1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

For a first order reaction, A $$ \to $$ P, t1/2 (half-life) is 10 days The time required for $${1 \over 4}$$th conversion of A (in days) is : (ln 2 = 0.693,    ln 3 = 1.1)
A
5
B
3.2
C
4.1
D
2.5

Explanation

For first order reaction,

The half life, t$${1 \over 2}$$ = $${{0.693} \over k}$$

Here given t$${1 \over 2}$$ = 10 days

$$\therefore\,\,\,$$ k = $${{0.693} \over {10}}$$ = 0.0693 days$$-$$1

Now, the time required for $${1 \over 4}$$th conversion of A is ,

t = $${{2.303} \over k}$$ log10 $$\left( {{a \over {a - x}}} \right)$$

= $${{2.303} \over {0.0693}}$$ log $$\left( {{1 \over {1 - {1 \over 4}}}} \right)$$

= $${{2.303} \over {0.0693}}$$ log $$\left( {{4 \over 3}} \right)$$

= 4.1 days.
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

If 50% of a reaction occurs in 100 second and 75% of the reaction occurs in 200 secod, the order of this reaction is :
A
Zero
B
1
C
2
D
3

Explanation

Assume initial concentration of the reactant = 1 M

After 100 second, concentration becomes of the reactant

= 1 $$ \times $$ $${{50} \over {100}}$$ = 0.5 M

After 200 second, concentration of the reactant becomes

= 1 $$ \times $$ $${{25} \over {100}}$$ = 0.25 M.

So, in first 100 second reactant concentration becomes half of initial and in the second 100 second concentration becomes 0.5 M to 0.25 M, which is also half of 0.5 M.

So, after each 100 second period, concentration of reactant becomes half of initial concentration.

So, 100 second is the half life period and it is independent of concentration of reactant. This is characteristic of first order reaction.
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

The following results were obtained during kinetic studies of the reaction ;
2A + B $$ \to $$ Products

Experiment [A] (in mol L$$-$$1) [b] (in mol L$$-$$1) Initial Rate of reaction
(In mol L$$-$$1 min$$-$$1)
I 0.10 0.20 6.93 G 10$$-$$3
II 0.10 0.25 6.93 G 10$$-$$3
III 0.20 0.30 1.386 G 10$$-$$2


The time (in minutes) required to consume half of A is :
A
5
B
10
C
1
D
100

Explanation

Rate (r) = K[A]x [B]y

From experienced (I),

6.93 $$ \times $$ 10$$-$$3 = K (0.1)x (0.2)y . . . . (1)

From experiment (II),

6.93 $$ \times $$ 10$$-$$3 = K(0.1)x (0.25)y . . . . . (2)

equation (1) and (2) satisfies when y = 0

From experiment (III),

1.386 $$ \times $$ 10$$-$$2 = K (0.2)x (0.30)y

$$ \Rightarrow $$   1.386 $$ \times $$ 10$$-$$2 = K(0.2)x . . . . . .(3)

Dividing (2) by (3) we get,

$${1 \over 2} = {\left( {{1 \over 2}} \right)^x}$$

$$ \Rightarrow $$   x = 1

$$ \therefore $$   Rate (r) = K (0.1)1 (0.2)0

$$ \Rightarrow $$  6.93 $$ \times $$ 10$$-$$3 = K (0.1)

$$ \Rightarrow $$   K = 6.93 $$ \times $$ 10$$-$$2

Now,

$${t_{{1 \over 2}}} = {{0.693} \over {{K}}}$$

= $${{0.693} \over {K}}$$

= $${{0.693} \over {6.93 \times {{10}^{ - 2}}}}$$

= 10

To consume half of A it becomes 10/2 = 5 min
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

For the reaction, 2A + B $$ \to $$ 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 mol L$$-$$1s$$-$$1 to 0.6 mol L$$-$$1s$$-$$1.
A
Total order of the reaction is 4
B
Order of the reaction with respect to B is 2
C
Order of the reaction with respect to B is 1
D
Order of the reaction with respect to A is 2

Explanation

$$r = K{\left[ A \right]^x}{\left[ B \right]^y}$$

$$ \Rightarrow 8 = {2^3} = {2^{x + y}}$$

$$ \Rightarrow x + y = 3\,...(1)$$

$$ \Rightarrow 2 = {2^x}$$

$$ \Rightarrow x = 1,y = 2$$

Order w.r.t. A = 1

Order w.r.t. B = 2

Questions Asked from Chemical Kinetics and Nuclear Chemistry

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