The half life of a first order reaction is one hour. During what time interval will the concentration be reduced to 7/8 of its initial value?

1 Answer
Jan 17, 2018

Answer:

8 min

Explanation:

For a 1st order reaction:

#sf(Ararr"products")#

#sf([A]_t=[A]_0e^(-kt))#

#sf(k)# is the rate constant which is related to the half - life:

#sf(k=0.693/t_(1/2)=0.693/1=0.693color(white)(x)"hr"^-1)#

#:.##sf(7/8cancel([A]_0)=cancel([A]_0)e^(-kt))#

Taking natural logs of both sides#rArr#

#sf(ln(7/8)=-kt)#

#sf(t=-ln(7/8)xx1/k)#

#sf(t=0.133color(white)(x)"hr")#

#sf(t=0.133xx60=8color(white)(x)"min")#