Question

2. Cell death due to hyperthermia has been described by first-order kinetics with an Arrhenius description of k(T). That is In(N/No) k(T)t where N is the number of living cells and No is the initial number of living cells. Using the data from Table 1: (a) Determine the Arrhenius constants, A and Ea using a curve fit of all of the data (b) Determine the Arrhenius constants, A and Ea using algebraic solution using only two of the temperatures (42 °C and 70 °C) (c) Calculate the fraction of cells remaining after 45 s at 320 K.
Solve for B and C

Answer 1

solution? - frest-order expression . en (A ) = -kt (6) All Joe siest ordeę reaction I tip = 0.693 so from here we can says- S katila = Carst. -a 1 Atasohening constant Here is too A é cada ] I Eat activation energy. using stating a data! - a ki (tina = k (+1/2) z a hexp ( 2 ) (Alle) = y exp ( 5 ) [fiel -) exp led t) = (tik),

Taking Ti= 42°C = 315k T BYSOC = 318 K (Ain), = 31000see (1/2)2 = 17000 see- Put all values! - esp (elety lata-sha)) - ( Ca = 166777.50 J 166.77 kg mol Now we know? = 0.693 for 480 - - 170vo = 0.693 T K - 4.076710s see Now = A exp (East) 4.076 .705 = A exp ( centare M=1.014870231 (1) fractim of celts remaining (~ at Asos and 320k fisst we have to find i at 320k 1-166777.5 1 = 6.045 X 10^-fseet Kosz (1.014 X10s) exp (13-31491320) = INO

New! em Condom --K220+ per = erp (- coussioʻx 40) = 0.997 NO