Chemical Kinetics

Decomposition of a hydrocarbon follows the equation k = (5.5 1011 s-1) e-28000 K/T. The activation energy of reaction is kJ mol-1. (Nearest Integer) Given : R = 8.3 J K-1 mol-1

For reaction A P, rate constant k = 1.5 103 s-1 at 27°C If activation energy for the above reaction is 60 kJ mol-1, then the temperature (in °C) at which rate constant, k = 4.5 103 s-1 is . (Nearest integer) Given : log 2 = 0.30, log 3 = 0.48, R = 8.3 J K-1 mol-1, ln 10 = 2.3

A → B (first reaction) C → D (second reaction) Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hours. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is 10-1 hour-1 (nearest integer).

The temperature at which the rate constants of the given below two gaseous reactions become equal is K. (Nearest integer) X Y, k1 = 106 e -30000T P Q, k2 = 104 e -24000T Given : 10 = 2.303

Consider A k1 B and C k2 D are two reactions. If the rate constant (k1) of the A B reaction can be expressed by the following equation 10 k = 14.34 - 1.5 104T/K and activation energy of C D reaction (Ea2) is 15th of the A B reaction (Ea1), then the value of (Ea2) is kJ mol-1. (Nearest Integer)

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceed the activation energy of second reaction by 20 kJ mol-1. If k1 and k2 are the rate constants of first and second reaction respectively at 300 K, then k2k1 will be . (nearest integer) [R = 8.3 J K-1 mol-1]