Chemical Kinetics

JEE Main 2024 (Online) 30th January Evening ShiftNumericalEasy📈 Topic in 5% papers+4

Ques. 1JEE Main 2024, 30 January (Evening Shift)

NO2 required for a reaction is produced by decomposition of N2O5 in CCl4 as by equation 2N2O5(g) 4NO2(g) + O2(g) The initial concentration of N2O5 is 3 mol L-1 and it is 2.75 mol L-1 after 30 minutes. The rate of formation of NO2 is x 10-3 mol L-1 min-1, value of x is . (nearest integer)

Answer

Solution

Rate of disappearance of N2O5 = 3 - 2.7530 = 0.2530 mol L-1 min-1

From the equation: [NO2] t = 42 (- [N2O5] t )

⇒ Rate of formation of NO2 = 2 0.2530 = 160

= 16.67 10-3 mol L-1 min-1

∴ x ≈ 17

Answer: 17

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Ques. 2JEE Main 2024, 29 January (Morning Shift)

For a reaction taking place in three steps at same temperature, overall rate constant K = K1 K2K3. If Ea1, Ea2 and Ea3 are 40, 50 and 60 kJ/mol respectively, the overall Ea is kJ/mol.

Ques. 3JEE Main 2024, 29 January (Evening Shift)

The half-life of radioisotope bromine - 82 is 36 hours. The fraction which remains after one day is 10-2. (Given antilog 0.2006 = 1.587)

Ques. 4JEE Main 2024, 27 January (Morning Shift)

Consider the following data for the given reaction 2HI(g) H2(g) + I2(g) [figure] The order of the reaction is .

Ques. 5JEE Main 2024, 27 January (Evening Shift)

Time required for completion of 99.9% of a First order reaction is times of half life (t1/2) of the reaction.

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