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**Q.1. During the nuclear explosion, one of the products is ^{90}Sr with half-life of 6.93 years. If 1 µg of ^{90}Sr 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)**

Given, half-life of ^{90}Sr = 6.93 year; amount = 1 × 10^{–6}g

From the first order kinetic

For 90% decay of ^{90}Sr,

⇒ t = 23.03 log 10

⇒ t = 23.03 years**Q.2. For the reaction2H _{2}(g) + 2NO(g) → N_{2}(g) + 2H_{2}O(g),the observed rate expression is, rate =. The rate expression for the reverse reaction is (2020)**

(2)

(3) k

(4)

Given, rate expression for reaction,

2H_{2} (g) + 2NO(g) → N_{2}(g) + 2H_{2}O(g)

r_{f}=k_{f}[NO]^{2}[H_{2}]

Now,

⇒

Rearranging the above equation,

=> r_{f} = r_{b}

**Q.3. The rate of a certain biochemical reaction at physiological temperature (T) occurs 10 ^{6 }**

From Arrhenius equation without enzyme

(1)

With enzyme

(2)

On Dividing Eq. (1) by (2), we get

⇒ (2)

Taking log of Eq. (3), we get

E'_{a} - E_{a }= - 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)****(1) E _{b} > E_{a} > E_{d} > E_{c}**

(2) E_{a} > E_{c} > E_{d} > E_{b}

(3) E_{c} > E_{a} > E_{d} > E_{b}

(4) E_{b} > E_{d} > E_{c} > E

From Arrhenius equation without enzyme

Taking log of Eq. (1), we get

By plotting log k against 1/T, the slope = and intercept = log A

From the graph of four different reaction,

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: E_{c} > E_{a} > E_{d} > E_{b}.**Q.5. For the following reactions**

Given, E_{al} = E_{a} and E_{a2} = E_{a} - 30

T_{1} = 700 K and T_{2} = 500 K

(1) [Given, Rate remains unchanged]

Substituting the values in Eq. (1), we get

⇒ 500E_{a} = 700E_{a} - 2100

⇒ 2100 = 700E_{a} - 500E_{a}

⇒ 2100 = 200E_{a}

⇒_{}

⇒ E_{a} = 105 KJ mol^{-1}

Thus, activation energy in presence of catalyst is Ea_{2} = E_{a }- 30

⇒ E_{a2} = 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)**

Using Arrhenius equation at two different temperature,

(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

⇒ E_{a} = 3.984 kJ mol^{−1}**Q.7. The following results were obtained during kinetic studies of the reaction;2A + B → ProductsThe time (in minutes) required to consume half of A is: (2019)(1) 5(2) 10(3) 1(4) 100Ans.** (1)

(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.

∴ y = 2

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.

From Arrhenius equation,

K= Ae

So, as E

(1) k

(2) 2k

(3) k

(4) 2k

Ans.

⇒

(1) y/R unit

(2) y unit

(3) yR unit

(4) - y unit

Ans.

From Arrhenius equation,

slope = -y (given)

-y = -Ea

⇒ E

(1) 9.0 h

(2) 12.0 h

(3) 18.0 h

(4) 7.2 h

Ans.

For zero order reaction

t=18 hrs

(1) 50

(2) 25

(3) 20

(4) 40

Ans.

Rate constant of decomposition of X= 0.05 mg/year.

For zero order kinetics,

(1) 10

(2) 2 x 10

(3) 10

(4) 4 x 10

Ans.

From Arrhenius equation,

The rate law for the reaction is: (2019)

(1) Rate = k[A][B]

(2) Rate = k[A]

(3) Rate = k[A][B]

(4) Rate = k[A]

Ans.

2A + B → C

Rate = k[A]

Exp-1, 0.045 = k[0.05]

Exp-2, 0.090 = k[0.1]

Exp-3, 0.72 = k[0.2]

Divide equation (i) by equation (ii)

Divide equation (i) by equation (iii)

Rate law = k[A]

(1) (k

(2) k

(3) (k

(4) (k

Ans.

**(1) 1, 0(2) 1, 1(3) 0, 1(4) 0, 2Ans.** (1)

For First order reaction

(1)

(2)

(3)

(4)

Ans.

When drug is administered bacterial growth is given by dN/dt = -5N

On integrating the above equation,

The above equation is similar to straight line equation with positive slope.

Thus N

(1) 166

(2) 150

(3) 72

(4) 59

Ans.

‘A' and ‘B’ respectively can be: (2019)

(1) n-Butane and Iso-butane

(2) C

(3) C

Ans.

xA → yB

Comparing this equation with the equation given in question. We get,

∴ x/y = 2

∴ The reaction is of type 2A → B.

Hence, option (3) is correct.

2N

Ans.

According to the question

(1) 2

(2) 3

(3) 1

(4) 0

(1) 106.25 mm Hg

(2) 116.25 mm Hg

(3) 136.25 mm Hg

(4) 175.0 mm Hg

2N

p − 2x 4x x

p

p

at t = 0, pt = p = 50 mm Hg

at t = 50 mm, p

p + 3x = 87.5

p = 87.5 − 3x

50 = 87.5 − 3x

12.5 = x

p − 2x = 50 − 2(12.5) = 25

Since K will remain same

50 = 50 × 4 − 8y

50 = 200 − 8y

8y = 150

y = 18.75

pt = p + 3y

= 50 + 3 (18.73) = 106.25 mm Hg

(1) 2

(2) 3

(3) Zero

(4) 1

(R = 8.314 J mole

(1) 8

(2) 12

(3) 6

(4) 4

(1) 4.92 K

(2) 9.84 K

(3) 19.67 K

(4) 2.45 K

Ans.

...(i)

...(ii)

= 304.92

(Assume activation energy and pre-exponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol

(1) 53.6 kJ mol

(2) 214.4 kJ mol

(3) 107.2 kJ mol

(4) 26.8 kJ mol

Ans.

= 107165.79 J = 107.165 KJ

(1) 6.93 x 10

(2) 6.93 x 10

(3) 2.66 L min

(4) 1.34 x 10

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

r = d[H

= 1.386 x 10

Now

O

k

ClO*(g) + O*(g) → O

[K

The closest rate constant for the overall reaction O

(1) 1.4 × 10

(2) 5.2 × 10

(3) 3.1 × 10

(4) 2.6 × 10

Ans.

The rate constant of overall reaction depends slowest step. Hence equation(i) is slowest step. Option(2) is correct.

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.

Rate constant is independent of concentration.

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