Henry Darius B.
Namoc 2-24-23
BSCHE-3 Problem Set 4
Solution:
Using the Arrhenius model:
T (deg
Running speed, v (m/hr) T (K) ln v 1/T (1/K)
C)
150 13 286.15 0.003494671 5.010635294
160 16 289.15 0.003458413 5.075173815
230 22 295.15 0.003388108 5.438079309
295 24 297.15 0.003365304 5.686975356
370 28 301.15 0.003320604 5.913503006
From the trendline equation:
𝐸𝑎
𝑚=− 𝑅
Frequency Factor (Ao) = 23376568232 Collisions/sec
𝑚 = −5417.9 R = 8.314 J/mol K
𝑏 = 23.875 Solving for 𝐸𝑎 :
𝐸𝑎 = −𝑚𝑅
𝐽
𝐸𝑎 = −(−5417.9(8.314)) 𝐸𝑎 = −45044.42 𝑚𝑜𝑙
For Fireflies
Flashes/min (r) T (C) T (K) 1/T(K) ln(r)
9 21 294.15 0.003399626 2.197224577
12.16 25 298.15 0.003354016 2.498151877
16.2 30 303.15 0.003298697 2.785011242
Fireflies
3
2.8 y = -5801.4x + 21.933
2.6
ln(r)
2.4
2.2
2
0.00328 0.0033 0.00332 0.00334 0.00336 0.00338 0.0034 0.00342
1/T(K)
From the trendline equation:
𝐸𝑎
𝑚=− 𝑅
Frequency Factor (A0) = 3352593291 Collisions/sec
𝑚 = −5801.4 Solving for 𝐸𝑎 :
𝑏 = 21.933 𝐸𝑎 = −𝑚𝑅
R = 8.314 J/mol K 𝐸𝑎 = −(−5801.4(8.314))
𝐸𝑎 = 48232.8396 𝐽/𝑚𝑜𝑙
For Ants
V (cm/s) T (C) T (K) 1/T(K) ln(r)
0.5 10 283.15 0.003531697 -0.693147181
2 20 293.15 0.003411223 0.693147181
3.4 30 303.15 0.003298697 1.223775432
6.5 38 311.15 0.003213884 1.871802177
Ants
2.5
2
y = -7769.2x + 26.909
1.5
1
ln(V)
0.5
0
0.00315 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345 0.0035 0.00355
-0.5
-1
1/T(K)
𝐸𝑎
𝑚=− 𝑅
𝑚 = −7769.2
b= 26.909
R = 8.314 J/mol.K
Frequency Factor (A0) = 4.85769E+11 Collisions/sec
Solving for 𝐸𝑎 :
𝐸𝑎 = −𝑚𝑅
𝐸𝑎 = −(−77692(8.314))
𝐽
𝐸𝑎 = 64593.13 𝑚𝑜𝑙
For Crickets
Chirps/min (r) T (C) T (K) 1/T(K) ln(r)
80 14.2 287.35 0.00348 4.382027
126 20.3 293.45 0.003408 4.836282
200 27 300.15 0.003332 5.298317
Crickets
5.4
5.2
y = -6173.2x + 25.868
5
4.8
ln(r)
4.6
4.4
4.2
4
0.00332 0.00334 0.00336 0.00338 0.0034 0.00342 0.00344 0.00346 0.00348 0.0035
1/T(K)
𝐸𝑎
𝑚=− 𝑅
𝑚 = −6173.2
b= 25.868
R = 8.314J/mol.K
Frequency Factor (A0) = 1.71526E+11 Collisions/sec
Solving for 𝐸𝑎 :
𝐸𝑎 = −𝑚𝑅
𝐸𝑎 = −(−6173.2(8.314))
𝐸𝑎 = 51323.9848 𝐽/𝑚𝑜𝑙
For Honeybees
V (cm/s) T (C) T (K) 1/T(K) ln(r)
0.7 25 298.15 0.003354 -0.35667
1.8 30 303.15 0.003299 0.587787
3 35 308.15 0.003245 1.098612
40 313.15 0.003193 #NUM!
Honeybees
1.4
1.2
y = -13391x + 44.625
1
0.8
0.6
ln(V)
0.4
0.2
0
0.00322 0.00324 0.00326 0.00328 0.0033 0.00332 0.00334 0.00336
-0.2
-0.4
-0.6
1/T(K)
𝐸𝑎
𝑚=− 𝑅
𝑚 = −13391
b= 44.625
R = 8.314 J/mol.K
Frequency Factor (A0) = 2.40099E+19 Collisions/sec
𝐸𝑎 = −𝑚𝑅
𝐸𝑎 = −(−13391(8.314))
𝐸𝑎 = 111332.774 𝐽/𝑚𝑜𝑙
(a) What do the firefly and cricket have in common? What are their differences?
The activation energy for firefly is 𝐸𝑎 = 48232.8396 𝐽/𝑚𝑜𝑙 while for the cricket, it is 𝐸𝑎 = 51323.9848 𝐽/𝑚𝑜𝑙.
They both have a graph of straight line. However, the firefly has a bit of a curvy part. The activation energy for
the two are also close to each other.
(b) What is the velocity of the honeybee at 40°C? At –5°C?
@40 deg C @5 deg C
1 1
ln(𝑉) = −13391 + 44.625 ln(𝑉) = −13391 + 44.625
𝑇 𝑇
1 1
𝑉 = 𝑒 −13391𝑇+44.625 𝑉 = 𝑒 −13391𝑇+44.625
1 1
−13391( )+44.625 −13391( )+44.625
𝑉=𝑒 40+273.15 𝑉=𝑒 5+273.15
𝑐𝑚 𝑐𝑚
𝑉 = 6.441397 𝑠
𝑉 = 0.0049 𝑠
(c) Nicolas wants to know if the bees, ants, crickets, and fireflies have anything in common. If so, what is it? You
may also do a pair-wise comparison.
INSECT ACTIVATION ENERGY(J/mol)
FIREFLIES 48232.8396 J/mol
CRICKETS 51323.9848 J/mol
ANTS 64593.1288 J/mol
HONEYBEES 1113332.774 J/mol
The fireflies and crickets both form a nearly perfect linear line. They also have close value for activation
energy. Ants and honeybees on the other hand have bigger value for the activation energy. Ants and honeybees
also form a curve in the graph.
(d) Would more data help clarify the relationships among frequency, speed, and temperature? If so, in what
temperature should the data be obtained? Pick an insect, and explain how you would carry out the
experiment to obtain more data.
Further information can help explain the connections between frequency, speed, and temperature. Since the
speed or rate is a function of temperature, the lower the temperature, the slower the speed. Consequently, the
lower the possibility of survival. It is best to get data depending on their survival temperature. Honeybees, for
instance, have a flying speed of 0.0049 cm/s at -5°C, below which the bees cannot survive. To obtain reliable and
precise results, I would suggest conducting an experiment at their range of survival temperatures. The data will
be more precise and accurate the smaller the temperature changes.
