A first order reaction is 40\; ^{o}/_{o} complete in 80 minutes. Calculate the value of rate constant (k). In what time will the reaction be 90\; ^{o}/_{o} completed ?

[ Given : log\; 2=0.3010,\; log\; 3=0.4771,\; log\; 4=0.6021,\; log\; 5=0.6771,\; log\; 6=0.7782,

 

 
 
 
 
 

Answers (1)

 Given : log\; 2=0.3010,\; log\; 3=0.4771,\; log\; 4=0.6021,\; log\; 5=0.6771,\; log\; 6=0.7782,

k = 2.303/t log [A]0/[A]  =  2.303/80 log 100/60 = 2.303/80 (1-0.7782)  

=  0.0064 min-1

t = 2.303/k log [A]0/[A] =  2.303/0.0064 log 100/10

= 360 min

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