1
JEE Main 2021 (Online) 26th February Evening Shift
Numerical
+4
-1
Change Language
If the activation energy of a reaction is 80.9 kJ mol$$-$$1, the fraction of molecules at 700 K, having enough energy to react to form
products is e$$-$$x. The value of x is __________. (Rounded off to the nearest integer) [Use R = 8.31 J K$$-$$1 mol$$-$$1]
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2
JEE Main 2021 (Online) 25th February Evening Shift
Numerical
+4
-1
Change Language
The rate constant of a reaction increases by five times on increase in temperature from 27$$^\circ$$C to 52$$^\circ$$C. The value of activation energy in kJ mol$$-$$1 is _________. (Rounded off to the nearest integer)

[R = 8.314 J K$$-$$1mol$$-$$1]
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3
JEE Main 2021 (Online) 25th February Morning Shift
Numerical
+4
-1
Change Language
For the reaction, aA + bB $$ \to $$ cC + dD, the plot of log k vs $${1 \over T}$$ is given below :

JEE Main 2021 (Online) 25th February Morning Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 87 English
The temperature at which the rate constant of the reaction is 10-4 s-1 is _________ K. (Rounded off to the nearest integer)

[Given : The rate constant of the reaction is 10-5 s-1 at 500 K.]
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4
JEE Main 2021 (Online) 24th February Evening Shift
Numerical
+4
-1
Change Language
Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25$$^\circ$$C. After 9 h, the fraction of sucrose remaining is f. The value of $${\log _{10}}\left( {{1 \over f}} \right)$$ is ________ $$\times$$ 10$$-$$2. (Rounded off to the nearest integer)

[Assume : ln 10 = 2.303, ln 2 = 0.693]
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