1
JEE Main 2023 (Online) 25th January Morning Shift
Numerical
+4
-1
Change Language

For the first order reaction A $$\to$$ B, the half life is 30 min. The time taken for 75% completion of the reaction is _________ min. (Nearest integer)

Given : log 2 = 0.3010

log 3 = 0.4771

log 5 = 0.6989

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2
JEE Main 2023 (Online) 24th January Morning Shift
Numerical
+4
-1
Change Language

The number of correct statement/s from the following is __________

A. Larger the activation energy, smaller is the value of the rate constant.

B. The higher is the activation energy, higher is the value of the temperature coefficient.

C. At lower temperatures, increase in temperature causes more change in the value of k than at higher temperature

D. A plot of $$\mathrm{\ln k}$$ vs $$\frac{1}{T}$$ is a straight line with slope equal to $$-\frac{E_a}{R}$$

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3
JEE Main 2022 (Online) 29th July Evening Shift
Numerical
+4
-1
Change Language

Assuming $$1 \,\mu \mathrm{g}$$ of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ______ $$\times\, 10^{-1} \mu \mathrm{g}$$.

[Given : ln 10 = 2.303; log 2 = 0.30]

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4
JEE Main 2022 (Online) 29th July Morning Shift
Numerical
+4
-1
Change Language

The reaction between X and Y is first order with respect to X and zero order with respect to Y.

Experiment $${{[X]} \over {mol\,{L^{ - 1}}}}$$ $${{[Y]} \over {mol\,{L^{ - 1}}}}$$ $${{Initial\,rate} \over {mol\,{L^{ - 1}}\,{{\min }^{ - 1}}}}$$
I 0.1 0.1 $$2 \times {10^{ - 3}}$$
I L 0.2 $$4 \times {10^{ - 3}}$$
III 0.4 0.4 $$M \times {10^{ - 3}}$$
IV 0.1 0.2 $$2 \times {10^{ - 3}}$$

Examine the data of table and calculate ratio of numerical values of M and L. (Nearest Integer)

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