1
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
2
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
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

Consider the following reversible chemical reactions :



The relation between K1 and K2 is :
A
K1K2 = $${1 \over 3}$$
B
K2 = K13
C
K2 = K1$$-$$3
D
K1K2 = 3

Explanation

4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface ?
A
B
C
D

Explanation

E = W + $${1 \over 2}$$mv2

K.E. = hv $$-$$ 4v0

K.E. = hv + ($$-$$ hv0)

y = mx + $$\underline C $$

Questions Asked from Chemical Kinetics and Nuclear Chemistry

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