NEET Chemistry Chemical Kinetics Questions Solved

BOARD

(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

(b) Rate constant 'k' of a reaction varies with temperature 'T' according to the equation :

log k = log A - Ea2.303 R1T

where Eis the activation energy. When a graph is plotted for log k vs.1T, a straight line wih a slope of -4250 K is obtained. Calculate 'E' for the reaction.(R = 8.314 JK-1 mol-1 )

(a) For first order reaction, t =2.303Klog aa-0.99a

99% completion means that x = 99% of a = 0.99a

  t99% =2.303Klog aa-0.99a

    = 2.303Klog 102 =2×2.303K

90% completion means that x = 90% of a = 0.90a

  t90% =2.303Klog aa-0.90a =2.303Klog 10 = 2.303K

 t99t90=2×2.303K2.303K=2

t99% = 2×t90%

(b) Slope of the line = -Ea2.303 R= -4250K

Ea = 2.303 ×8.314 (J K-1 mol-1×4250 K

 

Difficulty Level: