reactants, which is highly unstable and readily changes into product. Those reactants which possess activation energy and collide in proper orientation can form activated complex which can easily form products.
Lower the activation energy, more easily activated complex will be formed and faster will be the reaction.
Activation energy = energy of activated complex - energy of reactants
.
The rate of reaction
2NO + Cl2 → 2NOCl
is double when concentration of Cl2 is doubled and it becomes 8 times when concentrations of both NO and Cl2 are doubled. Deduce the order of this reaction.
Energy of activation. The minimum energy over average energy which must be gained by the molecules before they could react to form products is called the energy of activation. It is denoted by Ea.
According to the Arrhenius theory, activation energy is independent of temperature. However, precise measurements indicate that the activation energy tends to decrease slightly with a rise in temperature.
A photo-chemical rection may be defined as a process or reaction which is initiated by light absorption.
A reaction between H2 gas and chlorine gas is a photo-chemical reaction. The proposed mechanism for this reaction is given as follows:
Primary Process:
Secondary Process:
(step II)
(step III)
Step II and III are called chain propagation steps. The chain termination takes place on the walls of the reaction vessel by the reaction.
Reaction coordinate represents the profile of energy change when reactants change into products.
Order of reaction with respect to Cl2 = 1.
Order of reaction with respect to NO = 2.
Overall order of reaction = 1 + 2 + 3.
(a)The rate i.e., dx / dt depends on the concentration of the reactant. Let dx / dt = k[A]1.
On taking log
On plotting log dx / dt versus log A is always a straight line with slope equal to 1.
(b) Example of a first order reaction.
Decomposition of nitrogen pentoxide (N2O5)
Hydrogenation of ethene is an example of first order reaction.
C2H4(g) + H2 (g) → C2H6(g)
Rate = k [C2H4]
Rate of production of water
Rate of disappearance of oxygen
Rate of disappearance of HI
Rate of appearance of
From the chemical equation,
Rate of production
Therefore,
The rate of the reaction is proportional to the
first power of the concentration of the reactant R. For example,
R → P
Again, C is the constant of integration and its value can be determined
easily.
When t = 0, A = [A]0, where [A]0 is the initial concentration of the reactant.
Therefore,above equation can be written as
ln [A]0 = –k × 0 + C
ln [A]0 = C
putting the value of C in equation (1), we get
The rate of reaction can be express by Dividing the individual rate expression by the coefficients in the balanced chemical equation.
(a) Show graphically how the rate of first order reaction vary with only one reactant depends concentration of reactant.
(b) Give one example of first order reaction.
(a)
above graph show that by increasing the concentration of reactant the rate of reaction also increase.
(b) Example of first order reaction is Hydrogenation of ethene is an example of first order reaction.
C2H4(g) + H2 (g) → C2H6(g)
Rate = k [C2H4]
(i) Define specific reaction rate.
(ii) Define Half-life period of a chemical reaction. Also obtain the expression for half-life period.
Or
Derive the general for of the expression for the half-life of a first order reaction.
(i) Specific reaction rate. This is also called rate constant or velocity constant. Specific reaction rate may be defined as the rate of reaction under specific conditions, when the product of concentration of the reactants is unity.
(ii) Half-life period of a chemical reaction in which the concentration of reactant is reduced to half of the intial value of concentration.
It is denoted by tl/2 or t0.5.
The rate equation for a reaction of first order is expressed as:
or
When
On substituting these values, we get
The relation (i) is the required expression for half-life period of first order reaction.
Flash photolysis: When we pass a powerful flash of short duration or laser beam, through the reaction mixture to initiate a reaction the atoms, ions or free radicals are formed.
These atoms, ions or free radicals formed can be identified by passing a second, Flash of light through the mixture immediately after the first flash.
For this the absorption spectrum of the mixture is monitored continuously at small regular intervals after first flash and the changes in the spectrum with time indicate the various processes occurring in the system.
It can also be studied by observing some other property like electrical conductance or magnetic property of the reaction mixture.
Sponsor Area
(ii) Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants. Consider the reaction,
At t = 0, the concentration of the reactant R = [R]0, where [R]0 is
initial concentration of the reactant.
Substituting in equation (1)
[R]0 = –k × 0 + I
[R]0 = I
Substituting the value of I in the equation (1)
[R] = -kt + [R]0.............(2)
Comparing (2) with equation of a straight line,
y = mx + c, if we plot [R] against t, we get a straight
line (Figure) with slope = –k and intercept equal
to [R]0.
Rate of reaction in terms of disappearance of the reactants and appearance of product.
2NO(g) + O2(g) → 2NO2(g)
The rate of reaction in terms of disappearance of the reactant and appearance of product.
H2(g) + l2(g) → 2HI(g)
The rate of the reaction in terms of disappearance of the reactant and appearance of product.
CO(g) + NO2(g) → CO2(g) + NO(g)
The rate of the reaction in terms of disappearance of the reactant and appearance of product.
2NO2(g) + F2(g) → 2NO2 F(g)
The rate expression is derived by step II of the mechanism, as it is the slower one
rate = k[NOBr2][NO] ...(i)
However, NOBr2 is an intermediate and thus its concentration should be replaced from equation (i)
From step (i),
Equilibrium constant,
...(ii)
Then by equation (i) and (ii)
Order of the reaction is 2+1 =3
The slow step is the rate determining step. In the slow step in this reaction, 1 molecule of NO3 (intermediate) and 1 molecule of NO combine to form the products. Therefore, rate of this reaction depends upon 1 concentration term of NO3 and 1 concentration term of NO.
∴ Molecularity of the reaction = 1 + 1 = 2.
Rate = K [NO3] [NO]
NO3 is an intermediate which is formed rapidly by the collision of 1 molecule of NO and 1 molecule of O2.
[NO3] ∝ [NO][O2]
Rate = K1 [NO3] [NO]
= K [NO] [O2] [NO]
= K [NO]2[O2]
Order of reaction is 2 + 1 = 3.
|
Molecularity |
Order of a reaction |
|
1. It is the total number of molecules of the reactants taking part in a single step of the reaction. |
1. It is the sum of the indices to which the concentration terms are raised in rate equation, for the reaction. |
|
2. It is always a whole number. |
2. It may be a whole number, a fraction or zero. |
|
3. It is obtained from the simple balanced chemical equation. |
3. It is obtained through experimentation and is dependent upon the rate for the overall reactions.
|
|
4.It is theoretical concept. |
4. It is experimental quantity. |
A reaction is first order in A and second order in B.
(i)
Rate constant of a reaction is equal to the rate of reaction when concentration of each of the reactant is unity.
Whether a reaction is of first order or not can be verified by using the analytical methods based on the rate equations. The reagents are allowed to react, the change of concentration is measured at different time intervals. For a first order reaction:
(i) Plot of log [A0]/[A] versus t gives a straight line passing through the origin.
(ii) Plot of log[A] versus t gives a straight line with an intercept log [A]0 and a slope = – k /2.303.
(iii) Plot of [A] versus t is exponential.
(iv) The value of k calculated by using kt = 2.303 log [A]0 / [A] for different time intervals come out to
be same.
(v) The half life period is independent of the initial concentration of the reactant.
In the reaction
2A + B + C → A2B + C,
there is no change in C, therefore its conc. does not affect the rate of the reaction.
Initial rate = k[A] [B]2
But [A] = 0.1 M,
[B] = 0.2 M
and k = 2 x 10–6 M–2 s–1
Therefore initial rate
Rate= [k] x [A] x [B]2
= 2 x 10–6 M–2 s–1 x (0.1 M) (0.2 mol M)2 = 8 x 10–9 ms–1
From the equation:
2A + B + C → A2B + C,
it is clear that when 2 moles of A are used then 1 mol of B is used in the same time. Therefore, when A has been reduced to 0.06 M (due to its 0.04 M has been reacted to 0.02 of B). Thus,
Conc. of A left = [A] = 0.06 M
Conc. of B left = [B] = [0.02 M – 0.02 M]
= 0.018 M
Rate = k[A] [B]2
= 2 x 10–6 M2 S–1 x (0.06 M) (0.18 M)
= 3.89 x 10–9 Ms–1.
|
t / s |
0 |
30 |
60 |
90 |
|
Ester / mol L–1 |
0.55 |
0.31 |
0.17 |
|
|
A/mol L–1 |
0.20 |
0.20 |
0.40 |
|
B/mol L–1 |
0.30 |
0.10 |
0.05 |
|
r/mol L–1S–1 |
5.07 x 10–5 |
5.07 x 10–5 |
1.43 x 10–4 |
What is the order of the reaction with respect of A and B?
In a reaction A and B, Let order of reaction w.r.t. A is x and w.r.t. B is y. Then the rate of reaction can be written as
rate = k[A]x [B]y
From given table data, 1 and 2 we can write
5.07 x 10–5 = k[10.20]x [0.30]y ...(i)
5.07 x 10–5 = k[0.20]x [0.10]y ...(ii)
Dividing (ii) by (i), we get
or
|
exp.
|
[A]/ |
[B]/M |
Initial rate of formation |
|
I |
0.1 |
0.1 |
|
|
II |
0.3 |
0.2 |
7.2 x 10–2 |
|
III |
0.3 |
0.4 |
2.88 x 10–1 |
|
IV |
0.4 |
0.1 |
2.40 x 10–2 |
|
exp.
|
[A]/ |
[B]/M |
Initial rate of formation |
|
I |
0.1 |
0.1 |
|
|
II |
- |
0.2 |
4.0 x 10–2
|
|
III |
0.4 |
0.4 |
- |
|
IV |
- |
0.2 |
2.0 x 10–2 |
Rate law for the reaction is given by:
Rate = k [A]1 [B]0 = k[A]
2.0 x 10–2 mol L–1 min–1 = k[0.1 mol L–1]
Rate constant,
Rate constant = k = 0.2 min–1.
(i) In experiment II Rate = k[A]
(ii) In experiment III
Rate = k[A]
= 0.2 min–1 x 0.4 M
= 0.08 min–1
= 8.0 x 10–2 M min–1
(iii) In experiment IV
Rate = k[A]
Thus the completed table is
|
exp.
|
[A]/ |
[B]/M |
Initial rate of formation |
|
I |
0.1 |
0.1 |
|
|
II |
0.2 |
0.2 |
4.0 x 10–2
|
|
III |
0.4 |
0.4 |
8.0 x 10-2 |
|
IV |
0.1 |
0.2 |
2.0 x 10–2 |
(a) Plot [N2O5] again t.
(b) Find the half life period for the reaction.
(c) Draw a graph between log [N2O5] and t.
(d) What is rate law?
(e) Calculate the rate constant.
(f) Calculate the half life period from K and compare it with (ii).
(b) Time taken for the concentration of N2O5 to change from 1.63 x 10–2 mol L–1 to half the value.
t0.5 = 1420 s [from the graph] (c) Plot of log (N2O5) vs. time.
|
t |
log10[N2O5] |
|
0 400 800 1200 1600 2000 2400 2800 3200 3600 |
- 1.7918 - 1.8665 - 1.9431 - 2.0315 - 2.1079 - 2.1938 - 2.2757 - 2.3752 - 2.4559 - 2.5376 |
Rate constant of reaction,
k = 60 S–1
t15/16 = ?
Rate constant of first order reaction is given by.
where 15 / 16th the reaction is over the
For the first order reaction the half life can be shown written as :
t1/2 = 0.693/k
As we have given k= 2.31 x 10-3s-1
putting this value in above equation, we get
t1/2 = 0.693/2.31 x10-3s-1 = 300s
Since it is first order, thus
Half -life of first order reaction can be given by
t1/2 =0.693/k
for first order reaction :
let a=100
x= 25 (given)
time =5min
applying in above equation, we get
thus for half life
t1/2 =0.693/0.05704
Ans. 12.14 min.
to completed 60% reaction
Ans. 112.7 min
The first order reaction can be
Ans. 16 minutes
For a given reaction
A + B → 2C
we have given
the rate of disappearance of A is 10-2 moles L–1 sec–1
(ii)
rate of expression can be written as
since rate of disappearance of A is given to be 10-2 moles L–1 sec–1
thus
rate of disappearance B 10-2 moles L–1 sec–1
(iii)
for Ans. (i) 10–2 mol L–1 sec–1 (ii) 2 x 10–2mole L–1 sec–1
For the first order reaction for the gases phase given by
we have give p1 = 84mm
total pressure =110mm
at the time = 16 min
thus applying in above formula
Activated complex is the intermediate compound formed by reactants, which is highly unstable and readily changes into product. Those reactants which possess activation energy and collide in proper orientation can form activated complex which can easily form products.
Lower the activation energy, more easily activated complex will be formed and faster will be the reaction.
Activation energy = energy of activated complex - energy of reactants.
Energy of activation. The minimum energy over average energy which must be gained by the molecules before they could react to form products is called the energy of activation. It is denoted by Ea.
According to the Arrhenius theory, activation energy is independent of temperature. However, precise measurements indicate that the activation energy tends to decrease slightly with a rise in temperature.
A reaction in which the rate is independent of the concentration of the reactants.
Instantaneous rate expression can written as
we have given that
3 x 10–2 h–1
since unit of first order reaction is s-1
thus it is first order reaction
First half reaction gets completed within 2 hours and next half reaction gets completed within next 2 hours. That means, t1/2 is indepedent of concentration. Hence, the order of reaction is first order reaction.
Unit of second order reaction is L mol–1 s–1
Activation energy of a reaction is defined as the extra energy over and above the average potential energy of the reactants, which must be supplied to the reactants to enable them to cross-over the energy barrier between the reactants and products. It is denoted by EA.
It the time required for the concentration of a reactant to decrease to half of its initial value.
In some reactions, the rate is apparently independent of the reactant concentration. The rates of these zero-order reactions do not vary with increasing nor decreasing reactants concentrations. This means that the rate of thereaction is equal to the rate constant, k , of that reaction.
Reaction that are auto catalytic can have their rate increase with time i.e. as the product is formes the product start to work as catalyst and increase the reaction.
It is the expression, which shows how the reaction rate is related to concentration
Rate of equation
rate = k[A]0
any one of reactant is present in excess of amount in reaction.
CH3COOC2H5 + H2O ----> CH3COOH + C2H5OH
water is present in excess of amount.
The rate constant at 298 K = k.
When the temperature is raised from 298 k to 308 k.
Increase in rate constant
By using the equation
We have,
but
According to the Arrhenius equation
K = Ae–Ea/R
In absence of catalyst k1 = Ae–100/RT
In presence of catalyst k2 = Ae–25/RT
or
or
Since Rate k[ ]n at any temperature for a reaction n and concentration of reactants are same and temperature changes.
(a) The decomposition of N2O5(g) is a first order reaction with a rate constant of 5 x 10–4 sec–1 at 45°C.
i.e., 2N2O5(g) = 4NO2(g) + O(g)
If initial concentration of N2O5 is 0.25, calculate its concentration after two minutes. Also calculate half life for the decomposition of N2O5(g).
(b) For an elementary reaction: 2A + B → 3C The rate of appearance of C at time ‘t’ is 1.3 x 10–4 mol l–1 s–1. Calculate at this time:
(i) Rate of reaction (ii) Rate of disappearance of A.
Rate constant K = 5 x 10–4 sec. Initial concentration [A]0 = 0.25 M Final concentration [A]t =? Time taken by the reaction, t = 2 min.
For a first order reaction, rate constant (K) is given by
(b)
(i) The rate of appearance of C at time t
(ii) Rate of disappearance of A
We have given the slope value ,
Slope = -6670 k
Thus using the equation
The reaction:
SO2 + Cl2 SO2 + Cl2
is a first order reaction with k1 = 2.2 x 10–5 s–1 at 575 K. What percentage of a initial amount of SO2 Cl2 will get decomposed in 90 minutes when the reaction is carried out at 575 K?
The reaction
SO2Cl2 → SO2 + Cl2
is a first order reaction with k = 2.2 x 10–5 s–1 at 320° C. Calculate the percentage of SO2Cl2 that is decomposed on heating this gas for 30 minutes.
|
t/s |
0 |
100 |
200 |
300 |
|
p/pascal |
4.0 x 103 |
3.5 x 103 |
3.0 x 103 |
|
|
t(sec) |
P (mm of Hg) |
|
0 |
35.0 |
|
360 |
54.0 |
|
720 |
63.0 |
Calculate the rate constant.
|
Experiment |
Time/s-1 |
Total pressure/atm |
|
1 |
0 |
0.5 |
|
2 |
100 |
0.6 |
K1 = 4.5 x 103 S–1 , T1 = 273 + 10 = 283 K
Ea = 60 kJ mol–1 T2 = ?
K2 = 1.5 x 104S–1
We know that
The rate constant of reaction increases with increase of temperature. This increase is generally two fold to five fold for 10 K rise in temperature. This is explained on the basis of collision theory. The main parts of collision theory are as follows:
(i) For a reaction to occur, there must be collision between the reacting species.
(ii) Only a certain fraction of total collisions are effective in forming the products.
(iii) For effective collisions, the molecule must possess the sufficient energy (equal or greater
than threshold energy) as well as proper orientation.
On the basis above conclusions, the rate of reaction is given by
Rate =f x 2 (where f is the effective collisions and is total number of collisions per unit volume per second).
Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant k was proposed by Arrhenius. The equation, called Arrhenius equation is usually written in the form
...(i)
.
where A is a constant called frequency factor (because it gives the frequency of binary collisions of the reacting molecules per second per litre, E0 is the energy of activation, R is a gas constant and T is the absolute temperature. The factor e–Ea/RT gives the fraction of molecules having energy equal to or greater than the activation energy, Ea.
The energy of activation (Ea) is an important quantity and it is characteristic of the reaction. Using the above equation, its value can be calculated.
Taking logarithm or both sides of equation (i), we get,
If the value of the rate constant at temper-atures T1 and T2 are k1 and k2 respectively, then we have
Subtracting eqn. (i) from eqn. (ii), we get
or
Thus knowing the values of the constant k1 and k2 at two different temperature T1 and T2, the value of Ea can be calculated.
For a certain chemical reaction variation in the concentration in [R] vs time(s) plot is given below:
For the reaction write/draw
(i) What is the order of the reaction?
(ii) What are the units of rate constant K?
(iii) Give the relationship between K and t1/2 (half-life period).
(iv) What does the slope of the above line indicate?
(v) Draw the plot Vs. time(s).
Or
For a certain chemical reaction:
The experimentally obtained information is tabulated below:
The experimentally obtained information is tabulated below :
|
Exp. |
[A]0 |
[B]0 |
Initial rate of reaction |
|
1 2 3 4 |
0.30 0.60 0.30 0.60 |
0.30 0.30 0.60 0.60 |
0.096 0.384 0.192 0.768 |
For this reaction:
(i) Derive the order of reaction w.r.t. both the reactants A and B.
(ii) Write the rate law.
(iii) Calculate the value of rate constant K.
(iv) Write the expression for the rate of reaction in terms of A and C. 
For the reaction: A → B.
Let a is the initial concentration of A in g moles L–1 and (a – x) is the concentration in g moles L–1 after time t, then according to law of mass action
Rate of reaction (a - x)
Integrating the above equation, we get
when
or
or
or
Changing natural log to base 10, we get
Nature of the curve: Hypothetical variation of conc. of reactant [R] and product [P] during the course of reaction.
Fig. Instantaneous and average rate of a reaction.
Answer the following questions on the above curve for a first order reaction A → P.
(a) What is the relation between slope of this line and rate constant.
(b) (i) Calculate the rate constant of the above reaction if the slope is 2 x 10–4 s–1.
(ii) Derive the relationship between half life of a first order and its rate constant.
(a) Slope = – k
or k = – 2 x 10–4 mol–1 Ls–1.
(ii) Consider the following first order reaction
A → P
at t = 0 a 0
at t = t (a – x) x
Suppose a is the initial concentration of reactant. After time t, x gm mol lit–1 is changed to P. According to law of mass action the rate of reaction at time t is directly proportional to the concentration of A at that instant i.e., (a – x)
Hence
is the rate and K is the rate constant, after rearranging the above equation, we get
...(i)
On integrating the above equation
We get
When t = 0, x = 0 hence – In a = constant substituting this value of constant in eqn. (ii) we get
A.
Order of reaction may be fractional number.B.
All first order reactions are unimolecular.C.
All radio active distingegrations follow first order reaction.D.
For a first order reaction, the rate of the reaction doubles as the concentration of the reactant(s) doubles.E.
Molecularity and order of reaction can be 0, 1,2,3 etc.B.
Decreases as the reaction proceedsB.
In which concentration of the reactants does not change with time.B.
Rate is directly proportional to the number of collisions per secondNitrogen dioxide (NO2) reacts with fluorine (F2) to form nitryl fluoride (NOzF) according to the reaction
(i) Rate of formation of
(ii) Rate of disappearance of
(iii) Rate of disappearance of
Rate of reaction =
Also, Because
Dependence of rate on concentration of reactants: The rate of a chemical reaction at a given temperature may depend on the concentration of one or more reactants and sometimes on products. The representation of rate of a reaction in the terms of the concentration of the reactants is given by rate law. The rate for a given reaction is established by experimental study of the rate of reaction over a wide range of concentration of the reactants and products. Rate law expression differs for the same reaction under different experimental conditions. Rate constants and order of reaction:
(i) Rate constant or specific reaction rate: It is the rate of reaction when the concentration of each reactant is 1 mol/L. For a given reaction it is constant at a particular temperature and is independent of the concentration of reactants. The units of the rate constant of a reaction depends on the order of the reaction. For an nth order of reaction,
For zero order of reaction, units of k is mol L–1 time–1, for first order reaction, unit of k is time–1, for second order reaction, unit of k is L mol–1 time–1.
In terms of gaseous reactions, concentration is expressed terms of pressure having units of atmosphere. Let us consider the general reaction: aA + bB → Products
where A and B are the reactants and a and b are the stoichiometric coefficients in the balanced chemical equations.
The rate law is written as,
Rate = Δ[A] / Δ t = k[A]α [B]β
where k is called the rate constant. Rate constant (k) is the constant of proportionality within the empirical rate law linking the rate of reaction and concentration of reactants involved in the reaction. The rate law can be written in the form
Rate ∝ [A]α[B]β
The exponents ‘α’ and ‘β’ in the rate law indicate how sensitive the rate is to change in [A] and [B] and they are usually unrelated to the coefficients a and b in the balanced equation. In general, exponents are positive. But for complex reactions it can be negative, zero or even fractions. If exponent is one, it means rate depends linearly on the concentration of the reactant. If concentration of A is doubled, rate is also doubled. This means a = 1. If α = 2 and [A] is also doubled, rate increases by the factor of 4(z2). When exponent is zero {[A]0 = 1}, rate is independent of concentration.
For the reaction 3H2(g) + N2 (g) → 2NH3(g)
Rate of reaction =
The gas phase decomposition of acetaldehyde
CH3CHO(g) → CH4(g) + CO(g)
at 680 K is observed to follow the rate expression
Thus,
(i) Units of rate of reaction =
(ii)
Rate = k[H2] [NO]2
The order of each reactant
Order with respect to H2 = 1
Order with respect to NO = 2.
Overall order = 1 + 2 = 3.
|
S.No. |
A(M) |
B(M) |
Initial rate (M) |
|
1. |
1.00 |
1.00 |
1.2 x 10–2 |
|
2. |
1.00 |
2.00 |
4.8 x 10–2 |
|
3. |
1.00 |
4.00 |
1.9 x 10–1 |
|
4. |
4.00 |
1.00 |
4.9 x 10–2 |
Assuming that rate law can be written as
Determine the value of
We have to assume that law canbe written as
Thus
or
From reaction no. (iv) and (i)
or
The order of the reaction is first order with respect to A and second order with respect to B. Overall order of the reaction is 3.
We have given intial concentration 2x10-3and 1x10-3.
the intial concentration equal to 2.40 x 10-4 Ms-1 and 0.60 x10-4 MS-1
Let the rate be =
From the trial (i) and (ii), we get
or
Thus reaction is of second order
The decomposition of H2O2 in basic solution is first order in H2O2.
2H2O2(aq) → 2H2O2 (l) x O2(g)
the rate constant is 1.6 x 10–5 s–1 at 25°C and initial concentration of H2O2 is 0.20 M.
(a) What is the concentration of H2O2 after 2 hrs.
(b) How long will it take for H2O2 concentration to drop to 0.08 M.
(c) How long will it take for 90% of H2O2 to decompose?
(a)
or
or
(b)
Table 4.2. Integrated rate laws for the reactions of simpler order.
Total pressure P1 = PA + PB + Pc (pressure units)
pA, pB and pc are the partial pressures of A, B and C respectively. When x is the amount of A converted into products, when pi is the initial pressure at time t = 0
where
Pseudo first order reaction: Although in most reactions, order and molecularity are same, there are certain reactions whose order and molecularity differ. For example, hydrolysis of
sugarcane,
Molecularity of this reaction is 2 but its order 1 because its rate depends only on the concentration of surcrose. The concentration of water remains is very high and does not change during the reaction (i.e., concentration of water remains practically constant throughout the reaction). Such reactions are known as pseudo-unimolecular or pseudo first order reactions. Other examples, of pseudo-unimolecular reaction is the acidic hydrolysis of esters where water
remains in excess.
Although it is termolecular (molecularity = 3) reaction, its order is one as concentration of H+ and H2O+ remains constant during reaction. Hydrolysis of organic chlorides is also an example of first order reaction small water (one of the reactants) is again in large excess and its concentration remains constant throughout the reactions.
Thus when one of the reactants is present in large excess, the second order reaction conforms to the first order and is known as a pseudo-unimolecular reaction.
Reaction between acetic anhydride and excess of ethanol to form ester and conversion of N-Chloroacetanilide to p-chloroacetanilide are also examples of pseudo-unimolecular reactions.
|
t in seconds |
0 |
900 |
1800 |
|
Cone. of A |
50.8 |
19.7 |
7.62 |
Prove that the reaction is of first order of A to decompose to one-half.
For the first order reaction can given by
At t = 900 seconds,
(i)
(ii)
Since the value of k is both case is almost same thus it is first order reaction.
| t/sec | 1242 sec | 2745 sec | 4546 sec |
| At Conc. | -27.80ml | -29.70ml | -31.81ml |
Here [ A]0 is proportional to the alkali consumed between t = &&& and t = 339 sec, i.e., [A]0 = (39.81 mL – 26.34 mL) = 13.47 mL [A]t is proportional to the alkali consumed at &&& minus the alkali consmed at the specified time. Thus,
[A], at 1242 sec = 39.81 mL – 27.80 mL = 12.01 mL
[A], at 2745 sec = 39.81 mL – 29.70 mL = 10.11 mL
[A], at 4546 sec = 39.81 mL – 31.81 mL = 8.00 mL
Substituting the values of t, [A]0 and [A]t in the first-order rate equation
we get the following values of k
(i)
(ii)
(iii)
Since the three values of k derived from first-order reaction are close to being identical, hydrolysis of methyl acetate is a first order reaction.
|
t (in mitt) |
0 |
135 |
339 |
683 |
1680 |
|
C (mol L–1) |
2.08 |
1.91 |
1.68 |
1.35 |
0.57 |
Find the order of reaction and calculate its rate constant.
Catalytic decomposition of nitrous oxide by gold at 900°C at an initial pressure of 200 mm was 50% in 53 minutes and 73% in 100 minutes.
(i) What is the order of reaction?
(ii) How much will it decompose in 100 minutes at the same temperature but at an initial pressure of 600 mm?
(i) Let [A]0 = 100.
Then, [A]t at 53 minutes = (100 – 50) = 50 and [A], at 100 minutes = (100 – 73) = 27.
Substituting t and concentration values in the integrated rate equation for first-order reaction.
At t = 53 min,
At t = 100 min,
Since the value of k is constant, the order of reaction is 1.
(ii) For a first order reaction, the time required to complete any fraction is independent of the initial concentration of reactant.
∴ 73% of N2O will decompose when the initial concentration is 600 mm which corresponds to a pressure of
|
t/s |
0 |
100 |
200 |
300 |
|
p/pa |
4.00 x 103 |
3.50 x 103 |
3.00 x 103 |
2.5 x103 |
0–100 s, rate = – [3.50 – 400] x 103 Pa/100 s = 5 Pa / s
100-200 s, rate = – [3.00 – 3.50] x 103 Pa/100 s = 5 Pa / s
200-300 s, rate = – [2.50 – 3.00] x 103 Pa / 100 s = 5 Pa / s
We notice that the rate remains constant, therefore, reaction is of zero order.
k = rate = 5 Pa / s
t1 / 2 = initial concentration or pressure/2K
= 4.00 x 103 Pa / 2 x 5 Pa s–1 = 400 s.
Half-life of a reaction: The half-life of a reaction represented as t1/2 is the time required for the reactant concentration to drop to one half of its initial value. Consider the zeroth order reaction.
R → Products
Integrate Zeroth order rate equation.
[R] = – kt + [R]0
where at time t = t1/2, the fraction of [R] that remains [R] / [R]0, therefore above equation can be written
[R]0/2 = – k t1/2 + [ R]0
k1/2 = [R]0 /2
for the first order integrated rate equation
at time
The half life depends on reactant concentration in different order of reactions as follows.
For zero order raction t1/2 ∝ [R]0. For first order reaction t1/2 is independent of R0, for second order reaction t1/2 ∝ 1/[R]0.
For nth order reaction t1/2 ∝ 1/[R]0n–1.
For half- life first order reaction
k= 5years-1
or
For first order reaction
and
......(ii)
Dividing (i) by (ii), we get
or
For the first order reaction the Rate constant
is given 5 min, hence
Let initial concentration ‘a’ of A be 100 mol L–1. To reach 25% of initial concentration means, (a – x) = 25 mol L–1.
Or
or
Ostwald Isolation Method: In this method, the concentration of all the reactants are taken in large excess except that of one. The concentration change only for this reactant is significant as other are so much in excess that practically there is no change in their concentrations. The constant terms may be combined with the rate constant and we may write
The value of ‘a’, i.e., the order of reaction with respect to A can be determined by the methods given above.
When
when,
Substituting these values in the equation
We get
or
or
We have given that
Substituting these values in the equation
we get
or
by using of arrhenius of equation
or
or
or
|
Run |
Initial |
Initial |
Initial rate (mol s–1) |
|
concentration |
concentration |
||
|
1 |
0.10 M |
0.1 M |
2.1 x 10–3 |
|
2 |
0.20 M |
1.0 M |
8.4 x 10–3 |
|
3 |
0.20 M |
2.0 M |
8.4 x 10–3 |
Determine the order of reaction with respect to A and with respect to B and the overall order of reaction.
Order with respect to A = 2.
Order with respet to B = 0.
Overall order of reaction = 2.
Let rate k(CCH3COF)a (CH2O)b
CH2O >> CCH3COF and in the second case, CH2O << CCH3COF. In the first case we determine the order of the reaction with respect to CH3 COF. We note that the reaction is not of zero order as rate of reaction changes with time.
Therefore k(CH2O) = 0.0154 min–1 and we note the order of reaction with respect to CH3COF is 1.
Now we determine the order of raction with respect to water.
Again,
The reaction is first order in H2O and we have
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:
|
t/s |
0 |
30 |
60 |
|
[CH3COOCH3] / mol L–1 |
0.60 |
0.30 |
0.15 |
(i) Show that it follows pseudo-first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time intervals 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
For the hydrolysis of methyl acetate to be a pseudo first-order reaction, the reaction should be first order with respect to ester when [H2O] is constant. The rate constant k for a first order reaction is given by:
Where,
[R]0 = intial concentration of reactant
[R] =final concentration of reactant
At =30s
We have
: 
It can be seen that the rate constant k for the reaction has a constant value under any given time interval .Hence the given reaction follows pseudo-first order kinetics.
(ii) Average rate of reaction between the time interval 30-60 seconds is given by
Average rate = 
For a reaction A + B → P, the rate is given by Rate = k[A] [B]2
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
A+B
Rate = k [A][B]2
(i) Since the given reaction has order two with respect to reactant B, thus if the concentration of B is double in the given reaction, then the rate of reaction will become four times.
(ii) If A is present in large excess then the reaction will be independent of the concentration of A and will be independent only on the concentration of B.as [B]2 will be the only determining factor in the order of the reaction will be two.
A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010)
For the given first order reaction the rate constant for 50% completion is given by
Here t= time taken for 50% completion =30 min
[R]0 = initial concentration of the reactant
[R] =final concentration of the reactant
Let [R]0 be 100 and due to 50% completion of the reaction [R] will be 100-50 i.e. 50
Putting value in 1 we get
For same reaction the time required for 90% completion of the reaction can be computed using the expression 
Here [R] = final concentration of reactant=100-90 =10
So we have
Therefore the time required for 90% completion of the given first order reaction is 100.13min.
Write two differences between 'order of reaction' and 'molecularity of reaction?
|
Order |
Molecularity |
|
The overall order of reaction is the sum of all the exponents of all the reactants present in the rate law expression. |
It is the number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction. |
|
It is determined experimentally. |
It is a theoretical concept. |
|
It may be equal to zero or have fractional values. |
It cannot be equal to zero and it always has integral values( which cannot exceed 3) |
The following data were obtained during the first-order thermal decomposition of SO2Cl2 at a constant volume:
SO2Cl2(g)---> SO2(g) + Cl2(g)
|
Experiment |
Time/s-1 |
Total pressure/atm |
|
1 |
0 |
0.4 |
Calculate the rate constant. (Given : log 4 = 0.6021, log 2 = 0.3010)
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation:
A reaction is second order in A and first order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
A reaction is second order in A and first order in B.
Differential rate equation:- 
(ii) On increasing the concentration of A three times i.e. 3A:
Rate = k[3A]2[B]=9k[A]2[B]=9(Rate) , i.e. 9 times the initial rate.
(iii) On increasing the concentration of A and B as 2A and 2B:
Rate1= k[2A]2[2B]=k(4x2)[A]2[B]=8k[A]2[B]=8 (Rate) , i.e. 8 times the initial rate.
A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 for this reaction.
(Given log 1.428 = 0.1548)
A --> P
T=0 a 0
T=t (a-x) x
Now, it takes 40 min for 30% decomposition i.e. reactant left after 40 min is 70% of its initial concentration.
So,
Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
where Ea is the activation energy.
When a graph is plotted for 
a straight line with a slope of -4250 K is obtained. Calculate ‘Ea’ for the reaction.
(R = 8.314 JK-1 mol-1)
Ea --> Activation energy
The above equation is like y = mx + c where if we plot y v/s x we get a straight line with slope ‘m’ and intercept ‘c’.
What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) L-1 mol-1
(ii) L mol-1 s-1
The sum of the powers of the concentrations of the reactants of a chemical reaction in the rate law expression is called the order of that chemical reaction.
Rate = k [A]x [B]y
Order of reaction = x + y
The orders of reaction for the following units are:
(i) L-1 mol s-1: Zero order
(ii) L mol-1s-1: Second order
For the reaction
2NO(g) + Cl2(g) --> 2NOCl(g)
The following data were collected. All the measurements were taken at 263 K:
|
Experiment No. |
Initial [NO] (M) |
Initial [Cl2] (M) |
Initial rate of disappearance of Cl2(M/min) |
|
1 |
0.15 |
0.15 |
0.60 |
|
2 |
0.15 |
0.30 |
1.20 |
|
3 |
0.30 |
0.15 |
2.40 |
|
4 |
0.25 |
0.25 |
? |
(a) Write the expression for rate law.
(b) Calculate the value of rate constant and specify its units.
(c) What is the initial rate of disappearance of Cl2 in exp. 4?
The rate expression can be defined as the stoichiometric coefficients of reactants and products. An expression in which the rate of reaction is given in terms of the molar concentration of the reactants, with each term raised to some power, which may or may not is the stoichiometric coefficient of the reacting species in a balanced chemical equation.
The rate constant can be defined as the rate of the reaction when the concentration of each of the reactant is taken as unity.
Example: 2NO(g)+O2(g)--- 2NO2(g)
The rate expression for the above reaction can be written as follows:
Rate = k [NO]2 [O2] (Experimentally determined)
Now, if the concentration of NO and O2 is taken to be unity, then the rate constant is found to be equal to the rate of the reaction.
Nitrogen pentoxide decomposes according to equation: 2N2O5(g)---> 4NO2(g) + O2(g)
This first order reaction was allowed to proceed at 40° C and the data below were collected:
|
[N2O5] (M) |
Time (min) |
|
0.400 |
0.00 |
|
0.289 |
20.0 |
|
0.209 |
40.0 |
|
0.151 |
60.0 |
|
0.109 |
80.0 |
(a) Calculate the rate constant. Include units with your answer.
(b) What will be the concentration of N2O5 after 100 minutes?
(c) Calculate the initial rate of reaction.
|
[N2O5] (M) |
Time (min) |
log (N2O5] |
|
0.400 |
0.00 |
-0.3979 |
|
0.289 |
20.0 |
-0.5391 |
|
0.209 |
40.0 |
-0.6798 |
|
0.151 |
60.0 |
-0.8210 |
|
0.109 |
80.0 |
-0.9625 |
From the plot, log [N2O5] v/s t, we obtain by
The rate constant of a first order reaction increases from 2 — 10-2 to 4 — 10-2when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea).
(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
According to the Arrhenius equation.
K = Ae(-Ea/RT)
From this, we get
We are given that
initial temperature T1=300K
Final temperature T2=310 K
Rate constant at initial temperature, k 1 = 2 x 10-2
Rate constant at final temperature, k2 = 4 x 10-2
Gas constant, R = 8.314 J K-1
Substituting the value, we get
Therefore activation energy of the reaction, Ea = 
= 535985.94 J mol-
= 535.98 kJ mol-1
(a) For a reaction A + B --> P, the rate law is given by,
r = k [A]1/2 [B]2.
What is the order of this reaction?
(b) A first order reaction is found to have a rate constant k = 5·5 x 10-14 s-1. Find the half-life of the reaction.
(a)
For A + B--> P
r = k [A]1/2 [B]2
The order of the reaction = 
(b) For first order reaction
k = 5.5 x 10-14 s-1
Half-life period ( 
) for the first order reaction
The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature.
[R = 8·314 J K-1 mol-1, log 4 = 0·6021]
Given: T1 = 293 k
T2 = 313 k
R = 8.314 J k-1mol-1
k2 = 4k1
Ea =?
The formula used is;
A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Let the concentration of the reactant be [A] = a
Rate of reaction, R = k [A]2 = ka2
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k (2a)2
= 4ka
= 4R
Therefore, the rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half, i.e., 
then the rate of the reaction would be
Therefore, the rate of the reaction would be reduced to 
The reaction N2(g) + O2(g) 2NO(g), contributes to air pollution whenever a fuel is burnt in air at a high temperature. At 1500 K, equilibrium constant K for it is 1.0 x 10-5. Suppose in a case [N2] = 0.80 mol L-1and [O2] = 0.20 mol L-1 before any reaction occurs. Calculate the equilibrium concentrations of the reactants and the product after the mixture has been heated to 1500 K.
N2(g) + O2(g) 
2NO(g)
T=0 0.8 0.20
T=t 0.8-x 0.2-x 2x
Kc = 1.0 x 10-5
Kc = 
Or, 1.0 x 10-5 = (2x)2/ [(0.8 – x)(0.2 – x )]
If x is very small, then
0.8 – x 0.8
0.2 – x 0.2
1.0 x 10-5 = (2x)2 / [(0.80) (0.2)]
16 x (10-6) = 4 x2
X2 = 2 x 10-3
Therefore, the amount of reactant and product at equilibrium is as follows:
N2 = 0.8 - 0.002 = 0.798
O2 = 0.2 - 0.002 = 0.198
NO = 2x = 2 x 2 x 10-3 = 4 x 10-3
For a reaction : 
(i)Write the order and molecularity of this reaction.
(ii)Write the unit of k.
(i) This reaction is catalysed by Pt at high pressure. So, it is a zero-order reaction with molecularity 2.
(ii) The rate law expression for this reaction is Rate = k
Hence, the unit of k is mol L−1 s−1.
The rate constant for the first-order decomposition of H2O2 is given by the following equation:
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes. (Given: R = 8.314 JK–1mol–1).
Given:
Order of the reaction = First order
t1/2 = 200 minutes = 200 × 60 = 12,000 seconds The relation between t1/2
and k is given by t1/2 = 0.693/k
k = 0.693/12000 = 5.7 × 10−5
The rate constant for the first-order decomposition of H2O2 is given by
For a reaction: 
rate =k
i) Write the order and molecularity of this reaction.
ii) Write the unit of k.
(i) Zero-order reaction, Molecularity is 2. It is a bimolecular reaction.
(ii) The units of k is mol L-1 s-1Define activation energy of a reaction.
The energy required forming the intermediate called activated complex is known as activation energy. Activation energy = Threshold energy – Average energy of the reactants.
The thermal decomposition of HCO2H is a first-order reaction with a rate constant of 2.4 x 10-3 s-1 at a certain temperature. Calculate how long will it take for three-fourths of the initial quantity of HCO2H to decompose.
(log 0.25 = - 0.6021)
Given k = 2.4 x 10-3 s-1
According to the first order rate law,
What do you understand by the rate law and rate constant of a reaction?
Identify the order of a reaction if the units of its rate constant are:
(i) L-1 mol s-1
(ii) L mol-1 s-1The rate law can be defined as an expression containing the stoichiometric coefficients of reactants and products. It is an expression in which the rate of reaction is given in terms of the molar concentration of the reactants, with each term raised to some power, which may or may not is the stoichiometric coefficient of the reacting species in a balanced chemical equation. The rate constant can be defined as the rate of the reaction when the concentration of each of the reactant is taken as unity.
Example: 2NO(g) + O(g)---> 2NO2(g)
The rate expression for the above reaction can be written as follows:
Rate = k [NO]2 [O2] (Experimentally determined)
Now, if the concentration of NO and O2 is taken to be unity, then the rate constant is found to be equal to the rate of the reaction.
(i) Comparing power of mole in L-1 mol s-1 and (mol L-1)1-n s-1,
We get
1 = l – n => n = 0 i.e., zero order reaction
(ii) Again comparing power of mole in L mol-1 s-1 and (mol L-1)1-n s-1
We get
–1 = 1 – n => n = 2, i.e., second order reaction.What is the effect of temperature on chemisorption?
Chemisorption increases with increase in temperature at a certain level and then starts decreasing.
For a chemical reaction R → P, the variation in the concentration (R) vs. time (t) plot is given as,
(i) Predict the order of the reaction.
(ii) What is the slope of the curve?
(i) The variation in the concentration (R) vs. time (t) plot shown here represents a zero order reaction, for which the rate of the reaction is proportional to zero power of the concentration of the reactants.
(ii) For a zero-order reaction, rate constant is given as,
k = [R]o - [R]t
So, the slope of the curve for the variation in the concentration (R) vs. time (t) plot is equal to the negative of the rate constant for the reaction.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume :
SO2Cl2 (g) → SO2 (g) + Cl2 (g)
|
Experiment |
Time/s−1 |
Total pressure/atm |
|
1 |
0 |
0·4 |
|
2 |
100 |
0·7 |
Calculate the rate constant.
(Given : log 4 = 0·6021, log 2 = 0·3010)
The thermal decomposition of SO2Cl2 at a constant volume is represented by the following equation:
Following data are obtained for the reaction :
N2O5 → 2NO2 + ½O2
| t/s | 0 | 300 | 600 |
| [N2O5]/mol L–1 | 1.6 × 10–2 | 0.8 × 10–2 | 0.4 × 10–2 |
(a) Show that it follows first order reaction.
(b) Calculate the half-life.
(Given log 2 = 0.3010 log 4 = 0.6021)
Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:
6.93×10−4 mol min−1
2.66 L min−1 at STP
1.34×10−2 mol min−1
6.93×10−2 mol min−1
A.
6.93×10−4 mol min−1
For first order reaction
Higher order (>3) reactions are rare due to:
the increase in entropy and activation energy as more molecules are involved.
shifting of equilibrium towards reactants due to elastic collisions
loss of active species on a collision
low probability of simultaneous collision of all the reacting species
A.
the increase in entropy and activation energy as more molecules are involved.
Conditions for the occurrence of a reaction:
(i) Proper orientation and effective collision of the reactants.
(ii) the chances of simultaneous collision with proper orientation between more than 3 species are very rare, so reaction with order greater than 3 are rare.
The resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution of 0.5 M solution of the same electrolyte is 1.4 S m-1 and resistance of the same solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in Sm-2 mol-1 is
5 x 10-4
5 x 10-3
5 x 103
5 x 102
A.
5 x 10-4
For first solution,
k = 1.4 Sm-1, R =50 Ω , M =0.2
specific conductance
For the non- stoichiometric reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiment, all at 298 K.
| Initial concentration (A) | Initial concnetration (B) | Initial rate of formation of C (mol L-1 S-1) | |
| 1 | 0.1 M | 0.1 M | 1.2 x 10-3 |
| 2 | 0.1 M | 0.2 M | 1.2 x 10-3 |
| 3 | 0.2 M | 0.1 M | 2.4 x 10-3 |
D.
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be (R = 8.314 JK–1 mol–1 and log 2 = 0.301)
53.6 kJ mol-1
48.6 kJ mol-1
58.5 kJ mol-1
60.5 kJ mol-1
A.
53.6 kJ mol-1
By using Arrhenius equation,
Given, T2 = 310; T1 = 300K
On putting values in Eq (i), we get
For a first order reaction, (A) → products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is
1.73 x 10–5 M/ min
3.47 x 10–4 M/min
3.47 x 10–5 M/min
1.73 x 10–4 M/min
B.
3.47 x 10–4 M/min
By first order kinetic rate constant,
The rate of a chemical reaction doubles for every 10ºC rise of temperature. If the temperature is raised by 50ºC, the rate of the reaction increases by about
10 times
24 times
32 times
64 times
C.
32 times
For every 10o C rise of temperature, the rate is doubled. Thus, temperature coefficient of the reaction = 2 when temperature is increased by 50o rate becomes
The time for half life period of a certain reaction A → Products is 1 hour. When the initial concentration of the reactant ‘A’ is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1 if it is a zero order reaction?
1 h
4 h
0.25 h
0.5 h
C.
0.25 h
Given that [A]o = 2 mol L-1
The energies of activation for forward and reverse reactions for A2 + B2 ⇌ 2AB are 180 kJ mol–1 and 200 kJ mol–1 respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol–1. The enthalpy change of the reaction (A2 + B2 → 2AB) in the
presence of catalyst will be (in kJ mol–1) –
300
120
280
-20
D.
-20
So, ∆ =− HReaction Ef- Eb
= 80 – 100 = -20
Consider the reaction, 2A + B → Products. When the concentration of B alone was doubled, the half-life did not change. When the concentration of An alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is –
L mol–1 s–1
no unit
mol L–1s–1
s-1
A.
L mol–1 s–1
2A + B → Products
When conc. of B is doubled, the half-life did not change, hence reaction is of first order w.r.t. B. When the concentration of A is doubled, the reaction rate is doubled, hence reaction is of first order w.r.t. A. Hence over all order of reaction is 1 + 1 = 2. So, unit of rate constant L mol–1 s–1
Which of the following nuclear reactions will generate an isotope?
neutron particle emission
positron emission
α-particle emission
β-particle emission
A.
neutron particle emission
A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will
remain unchanged
triple
increase by a factor of 4
double
C.
increase by a factor of 4
R∝ [W]2
R'∝ [2CO]2
R∝ 4[W]2
R∝ 4M
In Langmuir’s model of adsorption of a gas on a solid surface
the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered
the adsorption at a single site on the surface may involve multiple molecules at the same time
the mass of gas striking a given area of the surface is proportional to the pressure of the gas
the mass of gas striking a given area of surface is independent of the pressure of the gas
C.
the mass of gas striking a given area of the surface is proportional to the pressure of the gas
Rate of a reaction can be expressed by Arrhenius equation as K= Ae-E/RT In this equation, E represents
the energy above which all the colliding molecules will react
the energy below which colliding molecules will not react
the total energy of the reacting molecules at a temperature, T
the fraction of molecules with energy greater than the activation energy of the reaction
B.
the energy below which colliding molecules will not react
The following mechanism has been proposed for the reaction of NO with Br2 to form NOBr
NO(g) Br2 (g) ⇌ NOBr2 (g)
NOBr2 (g) NO(g)→ 2NOBr(g)
If the second step is the rate determining step, the order of the reaction with respect to NO(g) is
1
0
3
2
D.
2
NO(g) Br2 (g) ⇌ NOBr2 (g)
NOBr2 (g) NO(g)→ 2NOBr(g)
R = K[NOBr2][NO]
= K.Kc[NO][Br2][NO], = 
=K' [NO]2 [Br2 ]
Consider an endothermic reaction, X → Y with the activation energies Eb and Ef for the backward and forward reactions, respectively. In general
Eb < Ef
Eb > EfEb > Ef
Eb = Ef
There is no definite relation between Eb and Ef
A.
Eb < Ef
∆H = Ef – Eb
For ∆H = Positive, Eb < Ef
In first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is
30 minutes
60 minutes
7.5 minutes
15 minutes
A.
30 minutes
Order = 1
Concentration changes from 0.8 M to 0.4 M(50%) in 15 min, thus half - life is 15 min = T50
A change from 0.1 M to 0.025 M is 75% and for first order reaction
T75 = 2 x T50
= 2 x 15 = 30 min
T50 = 15 min
The rate equation for the reaction 2A + B → C is found to be: rate k[A][B]. The correct statement in relation to this reaction is that the
unit of K must be s-1
values of k is independent of the initial concentration of A and B
rate of formation of C is twice the rate of disappearance of A
t1/2 is a constant
B.
values of k is independent of the initial concentration of A and B
values of k is independent of the initial concentration of A and B
The soldiers of Napolean army while at Alps during freezing winter suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transformation is related to
an interaction with nitrogen of the air at very low temperatures
an interaction with water vapour contained in the humid air
a change in the partial pressure of oxygen in the air
a change in the crystalline structure of tin
D.
a change in the crystalline structure of tin
As temperature decrease, white tin (beta-form) changes to grey tin (alpha form)
Alpha -Sn has a much lower density
The half-life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is
1.042 g
4.167 g
3.125 g
2.084 g
C.
3.125 g
If y = number of half-lives
The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is
Fe4[Fe(CN)6]3
Na4[Fe(CN)5NOS]
Fe(CN)3
Na3[Fe(CN)6]
A.
Fe4[Fe(CN)6]3
If nitrogen is present in the organic compound, then sodium extract contains Na4[Fe(CN)6].
At 5180C the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr s–1 when 5% had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is
0
2
3
1
B.
2
r1 = 1 torr sec–1 , when 5% reacted (95% unreacted)
r2 = 0.5 torr sec–1 , when 33% reacted (67% unreacted)
r ∝ (a – x)m
Where,
m = order of reaction
a–x = unreacted
The ionic radii of A+ and B- ions are 0.98 x 10-10 m and 1.81 x10-10 m. The coordination number of each ions in AB is
4
8
2
6
D.
6
Given, ionic radius of cation (A+) =0.98 x 10-10m
Ionic radius of anion (B-)=1.81 x 10-10m
therefore,
The coordination number of each ion in AB =?
Now we have.
If radius ratio range is in between 0.441-0.732 ion would have an octahedral structure with coordination number six.
The activation energy of a reaction can be determined from the slope of which of the following graphs?
InK vs T
C.
By Arrhenius equation
K = Ae-Ea/RT
where, Ea = energy of activation
Applying log on both side
In k= In A-Ea/RT
or log k = - Ea/2.303 RT +log A
This equation is of the form of y=mx + c i.e the equation of a straight line. Thus, if a plot of log vs 1/T is a straight line. the validity of the equation is confirmed.
slope of the line = - Ea/2.303 R
Thus, measuring the slope of the line, the value of Ea can be calculated.
When initial concentration of a reactant is doubled in a reaction, its half-life period os not affected the order of the reaction is?
Zero
first
second
more than zero but less than first
B.
first
For a zero order reaction t1/2 is directly proportional to the initial concentration of the reaction [R0]
i.e, half life period is independent of initial concentration of a reactant.
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200C to 350C ?(R=8.314 J mol-1 K-1).
342 kJ mol-1
269 kJ mol-1
34.7 kJ mol-1
15.1 kJ mol-1
C.
34.7 kJ mol-1
Given, initial temperature,
T1 = 20+273 =293 K
Final temperature
T2 =35+273 = 308 K
R= 8.314 J mol-1 K-1
Since, rate becomes double on raising temperature,
therefore,
r2 =2r1 or r2/r1 = 2
In a reaction, A+B → Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as
Rate = k[A][B]2
Rate = k[A]2[B]2
Rate = k[A][B]
Rate k[A]2[B]
D.
Rate k[A]2[B]
Let the order of reaction with respect to A and B is x and y respectively. so, the rate law can be given as,
R= k[A]x[B]y ......(i)
When the concentration of the only B is doubled, the rate is doubled, so
In Freundlich adsorption isotherm, the value of 1/n is
between 0 and 1 in all cases
between 2 and 4 in all cases
1 in case of physical adsorption
1 in case of chemisorption
A.
between 0 and 1 in all cases
In Freundlich adsorption isotherm
x/m = kp1/n
the value of n is always greater than 1. so the value of 1/n lies between 0 and 1 in all cases.
In a zero order reaction for every 10o rise of temperature, the rate is doubled. If the temperature is increased from 10oC to 100oC, the rate of the reaction will become
256 times
512
64 times
128 times
B.
512
For 10o rise in temperature, n = 1
so rate = 2n = 21 = 2
When temperature is increased from 10oC to 100o C change in temperature = 100-10 = 90oC i.e n=9 so, rate = 29 = 512 times.
Alternate method: With every 10o C rise in temperature, rate becomes double, so
Activation energy (Ea) and rate constants (k1 and k2) of chemical reaction at two different temperatures (T1 and T2) are related by
D.
According to Arrhenius equation, activation energy (Ea) and rate constants (k1 and k2) of chemical reaction at two different temperatures (T1 and T2) are related as,
Which one of the following statements for the order of a reaction is correct?
Order is not influenced by stoichiometric coefficient of the reactants
Order of reaction is sum of power to the concentration terms of reactants to express the rate of reaction
Order of reaction is always the whole number
Order can be determined only experimentally
C.
Order of reaction is always the whole number
Order of reaction may be zero, whole number or fraction number.
The rate of constant of the reaction A → B is 0.6 x 10-3 mole per second. If the concentration of A is 5 M then concentration of B after 20 min is
1.08 M
3.60 M
0.36 M
0.72 M
D.
0.72 M
For a zero -order reaction unit of the rate constant is mole per second. Hence, we can easily calculate concentration of B after 20 min by the following formula,
x = Kt
x = Kt = 0.6 x 10-3 x 20 x 60 = 0.72 M
The half -life of a substance in a certain enzyme catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L-1 is
414 s
552 s
690 s
276 s
C.
690 s
Enzyme -catalysed reactions follow first order kinectics.
therefore,
Fall of concentration from 1.28 mg L-1 to 0.04 mg L-1 involves five half - lives.
The unit of rate constant for a zero order reaction is
mol L-1 s-1
L mol-1 s-1
L2 mol-2 s-1
s-1
B.
L mol-1 s-1
For zero order reaction,
Rate = k [ reaction]o
therefore, Rate = k
and unit of k = mol L- s-1
For an endothermic reaction, energy of activation is Ea and enthalpy of reaction is ΔH (both of these in kJ/mol). Minimum value of Ea will be
less than ΔH
equal to ΔH
more than ΔH
equal to zero
C.
more than ΔH
In endothermic reactions, the energy of reactants is less than that of the products. Potential energy diagram for endothermic reactions is,
Where Ea = activation energy of forwarding reaction
Ea' = activation energy of backwards reaction
ΔH = enthalpy of the reaction
From the above diagram,
Ea = Ea' + ΔH
Thus, Ea > ΔH
During the kinetic study of the reaction, 2A + B --> C+ D, following results were obtained
|
|
[A]/mol L- |
[B]/ mol L- |
Initial rate of formation of D/ mol L- min- |
|
I |
0.1 |
0.1 |
6.0 x 10-3 |
|
II |
0.3 |
0.2 |
7.2 x 10-2 |
|
III |
0.3 |
0.4 |
2.88 x 10-1 |
|
IV |
0.4 |
0.1 |
2.40 x10-2 |
rate = k[A]2[B]
rate = k [A][B]
rate = k[A]2[B]2
rate = k[A][B]2
D.
rate = k[A][B]2
Let the order of reaction with respect to A is x and with respect to B is y. Thus,
rate = k[A]x[B]y
For the given cases,
The rate of the reaction,
2NO + Cl2 → 2NOCl is given by the rate equation, rate = k[NO]2[Cl2]
The value of the rate constant can be increased by
increasing the temperature
increasing the concentration of NO
increasing the concentration of the Cl2
doing all of the above
A.
increasing the temperature
Rate constant is only affected by temperature.
Half-life period of a first order reaction is 1386 s. The specific rate constant of the reaction is
5.0 x 10-3s-1
0.5 x 10-2 s-1
0.5 x 10-3 s-1
5.0 x 10-2s-1
C.
0.5 x 10-3 s-1
Specific rate constant,
For the reaction, A+B → Products, it is observed that
1) On doubling the initial concentration of A only, the rate of reaction is also doubled and
2) On doubling the initial concentrations of both A and B , there is a change by a factor of 8 in the rate of the reaction.
The rate of this reaction is, given by
rate = k [A]2[B]
rate = k[A][B]2
rate = k[A]2[B]2
rate = k[A][B]
B.
rate = k[A][B]2
For the reaction,
A +B → product
on doubling the initial concentration of An only the rate of reaction is also doubled, therefore,
..... (i)
Let initially rate law is
Rate = k[A][B]y
If the concentration of A and B both are doubled, the rate gets changed by a factor of 8.
In the reaction,
The rate of appearance of bromine (Br)2 is related to rate of disappearnace of bromide ions as following
A.
Rate of appearance/disappearance
For reaction,
The rate constant k1 and k2 for two different reactions are 1016. e-2000/T and 1015.e-1000/T respectively. The temperature at which k1 = k2
1000 K
2000/2.303 K
2000 K
1000/2.303 K
D.
1000/2.303 K
The Arrhenius equation is represented as;
k = Ae-Ea/RT
In the given equations first, take log and then compare them.
The bromination of acetone cytosine and guanine solution is represented by this equation.
CH3COCH3 (aq) + Br2 (aq) →CH3COCH2Br (aq) + H+ (aq) + Br- (aq)
These kinetic data were obtained for given reaction concentrations.
|
Initial Concentrations, M
|
||
| [CH3COOH] | [Br2] | [H+] |
| 0.30 | 0.05 | 0.05 |
| 0.30 | 0.10 | 0.05 |
| 0.30 | 0.10 | 0.10 |
| 0.40 | 0.05 | 0.20 |
Rate = k[CH3COCH3][H+]
Rate = k[CH=COCH3][Br2]
Rate = k [CH3COCH3][Br2][H+]
Rate = k [CH3COCH3][Br2][H+]
A.
Rate = k[CH3COCH3][H+]
By comparing the rate and concentration, the order of the reaction can be calculated.
Let the rate of the reaction wrt [CH3COCH3], [Br2] and [H+] are x,y and z respectively. Thus,
The reaction of hydrogen and iodine monochloride is given as:
H2 (g) + 2ICl (g) → 2 HCl (g) + I2 (g)
This reaction is of first order with respect to H2 (g) and ICI (g), following mechanisms were proposed:
Mechanism A:
H2 (g) + 2 ICl (g) → 2 HCl (g) + I2 (g)
Mechanism B:
H2 (g) + ICl (g) →HCl (g) + HI (g) ;slow
HI (g) + ICl (g) → HCl (g) + I2 (g); fast
Which of the above mechanism (s) can be consistent with the given information about the reaction?
B only
A and B both
Neither A nor B
A only
B.
A and B both
The rate of reaction always depends on slow reaction.
H2 (g) + ICl (g) →HCl (g) + HI (g) is a first order reaction respect to H2 and I2 can measure with the help of both mechanism A and B.
In a first order reaction A →B, if k israte constant and initial concentration of the reactant A is 0.5 A M then the half -life is :
0.693/0.5 k
log2/k
ln2 / k
D.
ln2 / k
For first order reaction.
Consider the reaction
N2 (g) + 3 H2 (g) → 2 NH3 (g)
The equality relationship between 
is:
B.
For the reaction,
N2 (g) + 3 H2 (g) → 2 NH3 (g)
Which one of the following statements is not correct?
Catalyst does not initiate any reaction
The value of equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium
Enzymes catalyse mainly bio-chemical reactions
Coenzymes increase the catalytic activity of enzyme
B.
The value of equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium
A catalyst decreases activation energies of both the forward and backward reaction by the same amount, therefore, it speeds up both forward and backward reaction by the same rate. The equilibrium constant is therefore not affected by the catalyst at a given temperature.
A first-order reaction has a specific reaction rate of 10–2 s–1. How much time will it take for 20 g of the reactant to reduce to 5 g?
238.6 second
138.6 second
346.5 second
693.0 second
B.
138.6 second
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in gram) of the remaining product at STP will be
1.4
3.0
4.4
2.8
D.
2.8
CO2 is absorbed by KOH.
So the remaining product is only CO.
moles of CO formed from both reaction
=
Let mass of CO = moles x molar mass
=
The correct difference between first and second order reactions is that
The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
The half-life of a first-order reaction does not depend on [A]0; the half-life of a second-order reaction does depend on [A]0
The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations
A first-order reaction can catalyzed; a second-order reaction cannot be catalyzed
B.
The half-life of a first-order reaction does not depend on [A]0; the half-life of a second-order reaction does depend on [A]0
The first-order reaction is given by, which is independent of the initial concentration of reactant.
Second order reaction, , which depends on the initial concentration of reactant.
For the chemical reaction,
2O3 ⇌ 3O2
The reaction proceeds as follows
O3 ⇌ O2 + O
O + O3 → 2O2 (slow)
The rate law expression will be
r = k' [O3]2
r = k' [O3]2[O2]-1
r = k'[O3][O2]
unpredictable
B.
r = k' [O3]2[O2]-1
As slowest step is the rate determining step. Hence, from slow reaction r = K[O][O3] ... (i)
From the fast reaction,
A first-order reaction is 50% completed in 1.26 x 1014s. How much time would it take for 100% completion?
1.26 x 1015 s
2.52 x 1014 s
2.52 x 1028 s
Infinite
D.
Infinite
K- for the first order,
For the chemical reaction, 2O3 ⇌ 3O2. The reaction proceeds as follows
O3 ⇌ O2 + O (fast)
O + O3 → 2O2 (slow)
The rate law expression will be
r = k'[O3]2
r = k'[O3]2[O2]-1
r = k'[O3][O2]
Unpredictable
B.
r = k'[O3]2[O2]-1
As slowest step is the rate determining step. Hence, from slow reaction, r = K[O][O3] .... (i)
From the fast reaction,
Sponsor Area
Sponsor Area
, if one emission is an α-particle, what should be the other emission(s)?
the value of 
would be
