Bihar Board - Class 12 Chemistry - Chapter 4: Chemical Kinetics Long Answer Question
Long Question Answer
Q. 1. Define ‘energy of activation of a reaction. How does it vary with a rise in temperature ?
Ans⇒ 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.
Q. 2. State and explain Arrhenius equation. How can we determine the activation energy of a reaction using this equation ?
Ans⇒ k = A eEaRT
where k is rate constant, A is Arrhenius of frequency factor, Ea is energy of activation, T is absolute temperature. This equation is called the Arrhenius equation.
Taking log, we get
In k = In A - EaRT
2.303 log k=2.303 log A A-EaRT
log k = log A - Ea2.303 RT
Plot of log k versus 1T is a straight line with a negative slope. The slope of the line is equal to
. slope =- Ea2.303 RT
∴ Ea= slope 2.303 R
Q. 3. Explain the relationship between the rate of production of iodine and rate of disappearance of hydrogen iodide in the following reaction ?
2 HI H2 +I2
Ans⇒ Rate of disappearance of HI =- d[HI]dt
Rate of appearance of I2 = d [I2]dt
From the chemical equation,
Rate of production I2 = 12 Rate of disappearance of HI
Therefore, d [I2]dt =12 d[HI]dt
Q. 4. Define Half-life period (t12) of a chemical reaction. Also obtain the expression for half-life period.
Ans⇒ Half-life period (t12) of a chemical reaction.
Half-life period of a chemical reaction is defined as a time period during which the concentration of a reactant is reduced to half of the initial value of concentration.
It is denoted by t12 or t0.5 .
The rate equation for a reaction of first order in expressed as :
k = 2.303t log aa-x
Or t = 2.303t log aa-x
When t = 12x= 0.5 a or a2
On substituting these values, we get
t12 2.303k log a(a-0.5 a) = 2.303k log 2
or , t12 = 0.693k
The relation (i) is the required expression for half life period of first order reaction.
Q. 5. Express the rate of each of the following reaction in terms of disappearance of the reactants and appearance of products :
(i) PCl5 (g) PCl3 (g)+ Cl2 (g)
(ii) 2NO (g) + O2 (g) 2NO2 (g)
(iii) H2 (g) + I2 (g) 2HI (g)
(iv) CO (g) + NO2 (g) CO2 (g) + NO (g)
(v) 2 NO2 (g) + F2 (g) 2 NO2 F (g)
Ans⇒ (i) PCl5 (g) PCl3 (g)+ Cl2 (g)
rate = -d [PCl5]dt= d [PCl3]dt = d [Cl2]dt
(ii) 2NO (g) + O2 (g) 2NO2 (g)
rate = -d [NO]2 dt= -d [O2]dt = +d [NO2]2 dt
(iii) H2 (g) + I2 (g) 2HI (g)
rate = -d [H2] dt= -d [I2]dt = 12d [HI] dt
(iv) CO (g) + NO2 (g) CO2 (g) + NO (g)
rate = -d [CO] dt= -d [NO2]dt =d [CO2]dt =d [NO] dt
(v) 2 NO2 (g) + F2 (g) 2 NO2 F (g)
rate = -12d [NO2] dt= -d [F2]dt = 12d [NO2 F] dt
Q. 6. The reaction :
2 NO + Br2 2 NOBr
is supposed to the following mechanism :
(i) NO + Br2
(ii) NOBr + NO 
Suggest the rate law expression.
Ans⇒ 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,Kc = NOBr2[NO2] [Br2]
∴ [NOBr2]= Kc [NO2] [Br2]
Then by equation (i) and (ii)
rate = k- - [NO2] [Br2]
Q. 7. (a) Define half life period of a chemical reaction. Write equation of t12 for a first order reaction.
(b) A first order reaction is 75 %complete in 60 minutes. Find the half life of this reaction.
Ans⇒ (a) The time during which half of the reactant is converted into the products is called the half life period of the reaction. It is represented by t12 .
t12 for a first order reaction is t12= 0.693K
(b) Let Initial concentration = [A]0
Final concentration=[A]t=[A]0- 75100[A]0=25100[A]0
Time required (t) = 60 min
Them for a first order reaction.
K = 2.303tlog [A]0[A]t=2.30360 min log [A]0 .10025 [A]0
K =2.30360 minlog4 = 2.303 0.602160 min=2.31 10-2 min-1
For a first order reaction
t12=0.693K=0.6932.31 10-2 min = 30 min
Q. 8. What is the effect of temperature on the rate constant of reaction ? How can this temperature effect on rate constant be represented quantitatively?
Ans⇒ The rate constant of reaction increases with increase of temperature. This increase in generally fold to five two fold for a 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 reactions to occur, there must be collisions 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 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 2 (where 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
K=A e- EaRT
where A is a constant called frequency factor (because it gives the frequency of binary collisions of the reacting molecules per second per liter. E0 is the energy of activation, R is a gas constant and T is the absolute temperature. The factor e- EaRT 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 (1), we get,
In K1 = in A- EaRT1 …………….(1)
In K2 = in A - EaRT2 …………….(2)
Substacting eqn. (1) from eqn. (2), we get
In k1-Ink1=-EaRT2+EaRT1=EaRT1+EaRT2
or In k2k1=EaR+1T1-1T2=EaRT2 -T1T1 T2
or, log k2k1=Ea2.303RT2 -T1T1 T2
Thus knowing the values of the constant k1 and k2 at two different temperature T1 and T2, the value of Ea can be calculated.
Q. 9. (a) What do you mean by zero order reaction ? How is the value of rate constant determined ? What is the relation between rate constant and half-life period ?
(b) Give two examples.
(c) What are the factors on which the rate of reaction depends ? Discuss each factor in brief.
Ans⇒ (a) When the rate of the reaction is independent of the concentration of the reactants, the reaction is known as zero order reaction. In zero order reaction, the concentration of reactant (R) remains unaltered the course of reaction.
It means rate= - d [R]d=k[R]0=k
or, d(R)=-kdt
On integration of equation we get
[R]=-kt + constant ……(i)
Since [R]=[R]0
where t=0 (i.e. on product is formed at the beginning of the reaction), the constant must be zero.
Thus, [R]0 =-K 0= constant
. constant=[R]0
Put the value of constant in eq. (i) we get
[R]=kt+[R]0 …. (ii)
When a plot is drawn between concentration of reactant (R) and time t , a straight line is obtained, slope of which gives the value of -k and intercept on y-axis is equal to the value of [R]0.
Alternatively, the value of k can be obtained from eq. (i) by putting the value of concentration [R] at any time t and initial concentration. [R]0
Hence , k =[R]0=[R] tt
Half life period is the time period
to reduce the initial concentration
of the reactant to half of its initial value.
Thus , t12 = [R] - 12[R]0k=[R]02k
Thus, the half life period is directly proportional to the initial concentration of the reactant, [R]0 and inversely proportional to the rate constant k.
(b) Example : (i) Decomposition of NH3 on the surface of platinum.
(ii) Decomposition of HI on the Gold surface.
(c) The rate of reaction depends upon the following factors :
(i) Concentration of the reactants : Greater the concentration faster is the reaction.
(ii) Presence of catalyst: It increases the speed of the reaction by lowering the activation energy.
(iii) Temperature : the rate of reaction increases with increase of temp, in most of cases, the rate of reaction become almost double for every 10 C rise in temp.
This is because of the increase in the number of effective collisions with rise in temp.
(iv) Surface area : With increase in surface area rate of reaction increases.
(v) Nature of reactants: With change in reactants rate of reaction changes.
Q. 10. Define order and molecularity of reaction. Derive a general expression for specific rate constant of first order reaction.
Ans⇒ It is the sum of the power of indicus of reacted terms no. of molecules which take part in a chemical rib known as molecularity of reaction.
Let us take a general except soln. A BA → B
to, -0, a g moles/liter O.
‘t’-(a-x)g moles liter g moles/liter.
dxdt (a-x) ; dxdt =K (a-x) ;
-In (a-x) =Kt+1 ; I=-In a
In (a-x)=kt-In a ; kt =In a - lk (a-x)
K= 2.303tlog aa-x
Q. 11. What is the difference between Order and Molecularity of a reaction ?
Ans⇒ Difference between Order & Molecularity :
Molecularity | Order |
1. It is the number of reacting species undergoing simultaneous collision in the reaction. | 1. It is the sum of the power of the concentration terms in the rate law expression. |
2. It is a theoretical concept. | 2. It is determined experi- mentally. |
3. It can have integral values only. | 3. It can have fractional values also. |
4. It cannot be zero. | 4. It can be zero. |
5. It does not tell us anything about the mechanism of the reaction. | 5. It tells us about the slowest step in the mechanism and hence gives some clue about the mechanism of the reaction. |
Q. 12. Rate of a reaction and Rate constant of a reaction ?
Ans⇒ Difference between Rate of a reaction and Rate constant :
Rate of Reaction | Rate constant of reaction |
1. It is the speed at which the reactants are converted into products at any moment of time. | 1. It is a constant of proportionality in the rate law expression. |
2. It depends upon the concentration of reactant species at that moment of time. | 2. It refers to the rate of the reaction at the specific point when concentration of every reacting species is unity. |
3. It decreases with the progress of reaction generally. | 3. It is constant and does not depend on the process of the reaction. |
4. It has same units for all reactions ML-1S-1 | 4. It has different units for different reactions. |
Q. 13. What are the factors which influence the rate of reaction? Discuss them in detail.
Ans⇒ 1. Nature of the reacting species : Consider the following two reactions :
2NO (g)+O2 (g) 2 NO2 (f) …fast
2 CO (g) + O2 (g) 2 CO2 (g) …slow
These reactions appear to be similar but the first is fast while the second is slow. This is because different amounts of energies are required for breaking of different bonds and different amounts of energies are released in the formation of different bonds.
2. Concentration of the reactants : Greater are the concentrations of the reactants, faster is the reaction. Conversely, if the concentrations of the reactants decrease, the rate of the reaction also decreases. A match stick which burns usually in air (O2:21 % by volume) burns with a flash in a jar containing O2 gas (volume of O2: 100 % )
3. Temperature : The rate of a reaction increases with the increase in temperature. In most of the reactions the rate of the reaction gets doubled for 10 C rise in temperature irrespective of the fact whether the reaction is exothermic or endothermic. In some cases reactions do not take place at room temperature but proceed only on heating
4. Surface area of the reactants : For a reaction involving a solid reactant or catalyst, the smaller is the particle size, i.e., greater is the surface area, the faster is the reaction. Wood shavings burn rapidly in air than a log of wood, because the surface area of wood shaving is much higher.
5. Pressure of a catalyst : A catalyst generally increases the speed of the reaction without itself being consumed in the reaction. In case of reversible reactions a catalyst helps attain chemical equilibrium quickly without disturbing the state of equilibrium. It speeds up the reaction both in the forward as well as backward direction so the equilibrium is attained earlier.
6. Presence of light : Some reactions do not take place in the dark, but take place in the presence of light.
e.g., H2 + Cl2 2HCl
Such reactions are called Photo-Chemical Reactions.
Q. 14. Calculate the half life of a first order reaction from their rate constants give below :
(a) 200 s-1 (b) 2 min-1 (c) 4 years s-1
Ans⇒ For the first order reaction
t12=0.693k
(a) K=200 s-1
t12=0.693200 = 3.465 10-3 s
(b) K=2 min-1
t12=0.6932 = 0.3465 min
(c) 4 years s-1
t12=0.6934 = 0.1732 years
Q. 15. The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its 116th value ?
Ans⇒ Rate constant of reaction,
k=60 s-1 ; t1516= ?
Rate constant of first order reaction is given by
k=2.303tlog aa-x ; t=2.203tlog aa-x
where 1516th the reaction is over then
a-x=1-116 Mifa=1 M
log1516 = 2.30360 S-1log11-1516
=2.30360 S-1log 16 = 2.303601.2041
=0.046 s = 4.6 10-2 sec
Q. 16. A first order reaction takes 40 min for 30 % decomposition. Calculate t12.
Ans⇒ (1) K= 2.303tlog aa-x=2.30340log 10070
=2.30340log100-log70 = 2.30340(2-1.8808)
=2.30340 0.1192 = 0.0068 min-1
(2) 0.0068=2.303tlog10050
0.0068=2.303tlog2
0.0068=2.303t0.3010
t=2.3030.30100.0068=101.94 min
Q.17. Differentiate between Absorption and Adsorption.
Ans⇒ Following are the main differences between Absorption and Adsorption :
Absorption | Adsorption |
(i) Absorption is a bulk phenomenon. | (i) Adsorption is a surface phenomenon. |
(ii) In absorption the con- centration remains the same throughout the material. | (ii) In absorption the rate concentration on the surface is different from that in the bulk. |
(ii) The absorption rate remains the same throughout the process. | (iii) The absorption rate is high in the beginning and then decreases. |
Q. 18. Explain Kohlrausch’s law of independent migration of ions. Mention one application of Kohlrausch’s law.
Ans⇒ It states that at infinite dilution, molar conductivity of an electrolyte is equal to sum of contribution due to cation as well as anion e.g.,
m NaCl = m Na+ + m Cl- …………(1)
m CH3COONa = m CH3COOH- + m Na+ ……………..(2)
m HCl = m H2 + m Cl ……………(3)
It helps to determine m CH3COOH by adding equation (1) and (3) and subtracting (1).
Q. 19. What is the initial rate method for the determination of a reaction rate when more than one reactant is present ?
Ans⇒R1 = dxdt=k [A1]a [B1]b …………(1)
R2=k [A2]a[B2]b …………..(2)
Dividing (2) by (1), we get
R2R1 = A2A1a
From this can be calculate
R3R1 = k[A1]a[B2]b
(3) Dividing (2) by (1), we get
R3R1 = B2B1b
From this we can calculate b .
Q. 20. How does the value of rate constant vary with reactant concentration ?
Ans⇒ For reaction of nth order = dxdt= Rate = k [conc.]n
Or k = dxdt. 1[conc.]n=conc.Time=1[conc.]n=1Time 1[conc.]n-1
i.e., k 1[conc.]n-1
Q. 21. A substance with initial concentration follows zero order kinetics. In how much time will the reaction go to completion ?
Ans⇒ dxdt=k or dx=k dt ∴ x=kt+I
when t=0 ; x=0 ∴ I=o
Hence x=kt or t= xK , For completion x=a ∴ t=aK .
Q. 22. Rate of a reaction is given by the equation : Rate = k [A]2 [B] . What are the units for the rate and rate constant for this reaction ?
Ans⇒ Units of rate = mol L-1 s-1
Units of k= Rate[A] [B]2
x=mol L-1 s-1(mol L-1) (mol L-1)2
= L2 mol-2 s-1
Q. 23. What is a salt bridge ? What is its significance ?
Ans⇒ Salt bridge : A salt bridge consists of a saturated solution of NH4NO3 or KCl mixed with gelatin or agar jelly filled in a glass tube bent according to the requirement of the experiment.
Significance :
(i) Salt bridge prevents mixing of two electrolytes.
(ii) Prevents junction potential.
(iii) Maintains electrical neutrality.
Q. 24. For the assumed reaction X2 + 3Y2 2 XY3 , write the rate equation in terms of rate of disappearance of Y2 .
Ans⇒ Rate = d [X2]dt= - 13d [Y2]dt=+12d [XY3]dt
Rate of disappearance of y2 = d [Y2]dt= -3 [X2]dt=+32d [XY3]dt
Q. 25. The rate Law for the decomposition of N2O5 is
Rate = k [N2O5]
What is the significance of k in this equation ?
Ans⇒If [N2O5]=1 ; then Rate = k
i.e., k is equal to the rate of the reaction when concentration of N2O5 is unity, i.e., 1 mol L-1
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