Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 11, Problem 91AE

Experiments during a recent summer on a number of fireflies (small beetles, Lampyridaes photinus) showed that the average interval between flashes of individual insects was 16.3 s at 21.0°C and 13.0 sat 27.8°C.

a. What is the apparent activation energy of the reaction that controls the flashing?

b. What would be the average interval between flashes of an individual firefly at 30.0°C?

c. Compare the observed intervals and the one you calculated in part b to the rule of thumb that the Celsius temperature is 54 minus twice the interval between flashes.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The average interval between flashes of individual insects at two different temperatures is given in an experiment. By using these values, the apparent activation energy of the reaction, average interval between flashes of a firefly at a given temperature is to be calculated and compare to the rule of thumb.

Concept introduction: A chemical reaction occurs inside the bodies of fireflies, which results in the flashing.  This reaction is called bioluminescence and follows first order kinetics.

The threshold energy needed to overcome to produce a chemical reaction is called activation energy.

The activation energy for bioluminescence reaction can be calculated by the following formula:

ln(k2k1)=EaR(1T11T2)

To determine: The activation energy of a chemical reaction that results in the flashing.

Answer to Problem 91AE

The activation energy is 24.1kJ/mol_ .

Explanation of Solution

Given

The average interval between the flashes of individual insects at 21°C is 16.3 s .

The average interval between the flashing of individual insects at 27.8°C is 13.0 s .

Rate constant at temperature 21°C is,

k1=1  flash16.3 s=6.13×102 s1

Rate constant at temperature 27.8°C is,

k2=1  flash13.0 s=7.69×102 s1

The activation energy is calculated using the formula,

ln(k2k1)=EaR(1T11T2)

Where,

  • k1 is rate constant at temperature T1 .
  • k2 is rate constant at temperature T2 .
  • R is universal gas constant (8.314J/Kmol) .
  • Ea is the activation energy.

Substitute the values of k1,k2,T1,T2 , and R in the above equation.

ln(7.69×102 s16.13×10=2 s1)=Ea8.314J/Kmol(1(21+273)K1(27.8+273)K)ln(7.69×102 s16.13×10=2 s1)=Ea8.314J/Kmol(1(294)K1(300.8)K)ln(1.25)=Ea8.314J/Kmol(7.69×105/K)0.223=Ea8.314J/Kmol(7.69×105/K)

Simplify the above equation.

Ea=0.223×8.314 J/Kmol7.69×105/K=24109.5J/mol

The conversion of joule (J) into kilojoules (kJ) is done as,

1J=11000kJ

Hence, the conversion of 24109.5J/mol into is,

24109.5J/mol=24109.51000kJ/mol=24.1kJ/mol_

Conclusion

The activation energy of the chemical reaction that occurs in the bodies of fireflies is 24.1kJ/mol .

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The average interval between flashes of individual insects at two different temperatures is given in an experiment. By using these values, the apparent activation energy of the reaction, average interval between flashes of a firefly at a given temperature is to be calculated and compare to the rule of thumb.

Concept introduction: A chemical reaction occurs inside the bodies of fireflies, which results in the flashing.  This reaction is called bioluminescence and follows first order kinetics.

The threshold energy needed to overcome to produce a chemical reaction is called activation energy.

The activation energy for bioluminescence reaction can be calculated by the following formula:

ln(k2k1)=EaR(1T11T2)

To determine: The average time interval between flashes of an individual firefly at 30°C .

Answer to Problem 91AE

The average time interval between flashes of an individual firefly at 30°C

is 12 s_ .

Answer: The rate constant of the chemical reaction at temperature 30°C is 8.21×10-2 s-1_ .

Explanation of Solution

Given

The average interval between the flashes of individual insects at 21°C is 16.3 s .

The activation energy is 24109.5J/mol .

Formula

The rate constant of chemical reaction is calculated using the formula,

ln(k2k1)=EaR(1T11T2)

Where,

  • k1 is rate constant at temperature T1 .
  • k2 is rate constant at temperature T2 .
  • R is universal gas constant (8.314J/Kmol) .
  • Ea is the activation energy.

Substitute the values of Ea,k2,T1,T2 , and R in the above equation.

ln(k26.13×102 s1)=24109.5J/mol8.314 J/Kmol(1(21+273)K1(30+273)K)ln(k26.13×102 s1)=24109.5J/mol 8.314 J/Kmol(1(294)K1(303)K)ln(k26.13×102 s1)=29.0×104×(1.01×104)ln(k26.13×102 s1)=0.293

Simplify the above equation.

(k26.13×102 s1)=e0.293(k26.13×102 s1)=1.340k=6.13×102 s1×1.340=8.21×10-2 s-1_

Conclusion

The rate constant for the chemical reaction at 30C is 8.21×102 s=1

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The average interval between flashes of individual insects at two different temperatures is given in an experiment. By using these values, the apparent activation energy of the reaction, average interval between flashes of a firefly at a given temperature is to be calculated and compare to the rule of thumb.

Concept introduction: A chemical reaction occurs inside the bodies of fireflies, which results in the flashing.  This reaction is called bioluminescence and follows first order kinetics.

The threshold energy needed to overcome to produce a chemical reaction is called activation energy.

The activation energy for bioluminescence reaction can be calculated by the following formula:

ln(k2k1)=EaR(1T11T2)

To determine: Comparison of observed and calculated interval and verify the value of temperature for each interval by using the rule of thumb.

Answer to Problem 91AE

All the average intervals follow the rule of thumb.

Explanation of Solution

The celsius temperature for each intervals are verified by the following formula:

T=(542(t))°C

Where,

  • T is temperature.
  • t is average time interval.

For T=21°C

Substitute the value of t in the above equation.

T=(542(16.3))°C=(5432.6)°C=21.4°C

For T=27.8°C

Substitute the value of t in the above equation.

T=(542(13.0))°C=(5426)°C=28°C

For T=30°C

Substitute the value of t in the above equation.

T=(542(12.0))°C=(5424)°C=30°C

Conclusion

All the average time intervals (observed and calculated) follow the rule of thumb as the value of celsius temperature for each interval comes to the same as given.

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Chapter 11 Solutions

Chemistry: An Atoms First Approach

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