1
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A bacterial infection in an internal wound grows as N'(t) = N0 exp(t), where the time t is in hours. A does of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as $${{dN} \over {dt}} = - 5{N^2}$$. What will be the plot of $${{{N_0}} \over N}$$ vs. t after 1 hour?
A
JEE Main 2019 (Online) 10th April Morning Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 104 English Option 1
B
JEE Main 2019 (Online) 10th April Morning Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 104 English Option 2
C
JEE Main 2019 (Online) 10th April Morning Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 104 English Option 3
D
JEE Main 2019 (Online) 10th April Morning Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 104 English Option 4
2
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Consider the given plot of enthalpy of the following reaction between A and B.
A+ B $$ \to $$ C + D
Identify the incorrect statement. JEE Main 2019 (Online) 9th April Evening Slot Chemistry - Chemical Kinetics and Nuclear Chemistry Question 105 English
A
Formation of A and B from C has highest enthalpy of activation.
B
D is kinetically stable product.
C
C is the thermodynamically stable product
D
Activation enthalpy to form C is 5kJ mol–1 less than that to form D.
3
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 106 English
A
0,1
B
1,0
C
0,2
D
1,1
4
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
$$({k_1} - {k_2})[A]$$
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