Question

a) For a first-order reaction, show that the time required for 99% completion is twice the time required for the completion of 90% of the reaction.
(b) Rate constant 'K' of a reaction varies with temperature 'T' according to the equation: log k = log A- Ea 1 2.303R T Where Ea is the activation energy? When a graph is plotted for log k Vs 1 T, a straight line with a slope of - 4250 K is obtained. Calculate 'Ed' for the reaction.

Answer 1

Answer: (A)For the first order reaction, time required for 99% completion. Hence, time required for 99% completion is twice for the time required for the completion of 90% reaction. (ii) Reduced to 1/2? Thus rate of reaction will become four times where concentration is doubled.

)(B)(Correct option is  B

127.71 kJmol  

−1

 Slope of the line =−  

2.303R

E  

a  =−6670K

E   =2.303×8.314(JK  

−1

mol  

−1  )×6670K =127711.4 J mol  

−1

=127.71 kJ mol  

−1

 

Explanation: