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Answer 1 · 2017-05-15T03:45:42

Solve the mass balance

For the first order reaction of #A\rightarrow Products#

#(dC_A)/dt = -kC_A#
#t = -1/k [ln(C_A)]|_(C_A0)^(C_A)#
#t = -1/k ln((C_A)/(C_(A0)))#

The given half time is when there is 50% of initial concentration present. Hence we have

#C_A = 0.5C_(A0)#

#3 = -1/k ln(0.5)#
#k = ln(2)/3#
#k = 0.231 hr^(-1)#

When 70 % of A is consumed the remaining concentration is

#C_A = (1-0.7)C_(A0)#
#C_A = 0.3C_(A0) #
#t = -1/0.231 ln(0.3)#
#t = 5.212 # hours.