A +B2+ 🡪 A2+ +B

0.0591 [𝐴2+]
Ecell = E0cell – 2
log [𝐵2+]
0.0591 10−4
Ecell = E0cell – 2
log 10−3
Ecell = 2.6805 + 0.02955x1
Ecell = 2.17V
ΔG0 = - n F E0
= -2 x 2.17 x96500
= - 523,030 j mol-1
= -523.030 KJ mol-1
Q.3. (i)State Kohlrausch’s law of independent migration of ions.
(ii) Calculate the degree of dissociation (α) of CH3COOH if λm and λ0m of CH3COOH are 48
Scm2 mol-1 and 400 Scm2 mol-1 respectively.
Ans. Kohlrausch law: It states that the limiting molar conductivity of an electrolyte is equal to
the sum of the individual contributions of the cations as well as anions of the electrolyte.
Because ions move independently at infinite dilution.
λ𝑚 48 𝑆𝑐𝑚2 𝑚𝑜𝑙−1
(ii)Degree of dissociation (α ) = λ0𝑚
= 400 𝑆 𝑐𝑚2 𝑚𝑜𝑙−1
= 0.12.
=0.12 x 100 = 12 %
Q.4. Out of the following pairs, predict with reason which pair will allow greater conduction of
electricity.
(i) Silver wire at 300C or silver wire at 600C.
(ii) 0.1M CH3COOH solution or 1 M CH3COOH solution.
(iii) KCl solution at 200C or KCl solution at 500C.
Ans.(i)Silver wire at 300C,therefore,conductance of metals increases with decrease in
temperature.
(ii) 0.1MCH3COOH, because, degree of ionisation increases with dilution.
(iii) KCl solution at 500C, therefore, mobility of ions increases at higher temperature.
Q.5. Calculate the maximum possible electrical work that can be obtained from the following cell
under the standard conditions at 250C. Zn|Zn2+ (aq) || Sn2+ (aq) |Sn
E0 Zn2+ |Zn= -0.76V E0Sn 2+ | Sn = - 0.14V F = 96500 C mol-1
Ans. E0cell = E0c – E0a
= - 0.14 V –(-0.76V) = 0.62V
Maximum work derived from the cell is = n F E0cell
=2 x 96500 x 0.62 J = 119660 J = 119.660 KJ
Q.6. Calculate the EMF of the following cell at 298K.
Sn|Sn2+ (0.1M) || Ag+ (0.1M) | Ag.
E0Sn2+ | Sn = -0.14V E0Ag+|Ag =0.80V R=8.314JK-1mol-1 F= 96500Cmol-1
Ans. E0cell = E0Ag+|Ag –E0Sn2+ |Sn
= 0.80V – (-0.14V) = 0.94V
0.0591 [𝑆𝑛2+]
Ecell = E0cell – 𝑛
log [𝐴𝑔+]2
0.0591 (0.1)
Ecell = 0.94V – 2
log (0.1)2

27
 = 0.94V -0.02955V
=0.94V-0.03V = 0.91V
Q.7. Calculate the standard free energy change for the reaction occurring in the cell.
Zn(s) |Zn2+ (1M)||Cu2+ (1M) |Cu(s) Given E0Zn2+ /Zn = -0.76V E0Cu2+ /Cu = 0.34V F=
96500C mol-1.How is it related to equilibrium constant for the reaction?
Ans. E0 cell = E0 cathode – E0 anode
= 0.34V – (-0.76V) = 1.10V
ΔG = -n F E0cell = -2x96500x1.10V
0

= - 212300J = -212.300 KJ
ΔG = 2.303RT log Kc
Q.8. (i) What is the necessity to use a salt bridge in a Galvanic cell?
(ii)What type of cell is a lead storage battery? Write the anode and the cathode reactions and
the overall cell reaction occurring in the use of a lead storage battery.
Ans.(i) To complete the inner circuit and to maintain the electrical neutrality of the
electrolytic solutions of the half cells we use a salt bridge in a galvanic cell.
(ii) It is a secondary cell.
Anode reaction – Pb + SO42- 🡪PbSO4 + 2e-
Cathode reaction- PbO2 + 4H+ + SO42- + 2e- 🡪PbSO4 + 2H2O
Net reaction - Pb + PbO2 + 2SO42- + 4 H+ 🡪 2PbSO4 + 2H2O
Q.9. (i) Calculate the Ecell for the following reaction at 298K
2Al(s) + 3 Cu2+ (0.01M) 🡪 2Al3+ (0.01M) + 3Cu(s)
Given E0cell = 1.98V
(ii)Give the advantages of fuel cell.
0.0591 [𝐴𝑙3+]2
Ans.(i)Ecell = E0cell – 𝑛
log [𝐶𝑢2+]3
0.0591 (0.01)2
= 1.98V – 6 log (0.01)3
0.0591
=1.98 – 6 log 102
0.0591
= 1.98 – 6 x 2log 10
=1.98 – 0.0197V = 1.961 V
(ii) Advantage of fuel cell
a) High efficiency
b) No harmful products are formed
c) No part of the cell creates environmental hazards.

28
Q.10. Write the Nernst equation and calculate the emf of the cell at 298K.
Zn|Zn2+ (0.001M) || H+(0.01M) |H2(g) (1bar)|Pt(s)
Given: E0Zn2+ /Zn = - 0.76V, E0H+/H2 = 0.00V [log 10=1]
Ans. Given cell is . Zn|Zn2+ (0.001M)|| H+(0.01M) |H2(g) (1bar)|Pt(s)
At anode : Zn 🡪 Zn2+ + 2e- At cathode : 2H+ + 2e- 🡪 H2
The net reaction: Zn + 2H+ 🡪 Zn2+ + H2
Nernst equation for the given cell is
0.0591 [𝑍𝑛2+]
𝐸𝑐𝑒𝑙𝑙 = 𝐸0 𝑐𝑒𝑙𝑙 – 2
𝑙𝑜𝑔[𝐻+]2
0.0591 0.0010
= (0. 00 + 0. 76)– 2 log 𝑙𝑜𝑔 0.01𝑥0.01
0.76 0.0591 1.52−0.059 1.461
= 1 – 2 = 2
= 2 = 0.7305V

5 MARKS QUESTION
Q1) (a)
i.State Kohlrausch law of independent migration of ions.
ii.Why does the conductivity of a solution decrease with dilution.
(b) Calculate emf of the following cell at 250C
Mg(s)/Mg2+ (0.01M) || Cu2+ (0.001M)| Cu(s)
Given E0cell = 2.71V
Write direction of flow of current when external opposing potential applied is
(i)less than 2.71 V (ii) 2.78 V
(c) The cell constant of a conductivity cell is 0.146cm-1 . What is conductivity of
0.01M
Solution of electrolyte at 298 K, if resistance of cell is 1000 ohm?
Ans : 1(a)
i.According to this law limiting molar conductivity of an electrolyte can be expressed as the
sum of individual contribution of cation and anion.
ii.Conductivity of a solution decreases with solution due decrease in number of ions per unit
volume.
(b) For the reaction, Mg(s) + Cu 2+ (0.01M) → Mg 2+ (0.001M) + Cu(s)
E0cell = 2.71V
Ecell = E0cell – (0.0591/n) (log Mg 2+/ Cu 2+)
= 2.71 – (0.0591/2) log (.001/.01)
= 2.71 + (0.0591/2) log 10
= 2.74 V

(i) current flows from Cu to Mg

(ii) current will flow from Mg to Cu.
(c) Conductivity = Cell constant / resistance
= 0.146/1000

29
 = 1.46 x 10-4 S cm-1

Q2)(a) Using E0values of X and Y given below, predict which is better for coating the surface
of
iron to prevent corrosion.
Given E0 X2+/X = -2.36V, E0Y2+/Y = - 0.14V , E0Fe2+/Fe = -0.44V
(b) Conductivity of .0002 M methanoic acid is 8x 10-5S cm-1. Calculate its molar
conductivity and degree of dissociation if, limiting molar conductivity for methanoic
acid
is 404 S cm2 mol-1.
(c) Calculate the 𝝙rG0 and log Kc for the given reaction at 298K.
Ni(s) + 2 Ag+(aq) --------> Ni 2+ + 2 Ag(s)
Given: E0 Ni2+/ Ni = -0.25 V, E0 Ag+/ Ag = 0.80 V

Ans 2) (a) X
κ 𝑥 1000
(b) Lm = 𝑀
Given κ = 8 x 10-5 S cm-1
M= 2 X 10-3 mol L-1
Lm = 8 𝑋 10 𝑋 1000
−5

−3
2 𝑋 10

= 40 S cm2 mol-1
Degree of dissociation (α) = Lm / L0m
Since, L0m = 404 Scm2 mol-1 (Given)
Therefore, α = 40/404 = 0.099
(c) 𝝙rG0 = -nFE0cell
E0cell = .80 – (-0.25) = 1.05 V
E0cell = (0.0591/n ) x log Kc
Log Kc = 1.05x2 / 0.0591 = 35.3

Q3) (a) Calculate the mass of Ag deposited at cathode when a current of 2A is passed through
a solution of AgNO3 for 15 min.(Given molar mass of Ag = 108 g/mol)
(b) What type of battery is Lead storage battery? Write the reaction taking place at the
cathode when the battery is in use.
(c) In the following graph

Predict the nature of two electrolytes A and B.

Ans 3 (a) I = 2A t = 15 min

30
 Q = It = 2x15x60
= 1800 C
Mass of Ag deposited = (108 x 1800 )/ 96500
= 2.014 g
(b) Secondary Cell
Reaction at cathode when battery is in use:
PbO2(s) + SO42- + 4H+ + 2 e- --------> PbSO4 + 2 H2O
(c) A= strong electrolyte B= weak electrolyte

CASE BASE QUESTIONS

1. Increase in concentration. Electrochemical cell converts chemical energy of redox

reaction into electricity. Mercury cell, Dry cells are primary cells where as Ni-Cd cell, lead
storage battery are secondary cells. Electroehemical series is arrangement of elements in
increasing order of their reduction potential. Electrolytic cell converts electrical energy into
chemical energy which is used in electrolysis. Amount of products formed are decided with
the help of Faraday's laws of Electrolysis. Kohlrausch law helps to determine limiting molar
conductivity of weak electrolyte, their degree of ionisation (α) and their dissociation
constants. Corrosion is electrochemical phenomenon. Metal undergoing corrosion acts as
anode, loses electrons to form ions which combine with substances present in atmosphere to
form surface compounds. More reactive metals are coated over less reactive metals to
prevent corrosions.

Questions:
(a) Out of 0.5 M, 0.01 M, 0.1 M and 1.0 M which solution of KCl will have highest value of
specific conductance? Why?
OR
Write the product of electrolysis of aq. NaCI on cathode.
(b) When does electrochemical cell behaves like electrolytic cell?

(c ) Calculate maximum work obtained from the cell Ni(s) + 2Ag+(aq) 🡪 Ni2+(aq) + 2Ag(s) Eocell =
1.05V.

Ans.(a) 0.01M because specific conductivity is inversely proportional to the molarity .So
lesser the value of molarity more will be the specific conductivity.
OR
Electrolysis of an aqueous NaCl solution produces H2 gas at the cathode
(b) An electrochemical cell can behave like an electrolytic cell when an external potential
difference is applied to the cell that is greater than the cell's potential. This causes the
reaction to proceed in the opposite direction, which is a non-spontaneous reaction that is
characteristic of an electrolytic cell. The electrochemical cell then converts the electrical
energy from the external cell into chemical energy.

31
(c) 𝝙rG0 = -nFE0cell
=-2*96500*1.05 J
=-202650 J

2. Electrochemistry plays a very important part in our daily life. Primary cells like dry cell
is used in torches, wall clock, mercury cell is used in hearing aids, watches. Secondary cells
Ni-Cd cell is used in cordless phones, lithium battery is used in mobiles, lead storage battery is
used in vehicle and inverter. Fuel cells like H2 -O2 cell was used in apollo space programme. A
38% solution of sulphuric and is used in lead storage battery. Its density is 1.30 g mL -1. The
battery holds 3.5 L of the acid. During the discharge of the battery, the density of H2 SO4 falls
to 1.14 g mL -1 (20% solution by mass) (Molar mass of H2 SO4 is 98 g mol -1).
3.
Questions:
1. Write the chemical reaction taking place at anode when lead storage battery is in use.
2. How much electricity in Faraday is required to carry out the reduction of one mole of PbO2?
3. What is molarity of sulphuric acid before discharge?
4. What is mass of sulphuric acid in solution after discharge?
5. Write the products of electrolysis when dilute sulphuric acid is electrolysed using platinum
electrodes.
Ans.1. Lead storage battery is a secondary cell that can be charged by passing a current.
Reactions occurring in the lead storage battery while operating (during discharge) are:
At cathode: PbO2+SO2−4+4H++2e−→PbSO4+2H2O
At anode: Pb+SO2−4→PbSO4+2e−
overall reaction : Pb+PbO2+2H2SO4→2PbSO4+2H2O.

3. According to Faraday's law of electrolysis, 2 Faraday (F) of electricity is required to reduce

one mole of lead dioxide (PbO2):
PbO2+SO42−+4H++2e−→PbSO4+2H2Ocap P b cap O sub 2 plus cap S cap O sub 4 raised to
the 2 minus power plus 4 cap H raised to the positive power plus 2 e raised to the negative
power right arrow cap P b cap S cap O sub 4 plus 2 cap H sub 2 cap O
𝑃𝑏𝑂2+𝑆𝑂2−4+4𝐻++2𝑒−→𝑃𝑏𝑆𝑂4+2𝐻2𝑂
3.The molarity of sulfuric acid in a lead storage battery before discharge is around 5.013
M. This is calculated by dividing the number of moles of sulfuric acid by the volume of the
solution.
4. During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294
to 1.139 g mL−1. Sulphuric acid of density 1.294 g mL−1 is 39 % H2SO4 by mass and that of
density 1.139 g mL−1 is 20 % H2SO4 by mass.
5. Therefore, in the electrolysis of dilute aqueous Sulphuric acid, using Platinum
electrodes, Oxygen gas is liberated at the anode and hydrogen gas is liberated at the cathode.

32
CASE BASED QUESTIONS
Electroplating Silver: A jewelry manufacturer wants to electroplate a silver layer onto a
copper bracelet. The electroplating process involves passing a direct current through an
electrolyte solution containing silver ions to deposit a thin layer of silver on the surface of the
copper bracelet.
a. Explain Faraday's First law.
b. Second Laws of Electrolysis as they apply to the electroplating process described above.
c. If the current passed through the electrolyte is 2.5 A and the electroplating process is
carried out for 4 hours, calculate the mass of silver deposited on the bracelet. Assume the
efficiency of the process is 100%.
OR
c. Silver is electrodeposited on a metallic vessel of surface area 800 cm2 by passing current of
0.2 ampere for 3 hrs. Calculate the thickness of silver deposited .Density of Ag = 10.47gcm-3
Atomic mass of Ag = 107.92 amu.
Ans.a. Faraday's First Law of Electrolysis states that the mass of a substance altered at an
electrode during electrolysis is directly proportional to the quantity of electricity (electric
charge) passed through the electrolyte.
b. Faraday's Second Law of Electrolysis states that the mass of a substance altered at an
electrode is proportional to its equivalent weight and the quantity of electricity passed. This
law essentially supports the first law by quantifying the relationship between the substance's
equivalent weight and the charge passed.
c. Calculation of Mass of Silver Deposited:
To find the mass of silver deposited, follow these steps:
Calculate the total charge (Q):
Q=I×t = 4x3600 = 14400 seconds = 2.5 A x 14400 s = 36000C
Determine the molar mass and valency of silver:
Molar mass of silver (Ag) = 107.87 g/mol,
Valency (z) = 1 (since silver ion Ag+ involves one electron).
calculate the mass of silver deposited:
𝑄𝑥𝑀 3600𝑥107.87
m= 𝐹𝑥𝑍
= 96500𝑥1
= 40.3gm.
OR
c. Ag+ +e- 🡪 Ag
Amount of charge passed(Q) =It =0.2Ax3x60x60 =2160C
96500C of charge deposite Ag =107.92 gm
107.92
In 1C of charge deposite Ag = 96500
107.92
2160 C of charge deposite Ag = 96500 x2160 =2.42gm
Mass = Volume x Density = Area x Thickness x Density
𝑀𝑎𝑠𝑠 2.42
Thickness = 𝐴𝑟𝑒𝑎 𝑥𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = 800 𝑥 10.47
=2.89x10-4cm
2. Rahul set up an experiment to find resistance of aqueous KCl solution for different
concentrations at 298K using a conductivity cell connected to a wheat-stone bridge. He fed
33
the Wheatstone bridge with the A.C. power in the audio frequency range 550 to 5000 cycles
per second . Once the resistance was calculated from null point he also calculated the
conductivity k and molar conductivity λm and recorded his readings in tabular form.
S.No. Conc.(M) K (S cm-1) Λm (S cm2mol-1)
1. 1.00 111.3 x 10-3 111.3
2. 0.10 12.9 x 10-3 129.0
3. 0.01 1.41 x10-3 141.0
Answer the following questions:
(i) Why does conductivity decrease with dilution?
(ii) If Λ0m KCl is 150.0 S cm2 mol-1 , calculate the degree of dissociation of 0.01 M KCl.
(iii) If Rahul had used HCl insted of KCl then would you expect the Λm values to be more or
less then those per KCl for a given concentration.Justify.
OR
(iii) Amit a classmate of Rahul repeated the same experiment with CH3COOH solution instead
of KCl solution . Give one point that would be similar and one that would be different in his
observations as compared to Rahul.
Ans. (i) It is because number of ions per unit volume decreases.
Λ𝑚 141
(ii) Degree of dissociation (α ) = Λ0𝑚
= 150
=0.94
(iii) HCl will have higher Λm value because mobility of H+ ions is more than K+ due to lesser
molar mass.
OR
(iii) Similarities: Λm increases with dilution in both cases gradually for KCl and rapidly for
CH3COOH.
Differences: Λ0m for KCl can be obtained by extrapolation for KCl but not for CH3COOH
because a curve cannot be extrapolated

34
 UNIT-III
CHEMICAL KINETICS

MCQs-
1. In the reaction A + B → product, if B is taken in excess, then it is an example of
a) Second order reaction b) Zero order reaction
c) Pseudo first order reaction d) First order reaction
2. The rate of reaction increase with increase in temperature because
a) Activation energy of reacting molecules decreases.
b) Kinetic energy of the product molecules increases.
c) The fraction of molecules possessing an energy equal to the activation energy are more.
d) The collision between molecules decreases.
3. Which is not a characteristic of a catalyst?
a) It changes the equilibrium constant. b) It alters the reaction path.
c) It increases the rate of reaction. d) It does not affect the Gibbs energy
4. Higher order (>3) reactions are rare due to
a) Low probability of simultaneous collision of all reacting species.
b) Increase in entropy and activation energy as more molecules are involved.
c) Shifting of equilibrium towards reactants due to elastic collisions.
d) Loss of active species on collision.
5. When initial concentration of reactant is double in a reaction, the half-life period is not
affected. The order of reaction is
a) Second b) Zero
c) First d) More than zero but less than first.
6. For which of the following reactions, the temperature co-efficient is maximum
a) A → B, Ea = 50 KJ b) P → Q, Ea = 40 KJ
c) X → B, Ea = 60 KJ d) W → Z, Ea = 80 KJ
7. A hypothetical reaction 2p + q → s + r, has rate constant as 2.0 × 10-3 mol lit-1 sec-1. The
order of the reaction is
a) Unpredictable b) Zero
c) One d) Two
1
8. If the graph plotted between ln k and 𝑇
for the first order reaction, the slope of the straight
line so obtained is given by
𝐸𝑎 𝐸𝑎
a) - 𝑅
b) - 2.303𝑅
2.303 𝐸𝑎
c) - 𝐸𝑎𝑅
d) - 2.303

9. For the reaction, NO2(g) + CO(g) → NO(g) + CO2(g), the correct expression for the rate of the
reaction is
𝑑[𝑁𝑂2] 𝑑[𝐶𝑂2]
a) rate = - 𝑑𝑡
b) rate = - 𝑑𝑡
[ ]
𝑑 𝑁𝑂2 −𝑑[𝐶𝑂] 𝑑[𝐶𝑂]
c) rate = 𝑑𝑡
d) rate = 𝑑𝑡

35
 1
10. The plot of log k vs 𝑇
helps to calculate
a) Energy of activation b) Rate constant of the reaction
c) Order of the reaction d) Energy of activation as well as frequency factor.
14
11. The half life of a radioactive decay of C is 5730 years. An archeological artifact containing
wood had only 80% of the 14C found in the living tree. The age of the wood is:
a) 1896 𝑦𝑒𝑎𝑟𝑠 b) 1847 years
c) 1803 years d) 1805 years
12. The rate law for the chemical reaction, 2NO2Cl → 2NO2 + Cl2 is r = k[NO2Cl]. Which of the
following is rate controlling steps?
a) 2NO2 + Cl → NO2Cl b) NO2Cl + Cl → NO2 + Cl2
c) NO2Cl(s) → NO2Cl (g) d) 2NO2Cl → 2NO2 + Cl2
13. Which of the following expression is correct for the rate of reaction given below?
5Br– (aq) + BrO3– (aq) + 6H+ (aq) → 3Br2(aq) + 3H2O(l)
− + − +
a) − [∆𝑡 ] = b) − [∆𝑡 ] =
∆ 𝐵𝑟 ∆[𝐻 ] ∆ 𝐵𝑟 5 ∆[𝐻 ]
− 5 ∆𝑡
− 6 ∆𝑡
− + − +
c) − [∆𝑡 ] = d) − [∆𝑡 ] =
∆ 𝐵𝑟 6 ∆[𝐻 ] ∆ 𝐵𝑟 ∆[𝐻 ]
− 5 ∆𝑡
− 6 ∆𝑡
14. The reaction , 3A → B + C would be a zero order reaction when:
a) The rate of reaction is proportional to square of concentration of A.
b) The rate of reaction remain same at any concentration of A.
c) The rate of reaction double if concentration of B is increased to double.
d) The rate remains unchanged at concentration of B and C
15. In a reversible reaction, the activation energies of the forward and the backward reaction are
equal. In such reactions
a) H= 0 b) S= 0
c) Order is zero d) H≠0 , S≠0
16. The rate of formation of SO3 in the reaction
2SO2 + O2 ⇌ 2SO3, is 100kg min-1
a) The rate of disappearance of SO2 is 50 kg min-1
b) The rate of disappearance of O2 is 50 kg min-1
c) The rate of appearance of SO2 cannot be increased by adding catalyst as reaction is in
equilibrium.
d) The overall order of reaction is three.
17. The rate law for the reaction
RCl + NaOH → ROH + NaCl, is given by r = k[RCl]. The rate of reaction is
a) Double by doubling concentration of NaOH
b) Is halved by reducing concentration of RCl by one half
c) Is increased by decreasing the temperature
d) Is unaffected by change in temperature.
18. The ratio of the time period for ¾ of the reaction of first order to complete to that required
for half of the reaction is:
a) 4 : 3 b) 3 : 2
c) 2 : 1 d) 1 : 2
36
19. Half of the substance ‘A’ following first order kinetic is 5 days. Starting with 100 g of A,
amount left after 15 days is
a) 25 g b) 50 g
c) 12.5 g d) 6.25 g
20. The ratio t1/8 : t1/2 for the first order reaction is
a) 3 b) 5
c) 2 d) 7
21. A first order reaction has a half life period of 34.65 second. Its rate constant is
a) 0.2 × 10-2 sec-1 b) 4 × 10-2 sec-1
c) 20 sec-1 d) 2 × 10-2 sec-1
22. The first order rate constant for the decomposition of N2O5 is 6.2 × 10-3 sec-1. The t1/2 of the
decomposition
a) 117.7 sec b) 111.7 sec
c) 228.4 sec d) 168.9 sec
23. The rate of a chemical reaction doubles for every 10° C rise to temperature. If the
temperature is raised by 50° C the rate of reaction increased by about
a) 10 times b) 24 times
c) 32 times d) 64 times
24. For a reaction 2N2O5 → 4NO2 + O2 rate and rate constant are 1.02 × 10-4 and 3.4 × 10-5 sec-1
respectively, if rate law is rate = k[N2O5] then concentration of N2O5 at that time will be
a) 1.732 b) 3
-4
c) 1.02 × 10 d) 3.4 × 10-5
25. Activation energy of a chemical reaction can be determined by _____________.
(a) determining the rate constant at standard temperature.
(b) determining the rate constants at two temperatures.
(c) determining probability of collision.
(d) using catalyst.
26. Consider the Arrhenius equation given below and mark the correct option.
k = A e–Ea/RT
(a) Rate constant increases exponentially with increasing activation energy and decreasing
temperature.
(b) Rate constant decreases exponentially with increasing activation energy and decreasing
temperature.
(c) Rate constant increases exponentially with decreasing activation energy and decreasing
temperature.
(d) Rate constant increases exponentially with decreasing activation energy and increasing
temperature.
27. During decomposition of an activated complex
(a) energy is always released (b) energy is always absorbed
(c) energy does not change (d) reactants may be formed
28. Which of the following statements are in accordance with the Arrhenius equation?
(a) Rate of a reaction increases with increase in temperature.

37