Chemical Kinetics

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with t1/2 = 3 hour. The percentage of sucrose remaining after 6 hours is . (Nearest integer) (Given : log 2 = 0.3010 and log 3 = 0.4771)

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

If the half life of a first order reaction is 6.93 minutes then the time required for completion of 99% of the reaction will be minutes. (Given : log 2 = 0.3010)

For a reaction A → P at T K, the half life (t1/2) is plotted as a function of initial concentration [A]0 of A as given below. [figure] The value of x in the given figure is s (Nearest integer)

For a first order reaction A B arrayll t/min & [A]/M \ 0 & 0.6500 \ x & 0.0650 \ 20 & 0.00065 array x = min. (Nearest integer)

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

Consider the following gas phase reaction being carried out in a closed vessel at 25°C. 2A(g) 4B(g) + C(g) array|c|c| time (min) & total pressure of the system (mm Hg) \ 30 & 300 \ & 600 \ array The pressure of C(g) at 30 minutes time interval would be mm Hg. (nearest integer)

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 half-life of 65Zn is 245 days. After x days, 75\% of original activity remained. The value of x in days is . (Nearest integer) (Given: 3 = 0.4771 and 2 = 0.3010)

For the thermal decomposition of reactant AB(g), the following plot is constructed. [figure] The half-life of the reaction is 'x' min. x = min. (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]

For the reaction A products. [figure] The concentration of A at 10 minutes is 10-3 mol L-1 (nearest integer). The reaction was started with 2.5 mol L-1 of A.

For the reaction A B the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to 2.5 g L-1 (if the initial concentration of A was 50 g L-1) is . (Nearest integer) Given : 2 = 0.3010 [figure]