CHEM 104 Homework 3 5. (25 points): Chapter 16.5 Fireflies flash in evening mole
ID: 1081261 • Letter: C
Question
CHEM 104 Homework 3 5. (25 points): Chapter 16.5 Fireflies flash in evening molecule, which cau the fireflies' tails flash, and it is dependent on the outside temperature. One set of experiments on a population of fireflies showed that the average interval between flashes of individual insects the evenings because of an enzyme-catalyzed oxidation reaction of the luciferin ses the luciferin to emit light. The rate of this reaction determines how often was 16.3 s at 21.0°C and 13.0 s at 27.8 C a. What is the apparent activation energy of the reaction that controls the flashing? (Hint: you can assume that the reaction obeys the Arrhenius equation and that the units of the rate constant k are s-1.) b. What would the average interval between flashes of an individual firefly be at 30.0 C? c. At what Celsius temperature would you expect the interval between flashes of an individual firefly to be 15.0 s?Explanation / Answer
population of flies at different temperatures can be explained using Arhenius equation which is
ln(K2/K1)= (E/R)*(1/T1-1/T2)
K2= rate constant at T2= 1/13= 0.077 and T2=27.8 deg.c, in K, T2= 27.8+273= 300.8 K
and K1= rate constant at T1= 1/16= 0.0625 at T1= 21 deg.c, T1= 21+273= 294
hence ln (0.077/0.0625)= (E/R)*(1/294-1/300.8), R= gas constant= 8.314 J/mole.K
E/R =2713.38 and E= 2713.38*8.314 J/mole= 22559 joules/mole
now when T3= 30 deg.c= 30+273= 303K, 1/303= 0.0033 K3 need to be determined
Hence the Arhenius equation becomes ln (K3/K2)= (E/R)*(1/T2-1/T3)= 2713.38*(1/300.8-0.0033)
K3/K2= 1.068, K3= 1.068*0.077= 0.0822, time =1/K3= 1/0.0822= 12.15 sec
Given T3= 300.8 and T1=?, K3= 0.0822 and K2= 1/15 =0.067
ln(K3/K2)= 2713.38*(1/T1-1/300.8)
ln(0.0822/0.067)= 2713.38*(1/T1-1/300.8), T1= 294 K, T1= 294-273= 21 deg.c