1
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The given plots represent 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 : JEE Main 2019 (Online) 9th April Morning Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 114 English
A
0,1
B
1,0
C
0,2
D
1,1
2
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For a reaction scheme $$A\buildrel {{k_1}} \over \longrightarrow B\buildrel {{k_2}} \over \longrightarrow C$$,

if the rate of formation of B is set to be zero then the concentration of B is given by :
A
$${k_1}{k_2}[A]$$
B
$$\left( {{{{k_1}} \over {{k_2}}}} \right)[A]$$
C
$$({k_1} + {k_2})[A]$$
D
$$({k_1} - {k_2})[A]$$
3
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For the reaction 2A + B $$ \to $$ C, the values of initial rate at diffrent reactant concentrations are given in the table below. The rate law for the reaction is :

[A] (mol L-1) [B] (mol L-1) Initial Rate (mol L-1s-1)
0.05 0.05 0.045
0.10 0.05 0.090
0.20 0.10 0.72
A
Rate = k[A][B]2
B
Rate = k[A]2[B]2
C
Rate = k[A]2[B]
D
Rate = k[A][B]
4
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For a reaction consider the plot of $$\ell $$n 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 –

JEE Main 2019 (Online) 12th January Evening Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 117 English
A
10$$-$$4 s$$-$$1
B
4 $$ \times $$ 10$$-$$4 s$$-$$1
C
10$$-$$6 s$$-$$1
D
2 $$ \times $$ 10$$-$$4 s$$-$$1
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