# 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 -

where Eis the activation energy. When a graph is plotted for log k vs.$\frac{1}{T}$, 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 =

99% completion means that $\mathcal{x}$ = 99% of a = 0.99a

$\therefore$   t99% =

=

90% completion means that $\mathcal{x}$ = 90% of a = 0.90a

t90% = =

$\therefore$ t99% = 2$×$t90%

(b) Slope of the line =

$\therefore$ Ea = 2.303 $×$8.314 (J K-1 mol-1$×$4250 K

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