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]

Consider a complex reaction taking place in three steps with rate constants k1, k2 and k3 respectively. The overall rate constant k is given by the expression k= k1k3k2. If the activation energies of the three steps are 60, 30 and 10 kJ mol-1 respectively, then the overall energy of activation in kJ mol-1 is . (Nearest integer)

For the thermal decomposition of N2O5(g) at constant volume, the following table can be formed, for the reaction mentioned below. 2N2O5(g) 2N2O4(g)+O2(g) [figure] x= 10-3 atm [nearest integer] Given : Rate constant for the reaction is 4.606 10-2 s-1.

A B The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ mol-1 and the frequency factor is 1020, the time required for 50\% molecules of A to become B is picoseconds (nearest integer). [R=8.314 J K-1 mol-1 ]

Given below are two statements : Statement I : The rate law for the reaction A + B C is rate (r) = k[A]2[B]. When the concentration of both A and B is doubled, the reaction rate is increased "x" times. Statement II : [figure] The figure is showing "the variation in concentration against time plot" for a "y" order reaction. The value of x + y is .

Consider the following first order gas phase reaction at constant temperature A(g) 2B(g) + C(g) If the total pressure of the gases is found to be 200 torr after 23 sec. and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is 10-2 s-1 (nearest integer) [Given : 10(2) = 0.301]

Consider the following reaction A + B C The time taken for A to become 1/4th of its initial concentration is twice the time taken to become 1/2 of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is .

Time required for 99.9\% completion of a first order reaction is times the time required for completion of 90\% reaction.(nearest integer)

Consider the two different first order reactions given below A + B C (Reaction 1) P Q (Reaction 2) The ratio of the half life of Reaction 1 : Reaction 2 is 5 : 2. If t1 and t2 represent the time taken to complete 2/3rd and 4/5th of Reaction 1 and Reaction 2, respectively, then the value of the ratio t1 : t2 is 10-1 (nearest integer). [Given : 10(3) = 0.477 and 10(5) = 0.699]

During kinetic study of reaction 2A + B C + D, the following results were obtained : [figure] Based on above data, overall order of the reaction is .

Consider the following single step reaction in gas phase at constant temperature. 2A(g) + B(g) C(g) The initial rate of the reaction is recorded as r1 when the reaction starts with 1.5 atm pressure of A and 0.7 atm pressure of B. After some time, the rate r2 is recorded when the pressure of C becomes 0.5 atm. The ratio r1 : r2 is 10-1. (Nearest integer)

Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below. [figure] Some details of the above reactions are listed below. [figure] If the overall rate constant of the above transformation (k) is given as k = k1k2k3 and the overall activation energy (Ea) is 400 kJ mol-1, then the value of Ea3 is kJ mol-1 (nearest integer)

Consider the following reaction, the rate expression of which is given below A + B C rate = k[A]1/2[B]1/2 The reaction is initiated by taking 1 M concentration of A and B each. If the rate constant (k) is 4.6 10-2 s-1, then the time taken for A to become 0.1 M is sec. (nearest integer)

The ratio of 14C12C in a piece of wood is 18 part that of atmosphere. If half life of 14C is 5730 years, the age of wood sample is years.

The following data were obtained during the first order thermal decomposition of a gas A at constant volume : A(g) 2B(g) + C(g) [figure] The rate constant of the reaction is 10-2 s-1 (nearest integer)

r = k[A] for a reaction, 50% of A is decomposed in 120 minutes. The time taken for 90% decomposition of A is minutes.