A researcher measured the decay rate of a pesticide in river that had a temperat

Question

A researcher measured the decay rate of a pesticide in river that had a temperature of 10.0 C. Under these conditions, the pesticide had a half-life of 18.0 days and its decay followed first order kinetics. The researcher then moved to Arizona where the streams in the deserts are much warmer. Assuming that the desert stream has a temperature of 16.0 C, then what is the half-life of the pesticide in the desert stream? You can assume: 1) the delta-H of reaction is -130 kJ; 2) delta-S is 97 J/K and 3) and standard activation energy of 100 kJ/mol. Enter a number below for the half-life (in days) without any units of symbols.

Explanation / Answer

t1/2 = 18 days ,T1=10^oC =273+10 = 283K , T2 = 273+16 = 289K

K1 = 0.693 / 18 = 0.0385/day

Ea = 100x10^3 =10^5 J/mol

ln K1/K2 = Ea /R (1/T2- 1/T1) = 10^5 /8.314 ( 1/289K - 1/ 283K) = 12027.905(3.46x10^-3 - 3.53x10^--3)

= 12027.905 x - 0.07x10^-3 = - 0.842

log K1/K2 = -0.842 /2.303 = - 0.3656

taking anti log both side

K1/K2 =0.4309

K2= K1/0.4309 =0.0385 / 0.4309=0.0893/day

K2= 0.693/t1/2

t1/2 = 0.693 / 0.0893 = 7.76 days

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