The missing concentration of C 2 H 5 I in the table for the given reaction should be calculated. Concept Introduction: For a first order reaction, A → B + C the rate constant is, ln [ A ] t [ A ] 0 = − k t w h e r e , [ A ] 0 = concentration of reactant a t s t a r t i n g t i m e ( t = 0 ) [ A ] t = concentration of reactant a f t e r t t i m e k = r ate constant t = time

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Answer 1

Question

Chapter 19.5, Problem 5PPB

Interpretation Introduction

Interpretation:

The missing concentration of C2H5I in the table for the given reaction should be calculated.

Concept Introduction:

For a first order reaction, A → B + C the rate constant is,

ln[A]t[A]0 = −ktwhere, [A]0 = concentration of reactant at starting time(t= 0)[A]t = concentration of reactant after t time k = rate constantt = time

Expert Solution & Answer

Answer to Problem 5PPB

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Explanation of Solution

Concentration of C2H5I after 10 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 10 min

Concentration of C2H5I after 10 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (10 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (10 min)= 0.869

Therefore,

Concentration of C2H5I after 10 min, [C2H5I]t is,

[C2H5I]t = (0.869) × (0.45)= 0.39 M

Concentration of C2H5I after 10 min, [C2H5I]t is 0.39 M

Concentration of C2H5I after 20 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 20 min

Concentration of C2H5I after 20 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (20 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (20 min)= 0.756

Therefore,

Concentration of C2H5I after 20 min, [C2H5I]t is,

[C2H5I]t = (0.756) × (0.45)= 0.34 M

Concentration of C2H5I after 20 min, [C2H5I]t is 0.34 M

Concentration of C2H5I after 30 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 30 min

Concentration of C2H5I after 30 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (30 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (30 min)= 0.657

Therefore,

Concentration of C2H5I after 30 min, [C2H5I]t is,

[C2H5I]t = (0.657) × (0.45)= 0.30 M

Concentration of C2H5I after 30 min, [C2H5I]t is 0.30 M

Concentration of C2H5I after 40 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 40 min

Concentration of C2H5I after 40 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (40 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (40 min)= 0.5712

Therefore,

Concentration of C2H5I after 40 min, [C2H5I]t is,

[C2H5I]t = (0.5712) × (0.45)= 0.26 M

Concentration of C2H5I after 40 min, [C2H5I]t is 0.26 M

Conclusion

The missing concentration of C2H5I in the table for the given reaction were calculated,

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

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Answer 2

Question

Chapter 19.5, Problem 5PPB

Interpretation Introduction

Interpretation:

The missing concentration of C2H5I in the table for the given reaction should be calculated.

Concept Introduction:

For a first order reaction, A → B + C the rate constant is,

ln[A]t[A]0 = −ktwhere, [A]0 = concentration of reactant at starting time(t= 0)[A]t = concentration of reactant after t time k = rate constantt = time

Expert Solution & Answer

Answer to Problem 5PPB

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Explanation of Solution

Concentration of C2H5I after 10 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 10 min

Concentration of C2H5I after 10 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (10 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (10 min)= 0.869

Therefore,

Concentration of C2H5I after 10 min, [C2H5I]t is,

[C2H5I]t = (0.869) × (0.45)= 0.39 M

Concentration of C2H5I after 10 min, [C2H5I]t is 0.39 M

Concentration of C2H5I after 20 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 20 min

Concentration of C2H5I after 20 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (20 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (20 min)= 0.756

Therefore,

Concentration of C2H5I after 20 min, [C2H5I]t is,

[C2H5I]t = (0.756) × (0.45)= 0.34 M

Concentration of C2H5I after 20 min, [C2H5I]t is 0.34 M

Concentration of C2H5I after 30 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 30 min

Concentration of C2H5I after 30 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (30 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (30 min)= 0.657

Therefore,

Concentration of C2H5I after 30 min, [C2H5I]t is,

[C2H5I]t = (0.657) × (0.45)= 0.30 M

Concentration of C2H5I after 30 min, [C2H5I]t is 0.30 M

Concentration of C2H5I after 40 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 40 min

Concentration of C2H5I after 40 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (40 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (40 min)= 0.5712

Therefore,

Concentration of C2H5I after 40 min, [C2H5I]t is,

[C2H5I]t = (0.5712) × (0.45)= 0.26 M

Concentration of C2H5I after 40 min, [C2H5I]t is 0.26 M

Conclusion

The missing concentration of C2H5I in the table for the given reaction were calculated,

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

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Answer 3

Question

Chapter 19.5, Problem 5PPB

Interpretation Introduction

Interpretation:

The missing concentration of C2H5I in the table for the given reaction should be calculated.

Concept Introduction:

For a first order reaction, A → B + C the rate constant is,

ln[A]t[A]0 = −ktwhere, [A]0 = concentration of reactant at starting time(t= 0)[A]t = concentration of reactant after t time k = rate constantt = time

Expert Solution & Answer

Answer to Problem 5PPB

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Explanation of Solution

Concentration of C2H5I after 10 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 10 min

Concentration of C2H5I after 10 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (10 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (10 min)= 0.869

Therefore,

Concentration of C2H5I after 10 min, [C2H5I]t is,

[C2H5I]t = (0.869) × (0.45)= 0.39 M

Concentration of C2H5I after 10 min, [C2H5I]t is 0.39 M

Concentration of C2H5I after 20 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 20 min

Concentration of C2H5I after 20 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (20 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (20 min)= 0.756

Therefore,

Concentration of C2H5I after 20 min, [C2H5I]t is,

[C2H5I]t = (0.756) × (0.45)= 0.34 M

Concentration of C2H5I after 20 min, [C2H5I]t is 0.34 M

Concentration of C2H5I after 30 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 30 min

Concentration of C2H5I after 30 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (30 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (30 min)= 0.657

Therefore,

Concentration of C2H5I after 30 min, [C2H5I]t is,

[C2H5I]t = (0.657) × (0.45)= 0.30 M

Concentration of C2H5I after 30 min, [C2H5I]t is 0.30 M

Concentration of C2H5I after 40 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 40 min

Concentration of C2H5I after 40 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (40 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (40 min)= 0.5712

Therefore,

Concentration of C2H5I after 40 min, [C2H5I]t is,

[C2H5I]t = (0.5712) × (0.45)= 0.26 M

Concentration of C2H5I after 40 min, [C2H5I]t is 0.26 M

Conclusion

The missing concentration of C2H5I in the table for the given reaction were calculated,

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

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Answer 4

Question

Chapter 19.5, Problem 5PPB

Interpretation Introduction

Interpretation:

The missing concentration of C2H5I in the table for the given reaction should be calculated.

Concept Introduction:

For a first order reaction, A → B + C the rate constant is,

ln[A]t[A]0 = −ktwhere, [A]0 = concentration of reactant at starting time(t= 0)[A]t = concentration of reactant after t time k = rate constantt = time

Expert Solution & Answer

Answer to Problem 5PPB

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Explanation of Solution

Concentration of C2H5I after 10 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 10 min

Concentration of C2H5I after 10 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (10 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (10 min)= 0.869

Therefore,

Concentration of C2H5I after 10 min, [C2H5I]t is,

[C2H5I]t = (0.869) × (0.45)= 0.39 M

Concentration of C2H5I after 10 min, [C2H5I]t is 0.39 M

Concentration of C2H5I after 20 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 20 min

Concentration of C2H5I after 20 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (20 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (20 min)= 0.756

Therefore,

Concentration of C2H5I after 20 min, [C2H5I]t is,

[C2H5I]t = (0.756) × (0.45)= 0.34 M

Concentration of C2H5I after 20 min, [C2H5I]t is 0.34 M

Concentration of C2H5I after 30 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 30 min

Concentration of C2H5I after 30 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (30 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (30 min)= 0.657

Therefore,

Concentration of C2H5I after 30 min, [C2H5I]t is,

[C2H5I]t = (0.657) × (0.45)= 0.30 M

Concentration of C2H5I after 30 min, [C2H5I]t is 0.30 M

Concentration of C2H5I after 40 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 40 min

Concentration of C2H5I after 40 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (40 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (40 min)= 0.5712

Therefore,

Concentration of C2H5I after 40 min, [C2H5I]t is,

[C2H5I]t = (0.5712) × (0.45)= 0.26 M

Concentration of C2H5I after 40 min, [C2H5I]t is 0.26 M

Conclusion

The missing concentration of C2H5I in the table for the given reaction were calculated,

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

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Answer 5

Chapter 19.5, Problem 5PPB

Interpretation Introduction

Interpretation:

The missing concentration of C2H5I in the table for the given reaction should be calculated.

Concept Introduction:

For a first order reaction, A → B + C the rate constant is,

ln[A]t[A]0 = −ktwhere, [A]0 = concentration of reactant at starting time(t= 0)[A]t = concentration of reactant after t time k = rate constantt = time

Expert Solution & Answer

Answer to Problem 5PPB

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Explanation of Solution

Concentration of C2H5I after 10 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 10 min

Concentration of C2H5I after 10 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (10 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (10 min)= 0.869

Therefore,

Concentration of C2H5I after 10 min, [C2H5I]t is,

[C2H5I]t = (0.869) × (0.45)= 0.39 M

Concentration of C2H5I after 10 min, [C2H5I]t is 0.39 M

Concentration of C2H5I after 20 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 20 min

Concentration of C2H5I after 20 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (20 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (20 min)= 0.756

Therefore,

Concentration of C2H5I after 20 min, [C2H5I]t is,

[C2H5I]t = (0.756) × (0.45)= 0.34 M

Concentration of C2H5I after 20 min, [C2H5I]t is 0.34 M

Concentration of C2H5I after 30 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 30 min

Concentration of C2H5I after 30 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (30 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (30 min)= 0.657

Therefore,

Concentration of C2H5I after 30 min, [C2H5I]t is,

[C2H5I]t = (0.657) × (0.45)= 0.30 M

Concentration of C2H5I after 30 min, [C2H5I]t is 0.30 M

Concentration of C2H5I after 40 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 40 min

Concentration of C2H5I after 40 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (40 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (40 min)= 0.5712

Therefore,

Concentration of C2H5I after 40 min, [C2H5I]t is,

[C2H5I]t = (0.5712) × (0.45)= 0.26 M

Concentration of C2H5I after 40 min, [C2H5I]t is 0.26 M

Conclusion

The missing concentration of C2H5I in the table for the given reaction were calculated,

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Answer 6

Chapter 19.5, Problem 5PPB

Interpretation Introduction

Interpretation:

The missing concentration of C2H5I in the table for the given reaction should be calculated.

Concept Introduction:

For a first order reaction, A → B + C the rate constant is,

ln[A]t[A]0 = −ktwhere, [A]0 = concentration of reactant at starting time(t= 0)[A]t = concentration of reactant after t time k = rate constantt = time

Expert Solution & Answer

Answer to Problem 5PPB

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Explanation of Solution

Concentration of C2H5I after 10 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 10 min

Concentration of C2H5I after 10 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (10 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (10 min)= 0.869

Therefore,

Concentration of C2H5I after 10 min, [C2H5I]t is,

[C2H5I]t = (0.869) × (0.45)= 0.39 M

Concentration of C2H5I after 10 min, [C2H5I]t is 0.39 M

Concentration of C2H5I after 20 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 20 min

Concentration of C2H5I after 20 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (20 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (20 min)= 0.756

Therefore,

Concentration of C2H5I after 20 min, [C2H5I]t is,

[C2H5I]t = (0.756) × (0.45)= 0.34 M

Concentration of C2H5I after 20 min, [C2H5I]t is 0.34 M

Concentration of C2H5I after 30 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 30 min

Concentration of C2H5I after 30 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (30 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (30 min)= 0.657

Therefore,

Concentration of C2H5I after 30 min, [C2H5I]t is,

[C2H5I]t = (0.657) × (0.45)= 0.30 M

Concentration of C2H5I after 30 min, [C2H5I]t is 0.30 M

Concentration of C2H5I after 40 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 40 min

Concentration of C2H5I after 40 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (40 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (40 min)= 0.5712

Therefore,

Concentration of C2H5I after 40 min, [C2H5I]t is,

[C2H5I]t = (0.5712) × (0.45)= 0.26 M

Concentration of C2H5I after 40 min, [C2H5I]t is 0.26 M

Conclusion

The missing concentration of C2H5I in the table for the given reaction were calculated,

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Answer 7

Chapter 19.5, Problem 5PPB

Interpretation Introduction

Interpretation:

The missing concentration of C2H5I in the table for the given reaction should be calculated.

Concept Introduction:

For a first order reaction, A → B + C the rate constant is,

ln[A]t[A]0 = −ktwhere, [A]0 = concentration of reactant at starting time(t= 0)[A]t = concentration of reactant after t time k = rate constantt = time

Answer 8

Expert Solution & Answer

Answer to Problem 5PPB

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Concentration of C2H5I after 10 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 10 min

Concentration of C2H5I after 10 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (10 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (10 min)= 0.869

Therefore,

Concentration of C2H5I after 10 min, [C2H5I]t is,

[C2H5I]t = (0.869) × (0.45)= 0.39 M

Concentration of C2H5I after 10 min, [C2H5I]t is 0.39 M

Concentration of C2H5I after 20 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 20 min

Concentration of C2H5I after 20 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (20 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (20 min)= 0.756

Therefore,

Concentration of C2H5I after 20 min, [C2H5I]t is,

[C2H5I]t = (0.756) × (0.45)= 0.34 M

Concentration of C2H5I after 20 min, [C2H5I]t is 0.34 M

Concentration of C2H5I after 30 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 30 min

Concentration of C2H5I after 30 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (30 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (30 min)= 0.657

Therefore,

Concentration of C2H5I after 30 min, [C2H5I]t is,

[C2H5I]t = (0.657) × (0.45)= 0.30 M

Concentration of C2H5I after 30 min, [C2H5I]t is 0.30 M

Concentration of C2H5I after 40 min,[C2H5I]t is calculated below

Given,

Initial concentration of C2H5I, [C2H5I]0 = 0.45 MRate constant, k = 1.4 × 10−2 min−1Time, t = 40 min

Concentration of C2H5I after 40 min, [C2H5I]t is calculated

ln[C2H5I]t[C2H5I]0 = −ktln[C2H5I]t0.45 M = −(1.4 × 10−2 min−1 )× (40 min)[C2H5I]t0.45 M = e−(1.4 × 10−2 min−1 )× (40 min)= 0.5712

Therefore,

Concentration of C2H5I after 40 min, [C2H5I]t is,

[C2H5I]t = (0.5712) × (0.45)= 0.26 M

Concentration of C2H5I after 40 min, [C2H5I]t is 0.26 M

Answer 12

Conclusion

The missing concentration of C2H5I in the table for the given reaction were calculated,

Time (min)[C2H5I] (M)00.45100.39200.34300.30400.26

Answer 13

Ch. 19.3 - Prob. 19.1WE Ch. 19.3 - Write the rate expressions for each of the...Ch. 19.3 - Write the balanced equation corresponding to the...Ch. 19.3 - The diagrams represent a system that initially...Ch. 19.3 - Consider the reaction 4NO2(g)+O2(g)2N2O5(g) At a...Ch. 19.3 - Consider the reaction 4PH3(g)P4(g)+6H2(g) At a...Ch. 19.3 - Prob. 2PPB Ch. 19.3 - Prob. 2PPC Ch. 19.3 - Prob. 19.3.1SR Ch. 19.3 - Prob. 19.3.2SR

Answer to Problem 5PPB

Explanation of Solution

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