The rate constant of a reaction at 293 K is 1 . 7 × 10 5 s - 1 . When the temperature is increased by 20 K , the rate constant is increased to 2 . 57 × 10 6 s - 1 . Calculate E a and A of the reaction.

The Arrhenius equation is a formula for the temperature dependence of reaction rates. The Arrhenius equation at two different temperature is log 10 k 2 k 1 = E a 2 . 303 R 1 T 1 - 1 T 2 log 10 2 . 57 × 10 6 1 . 7 × 10 5 = E a 2 . 303 × 8 . 314 1 293 - 1 313 1 . 179 = E a × 20 1755965 . 24 E a = 103514 J / mol We know that, k = A e - E a / RT log k 1 = log A - E a 2 . 303 RT 1 A = 4 . 89 × 10 23 s - 1

The rate of a chemical reaction doubles for an increase of 10 K in absolute temperature from 300 K . Calculate the activation energy ( E a ). R = 8 . 314 J K - 1 mol - 1 , log 2 = 0 . 3010

Given, Initial temperature T 1 = 300 K Final temperature T 2 = 300 + 10 K = 310 K We know that the rate of the reaction doubles when temperature is increased by 10 K . Therefore, let assume value of k 1 = k and that of k 2 = 2 k Also, R = 8.314 J K - 1 m o l - 1 From Arrhenius equation, l o g k 2 k 1 = E a 2.303 R T 2 - T 1 T 1 T 2 l o g 2 k k = E a 2.303 × 8 . 314 310 - 300 310 × 300 ⇒ l o g 2 = E a 2.303 × 8.314 10 300 × 310 ⇒ E a = 2.303 × 8.314 × 300 × 310 × 0 . 3010 10 = 53598 . 6 J m o l - 1 Or E a = 53 . 598 kJ mol - 1 .

The temperature dependence of the rate of a chemical reaction can be explained by Arrhenius equation. Give Arrhenius equation.

Arrhenius proposed a relationship between the rate constant k and absolute temperature T as k = A e - E a / RT , and this equation is called Arrhenius equation. In this equation, A is called Arrhenius constant or Arrhenius factor, R is ideal gas constant and E a is the activation energy for the reaction.

For a reaction A + B → C + D , the rate equation is Rate = k A 3 / 2 B 1 / 2 . Give overall order and molecularity of reaction.

Consider the reaction, A + B → C + D . The rate equation is Rate = k A 3 / 2 B 1 / 2 . Overall order of the reaction is equal to the sum of the powers of the concentration terms in the rate equation. Therefore, overall order = 3 2 + 1 2 = 2 . The molecularity of a reaction is equal to the number of reacting species taking part in an elementary reaction. Therefore, the molecularity of this reaction 1 + 1 = 2 .

For hydrolysis of methyl acetate in aqueous solution, the following results were observed. t / s 0 30 60 CH 3 COOCH 3 C / mol L - 1 0 . 60 0 . 30 0 . 15 Show that it follows pseudo first order reaction as the concentration of water remains constant.

In pseudo first order reactions, the molecularity of the reaction is two but the order of reaction is one. t / s 0 30 60 CH 3 COOCH 3 mol L - 1 0 . 60 0 . 30 0 . 15 Integrated rate equation for first order reactions : k = 2 . 303 t log A ∘ A t So, k 1 = 2 . 303 30 log 0 . 60 0 . 30 k 1 = 2 . 303 30 log 2 = 0 . 0767 × 0 . 301 = 0 . 02308 s - 1 and, k 2 = 2 . 303 60 log 0 . 60 0 . 15 k 2 = 2 . 303 60 log 4 = 0 . 0384 × 0 . 602 = 0 . 0231 s - 1 . In both the time intervals, the value of rate constant is almost same. Hence, the given reaction is a pseudo first order reaction.

Identify the order of reaction if the unit of the rate constant is mol L - 1 s - 1 .

Given: Unit of k = mol L - 1 s - 1 Rate = k A x , Where x is the order of the reaction. Unit of rate = mol L - 1 s - 1 unit of rate constant = mol L - 1 s - 1 unit of reactant concentration = mol L - 1 A x = r k = mol L - 1 s - 1 mol L - 1 s - 1 = 1 . A x = 1 . Therefore, x= 0. Hence, the given rate constant is of zero-order reaction.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Chemical Kinetics>Kerala Board-2019>Q 4

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12th Kerala Board

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Examine the graph given below. Identify the integrated rate equation and the order of the reaction corresponding to it.

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