JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

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JEE : JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

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Q.1. During the nuclear explosion, one of the products is 90Sr with half-life of 6.93 years. If 1 µg of 90Sr was absorbed in the bones of a newly bom baby in place of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically?    (2020)
Ans. (23.03) 

Given, half-life of 90Sr = 6.93 year; amount = 1 × 10–6g
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
From the first order kinetic
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev 
For 90% decay of 90Sr,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
⇒ t = 23.03 log 10
⇒ t = 23.03 years

Q.2. For the reaction
2H2(g) + 2NO(g) → N2(g) + 2H2O(g),
the observed rate expression is, rate =JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev. The rate expression for the reverse reaction is    (2020)

(1) kb[N2][H2O]2
(2) JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
(3) kb [N2][H2O]
(4) JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Ans. (4) 

Given, rate expression for reaction,
2H2 (g) + 2NO(g) → N2(g) + 2H2O(g)

rf=kf[NO]2[H2]

Now,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Rearranging the above equation,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

=> rf = rb

JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev


Q.3. The rate of a certain biochemical reaction at physiological temperature (T) occurs 10times faster with enzyme than without it. The change in the activation energy upon adding enzyme is    (2020)
(1) −6(2.303)RT
(2) −6RT
(3) +6(2.303)RT
(4) +6RT
Ans. (1) 

From Arrhenius equation without enzyme
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev    (1)
With enzyme
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev    (2)
On Dividing Eq. (1) by (2), we get
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev    (2)
Taking log of Eq. (3), we get
E'a - E= - 6(2.303)RT

Q.4.Consider the following plots of rate constant versus 1/T for four different reactions. Which of the following orders is correct for the activation energies of these reactions?    (2020)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
(1) Eb > Ea > Ed > Ec
(2) Ea > Ec > Ed > Eb
(3) Ec > Ea > Ed > Eb
(4) Eb > Ed > Ec > E
a
Ans. (3) 

From Arrhenius equation without enzyme
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
Taking log of Eq. (1), we get
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
By plotting log k against 1/T,  the slope =JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev and intercept = log A
From the graph of four different reaction,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
The plot having high value of slop will have high value of activation energy. Thus, the correct order for activation energy of the reactions is: Ec > Ea > Ed > Eb.

Q.5. For the following reactions
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

it was found that the Ea is decreased by 30 kJ mol–1 in the presence of catalyst. If the rate remains unchanged, the activation energy for catalyzed reaction is (Assume pre exponential factor is same)    (2020)
(1) 75 kJ mol–1
(2) 105 kJ mol–1
(3) 135 kJ mol–1
(4) 198 kJ mol–1
Ans. (1)

 Given, Eal = Ea and Ea2 = Ea - 30
T1 = 700 K and T2 = 500 K
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev    (1) [Given, Rate remains unchanged]
Substituting the values in Eq. (1), we get
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
⇒ 500Ea = 700Ea - 2100
⇒ 2100 = 700Ea - 500Ea
⇒ 2100 = 200Ea
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
⇒ Ea = 105 KJ mol-1
Thus, activation energy in presence of catalyst is Ea2 = E- 30
⇒ Ea2 = 105 - 30 = 75 kJ mol-1

Q.6. A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population of lactobacillus acidophilus in it doubles. The activation energy (in kJ mol−1) for this process is closest to ________.    (2020)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
Ans. (3.98) 

Using Arrhenius equation at two different temperature,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev    (1)
Since, milk splits after 60 min. at 300 K and after 40 min. at 400 K.
Substituting the values in Eq. (1), we get
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev    JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
⇒ Ea = 3.984 kJ mol−1

Q.7. The following results were obtained during kinetic studies of the reaction;
2A + B → Products
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
The time (in minutes) required to consume half of A is:    (2019)
(1) 5
(2) 10
(3) 1
(4) 100
Ans.
(1)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.8. For the reaction, 2A + B → products, when the concentrations of A and B both were doubled, the rate of the reaction increased from 0.3 mol L-1s-1 to 2.4 mol L-1s-1. When the concentration of A alone is doubled, the rate increased from 0.3 L-1s-1 to 0.6 mol L-1s-1. Which one of the following statements is correct?    (2019)
(1) Total order of the reaction is 4
(2) Order of the  reaction with respect to B is 2
(3) Order of the reaction with respect to B is 1
(4) Order of the reaction with respect to A is 2
Ans.
(2)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
∴ y = 2

Q.9. Consider the given plots for a reaction obeying Arrhenius equation (0°C < T < 300°C): (K and Ea are rate constant and activation energy, respectively)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
Choose the correct option:    (2019)
(1) I is right but II is wrong
(2) Both I and II are correct
(3) I is wrong but II is right
(4) Both I and II are wrong
Ans. 
(2)
From Arrhenius equation,
K= Ae-Ea/RT
So, as Ea increases, e-Ea/RT decreases, K decreases and as T increases, Ea/RT decreases, e-Ea/RT increases,

Q.10. For an elementary chemical reaction, JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev the expression for d[A]/dt is:    (2019)
(1) k1[A2] - k-1[A]2
(2) 2k1[A2] - k-1[A]2
(3) k1 [A2] + k-1[A]2
(4) 2k1[A2] - 2k-1[A]2
Ans. 
(4)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.11. If a reaction follows the Arrhenius equation, the plot lnk vs 1/(RT) gives straight line with a gradient (-y) unit. The energy required to activate the reactant is:    (2019)
(1) y/R unit
(2) y unit
(3) yR unit
(4) - y unit
Ans.
(2)
From Arrhenius equation,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
slope = -y (given)
-y = -Ea
⇒  Ea = y

Q.12. The reaction 2X → B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the half-life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be:    (2019)
(1) 9.0 h
(2) 12.0 h
(3) 18.0 h
(4) 7.2 h
Ans.
(3)
For zero order reaction
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
t=18 hrs

Q.13. Decomposition of X exhibits a rate constant of 0.05  μg/year. How many years are required for the decomposition of 5 μg of X into 2.5  μg?    (2019)
(1) 50
(2) 25
(3) 20
(4) 40
Ans.
(1)
Rate constant of decomposition of X= 0.05 mg/year. Unit of rate constant confirms that the decomposition of X is a zero order reaction.
For zero order kinetics,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.14. For a reaction, consider the plot of In k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10-5 s-1, then the rate constant at 500 K is:    (2019)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
(1) 10-6 s-1
(2) 2 x 10-4 s-1
(3) 10-4 s-1
(4) 4 x 10-4 s-2
Ans.
(3)
From Arrhenius equation,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.15. For the reaction 2A + B → C, the values of initial rate at different reactant concentrations are given in the table below.
The rate law for the reaction is:    (2019)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
(1) Rate = k[A][B]2
(2) Rate = k[A]2[B]2
(3) Rate = k[A][B]
(4) Rate = k[A]2[B]
Ans. 
(1)
2A + B → C
Rate = k[A]x [B]y
Exp-1, 0.045 = k[0.05]x [0.05]y    ...(i)
Exp-2, 0.090 = k[0.1]x [0.05]y    ...(ii)
Exp-3, 0.72 = k[0.2]x [0.1]y    ...(iii)
Divide equation (i) by equation (ii)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
Divide equation (i) by equation (iii)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
Rate law = k[A]1 [B]2.

Q.16. For a reaction scheme JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev if the rate of formation of B is set to be zero then the concentration of B is given by:    (2019)
(1) (k1 - k2) [A]
(2) k1k2 [A]
(3) (k1 + k2) [A]
(4) (k1/k2) [A]
Ans. 
(4)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.17. The given plots represents the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reactions are:    (2019)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev 

JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
(1) 1, 0
(2) 1, 1
(3) 0, 1
(4) 0, 2
Ans.
(1)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
For First order reaction
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.18. A bacterial infection in an internal wound grows as N2(t) = N0 exp(t), where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as dN/dt = -5N2. What will be the plot of N0/N vs. t after 1 hour?    (2019)
(1) JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
(2) JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
(3) JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
(4) JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
Ans.
(3)
When drug is administered bacterial growth is given by dN/dt = -5N2
On integrating the above equation,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
The above equation is similar to straight line equation with positive slope.
Thus N0/Nt increases linearly with t.

Q.19. For the reaction of H2 with I2, the rate constant is 2.5 x 10-4 dm3 mol-1 s-1 at 327°C and 1.0 dm3 mol-1 s-1 at 527°C. The activation energy for the reaction, in kJ mol-1 is: (R = 8.314J K-1 mol-1)    (2019)
(1) 166
(2) 150
(3) 72
(4) 59
Ans.
(1)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.20. In the following reaction: xA → yB
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
‘A' and ‘B’ respectively can be:    (2019)
(1) n-Butane and Iso-butane
(2) C2H2 and C6H6
(3) C2H4 and C4H8

(4) N2O4 and NO2
Ans.
(3)
xA → yB
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
Comparing this equation with the equation given in question. We get,
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
∴ x/y = 2
∴ The reaction is of type 2A → B.
Hence, option (3) is correct.

Q.21. NO, required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation,
2N2O5(g) → 4NO2(g) + O2(g).

The initial concentration of N2O5 is 3.00 mol L-1 and it is 2.75 mol L-1 after 30 minutes. The rate of formation of NO2 is:    (2019)
(1) 4.167 x 10-3 mol L-1 min-1
(2) 1.667 x 10-2 mol L-1 min-1
(3) 8.333 x 10-3 mol L-1 min-1
(4) 2.083 x 10-3 mol L-1 min-1
Ans.
(2)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
According to the question
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev 

Q.22. 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:    (2018)
(1) 2
(2) 3
(3) 1
(4) 0
Ans. (1)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.23. N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mm Hg to 87.5 mm Hg. The pressure of the gaseous mixture after 100 minute at constant temperature will be:    (2018)
(1) 106.25 mm Hg 
(2) 116.25 mm Hg 
(3) 136.25 mm Hg 
(4) 175.0 mm Hg 
Ans. (1)
2N2O5 → 4NO2 + O2
p − 2x         4x         x
pt = p − 2x + 4x + x
pt = p + 3x 

JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
at  t = 0,   pt = p = 50 mm Hg
at  t = 50 mm,   pt = 87.5 mm Hg
p + 3x = 87.5
p = 87.5 − 3x
50 = 87.5 − 3x
12.5 = x 

JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
p − 2x = 50 − 2(12.5) = 25 
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
Since K will remain same  
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
50 = 50 × 4 − 8y
50 = 200 − 8y
8y = 150
y =  18.75
pt = p + 3y
= 50 + 3 (18.73) = 106.25 mm Hg 


Q.24. If 50 % of a reaction occurs in 100 second and 75 % of the reaction occurs in 200 second, the order of this reaction is:    (2018)
(1) 2
(2) 3
(3) Zero
(4) 1
Ans. (4)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
First order reaction as half life is constant. 

Q.25. Two reactions R1 and R2 have identical pre - exponential factors. Activation energy of Rexceeds that of Rby 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln (k1/ k2) is equal to
(R = 8.314 J mole–1 K–1)    (2017)
(1) 8
(2) 12
(3) 6
(4) 4
Ans.  (4)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.26. The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.    (2017)
(1) 4.92 K
(2) 9.84 K
(3) 19.67 K
(4) 2.45 K
Ans.
(1)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev ...(i)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev ...(ii)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
= 304.92

Q.27. The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is:
(Assume activation energy and pre-exponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol-1 K-1)    (2017)
(1) 53.6 kJ mol-1 
(2) 214.4 kJ mol-1 
(3) 107.2 kJ mol-1 
(4) 26.8 kJ mol-1
Ans.
(3)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
= 107165.79 J = 107.165 KJ

Q.28. 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:    (2016)
(1) 6.93 x 10-2 mol min-1 
(2) 6.93 x 10-4 mol min-1 
(3) 2.66 L min-1 at STP 
(4) 1.34 x 10-2 mol min-1 
Ans. (2)
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev
For a first order reaction
K = 2.303/t log a/(a-x)
Given a = 0.5, (a-x) = 0.125, t = 50 min
∴ k = 2.303/50 log 0.5/0.125
= 2.78 x 10-2 min-1
r = d[H2O2] = 2.78 x 10-2 x 0.05
= 1.386 x 10-3 mol min-1
Now
JEE Main Previous year questions (2016-20): Chemical Kinetics & Nuclear Chemistry Notes | EduRev

Q.29. The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process show below:
O3(g) + Cl*(g) → O2(g) + ClO*(g) ...(i)
ki = 5.2 × 109 L mol-1 s-1 
ClO*(g) + O*(g) → O2(g) + Cl*(g) ...(ii)
[Kii = 2.6 x 1010 L mol-1s-1]
The closest rate constant for the overall reaction O3(g) + O*(g) → 2O2(g) is:    (2016)
(1) 1.4 × 1020 L mol-1 s-1
(2) 5.2 × 109 L mol-1 s-1
(3) 3.1 × 1010 L mol-s-1
(4) 2.6 × 1010 L mol-1 s-1
Ans.
(2)
The rate constant of overall reaction depends slowest step. Hence equation(i) is slowest step. Option(2) is correct.

Q.30. The rate law for the reaction below is given by the expression k [A][B]
A + B → Product
If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be:    (2016)
(1) 9 k
(2) 3 k
(3) k/3
(4) k
Ans.
(4)
Rate constant is independent of concentration.


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