(a) Interpretation: The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated. Concept introduction: The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by, rate = 10 − Δ G ° ∗ / 2.3 R T Where, • Δ G ° ∗ represents the activation energy for reaction. • R is the gas constant. • T is the temperature.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 4, Problem 4.28P

Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of 429.18.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The relative rates of two reactions are expressed as,

log(rateArateB)=ΔG°B∗−ΔG°A∗2.3RT

Where,

• ΔG°B∗ and ΔG°A∗ represents the activation energy for reaction A and B.

• R is the gas constant.

• T is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

log(rateArateB)=75 kJ mol−1−90 kJ mol−12.3(8.314 J K−1 mol−1)(298 K)log(rateArateB)=−15 kJ mol−15698.41 Jlog(rateArateB)=−15×103J mol−15698.41 J mol−1log(rateArateB)=−2.632

Inverse of the logarithm is taken on both the sides of the equation.

(rateArateB)=0.00233(rateBrateA)=429.18

rateB=429.18(rateA) …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of 429.18.

Conclusion

The reaction B is faster than the reaction A by a factor of 429.18.

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The rate of the reaction is expressed as,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by T'. The calculation for the T' is as follows,

rateA=rateA10−ΔG°A∗/2.3RT'=10−ΔG°B∗/2.3RT

Take log on both sides of equation.

−ΔG°A∗2.3RT'=−ΔG°B∗2.3RT

ΔG°A∗T'=ΔG°B∗T …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

90 kJ mol−1T'=75 kJ mol−1298 KT'=90 kJ mol−1×298 K75 kJ mol−1T'=357.6 K

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

1) Calculate the longest and shortest wavelengths in the Lyman and Paschen series. 2) Calculate the ionization energy of He* and L2+ ions in their ground states. 3) Calculate the kinetic energy of the electron emitted upon irradiation of a H-atom in ground state by a 50-nm radiation.

Calculate the ionization energy of He+ and Li²+ ions in their ground states. Thannnxxxxx sirrr Ahehehehehejh27278283-4;*; shebehebbw $+$;$-;$-28283773838 hahhehdva

Plleeaasseee solllveeee question 3 andd thankss sirr, don't solve it by AI plleeaasseee don't use AI

Answer 2

Question

Chapter 4, Problem 4.28P

Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of 429.18.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The relative rates of two reactions are expressed as,

log(rateArateB)=ΔG°B∗−ΔG°A∗2.3RT

Where,

• ΔG°B∗ and ΔG°A∗ represents the activation energy for reaction A and B.

• R is the gas constant.

• T is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

log(rateArateB)=75 kJ mol−1−90 kJ mol−12.3(8.314 J K−1 mol−1)(298 K)log(rateArateB)=−15 kJ mol−15698.41 Jlog(rateArateB)=−15×103J mol−15698.41 J mol−1log(rateArateB)=−2.632

Inverse of the logarithm is taken on both the sides of the equation.

(rateArateB)=0.00233(rateBrateA)=429.18

rateB=429.18(rateA) …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of 429.18.

Conclusion

The reaction B is faster than the reaction A by a factor of 429.18.

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The rate of the reaction is expressed as,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by T'. The calculation for the T' is as follows,

rateA=rateA10−ΔG°A∗/2.3RT'=10−ΔG°B∗/2.3RT

Take log on both sides of equation.

−ΔG°A∗2.3RT'=−ΔG°B∗2.3RT

ΔG°A∗T'=ΔG°B∗T …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

90 kJ mol−1T'=75 kJ mol−1298 KT'=90 kJ mol−1×298 K75 kJ mol−1T'=357.6 K

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 4, Problem 4.28P

Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of 429.18.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The relative rates of two reactions are expressed as,

log(rateArateB)=ΔG°B∗−ΔG°A∗2.3RT

Where,

• ΔG°B∗ and ΔG°A∗ represents the activation energy for reaction A and B.

• R is the gas constant.

• T is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

log(rateArateB)=75 kJ mol−1−90 kJ mol−12.3(8.314 J K−1 mol−1)(298 K)log(rateArateB)=−15 kJ mol−15698.41 Jlog(rateArateB)=−15×103J mol−15698.41 J mol−1log(rateArateB)=−2.632

Inverse of the logarithm is taken on both the sides of the equation.

(rateArateB)=0.00233(rateBrateA)=429.18

rateB=429.18(rateA) …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of 429.18.

Conclusion

The reaction B is faster than the reaction A by a factor of 429.18.

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The rate of the reaction is expressed as,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by T'. The calculation for the T' is as follows,

rateA=rateA10−ΔG°A∗/2.3RT'=10−ΔG°B∗/2.3RT

Take log on both sides of equation.

−ΔG°A∗2.3RT'=−ΔG°B∗2.3RT

ΔG°A∗T'=ΔG°B∗T …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

90 kJ mol−1T'=75 kJ mol−1298 KT'=90 kJ mol−1×298 K75 kJ mol−1T'=357.6 K

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 4, Problem 4.28P

Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of 429.18.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The relative rates of two reactions are expressed as,

log(rateArateB)=ΔG°B∗−ΔG°A∗2.3RT

Where,

• ΔG°B∗ and ΔG°A∗ represents the activation energy for reaction A and B.

• R is the gas constant.

• T is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

log(rateArateB)=75 kJ mol−1−90 kJ mol−12.3(8.314 J K−1 mol−1)(298 K)log(rateArateB)=−15 kJ mol−15698.41 Jlog(rateArateB)=−15×103J mol−15698.41 J mol−1log(rateArateB)=−2.632

Inverse of the logarithm is taken on both the sides of the equation.

(rateArateB)=0.00233(rateBrateA)=429.18

rateB=429.18(rateA) …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of 429.18.

Conclusion

The reaction B is faster than the reaction A by a factor of 429.18.

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The rate of the reaction is expressed as,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by T'. The calculation for the T' is as follows,

rateA=rateA10−ΔG°A∗/2.3RT'=10−ΔG°B∗/2.3RT

Take log on both sides of equation.

−ΔG°A∗2.3RT'=−ΔG°B∗2.3RT

ΔG°A∗T'=ΔG°B∗T …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

90 kJ mol−1T'=75 kJ mol−1298 KT'=90 kJ mol−1×298 K75 kJ mol−1T'=357.6 K

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 4, Problem 4.28P

Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of 429.18.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The relative rates of two reactions are expressed as,

log(rateArateB)=ΔG°B∗−ΔG°A∗2.3RT

Where,

• ΔG°B∗ and ΔG°A∗ represents the activation energy for reaction A and B.

• R is the gas constant.

• T is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

log(rateArateB)=75 kJ mol−1−90 kJ mol−12.3(8.314 J K−1 mol−1)(298 K)log(rateArateB)=−15 kJ mol−15698.41 Jlog(rateArateB)=−15×103J mol−15698.41 J mol−1log(rateArateB)=−2.632

Inverse of the logarithm is taken on both the sides of the equation.

(rateArateB)=0.00233(rateBrateA)=429.18

rateB=429.18(rateA) …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of 429.18.

Conclusion

The reaction B is faster than the reaction A by a factor of 429.18.

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The rate of the reaction is expressed as,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by T'. The calculation for the T' is as follows,

rateA=rateA10−ΔG°A∗/2.3RT'=10−ΔG°B∗/2.3RT

Take log on both sides of equation.

−ΔG°A∗2.3RT'=−ΔG°B∗2.3RT

ΔG°A∗T'=ΔG°B∗T …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

90 kJ mol−1T'=75 kJ mol−1298 KT'=90 kJ mol−1×298 K75 kJ mol−1T'=357.6 K

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Answer 6

Chapter 4, Problem 4.28P

Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of 429.18.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The relative rates of two reactions are expressed as,

log(rateArateB)=ΔG°B∗−ΔG°A∗2.3RT

Where,

• ΔG°B∗ and ΔG°A∗ represents the activation energy for reaction A and B.

• R is the gas constant.

• T is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

log(rateArateB)=75 kJ mol−1−90 kJ mol−12.3(8.314 J K−1 mol−1)(298 K)log(rateArateB)=−15 kJ mol−15698.41 Jlog(rateArateB)=−15×103J mol−15698.41 J mol−1log(rateArateB)=−2.632

Inverse of the logarithm is taken on both the sides of the equation.

(rateArateB)=0.00233(rateBrateA)=429.18

rateB=429.18(rateA) …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of 429.18.

Conclusion

The reaction B is faster than the reaction A by a factor of 429.18.

Answer 7

Chapter 4, Problem 4.28P

Interpretation Introduction

(a)

Interpretation:

The faster reaction from the given two reactions is to be predicted and the factor by which the one reaction is faster than the other is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Answer 8

Expert Solution

Answer to Problem 4.28P

The reaction B is faster than the reaction A by a factor of 429.18.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The relative rates of two reactions are expressed as,

log(rateArateB)=ΔG°B∗−ΔG°A∗2.3RT

Where,

• ΔG°B∗ and ΔG°A∗ represents the activation energy for reaction A and B.

• R is the gas constant.

• T is the temperature.

Substitute the activation energy for reaction A and B, gas constant and temperature in the given formula.

log(rateArateB)=75 kJ mol−1−90 kJ mol−12.3(8.314 J K−1 mol−1)(298 K)log(rateArateB)=−15 kJ mol−15698.41 Jlog(rateArateB)=−15×103J mol−15698.41 J mol−1log(rateArateB)=−2.632

Inverse of the logarithm is taken on both the sides of the equation.

(rateArateB)=0.00233(rateBrateA)=429.18

rateB=429.18(rateA) …(1)

Thus, from the equation (1), the reaction B is faster than the reaction A by a factor of 429.18.

Answer 13

Conclusion

The reaction B is faster than the reaction A by a factor of 429.18.

Answer 14

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Expert Solution

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The rate of the reaction is expressed as,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by T'. The calculation for the T' is as follows,

rateA=rateA10−ΔG°A∗/2.3RT'=10−ΔG°B∗/2.3RT

Take log on both sides of equation.

−ΔG°A∗2.3RT'=−ΔG°B∗2.3RT

ΔG°A∗T'=ΔG°B∗T …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

90 kJ mol−1T'=75 kJ mol−1298 KT'=90 kJ mol−1×298 K75 kJ mol−1T'=357.6 K

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Answer 15

Interpretation Introduction

(b)

Interpretation:

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is to be calculated.

Concept introduction:

The rate of the reaction is affected by the free energy of activation of the reaction. The relationship between the free energy of activation and rate of reaction is given by,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

Answer 16

Expert Solution

Answer to Problem 4.28P

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

It is given that standard free energy of activation of reaction A is 90 kJ mol−1(21.5 kcal mol−1) and for reaction B is 75 kJ mol−1(17.9 kcal mol−1).

The rate of the reaction is expressed as,

rate=10−ΔG°∗/2.3RT

Where,

• ΔG°∗ represents the activation energy for reaction.

• R is the gas constant.

• T is the temperature.

The temperature for reaction A at which rate of reaction of A and B are equal is given by T'. The calculation for the T' is as follows,

rateA=rateA10−ΔG°A∗/2.3RT'=10−ΔG°B∗/2.3RT

Take log on both sides of equation.

−ΔG°A∗2.3RT'=−ΔG°B∗2.3RT

ΔG°A∗T'=ΔG°B∗T …(1)

Substitute the activation energy for reaction A and B, and temperature for reaction B in the given equation (2).

90 kJ mol−1T'=75 kJ mol−1298 KT'=90 kJ mol−1×298 K75 kJ mol−1T'=357.6 K

Thus, the increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Answer 21

Conclusion

The increase in temperature of the slower reaction such that its rate becomes equal to that of faster reaction is 357.6 K.

Answer 22

Ch. 4 - Prob. 4.1P Ch. 4 - Prob. 4.2P Ch. 4 - Prob. 4.3P Ch. 4 - Prob. 4.4P Ch. 4 - Prob. 4.5P Ch. 4 - Prob. 4.6P Ch. 4 - Prob. 4.7P Ch. 4 - Prob. 4.8P Ch. 4 - Prob. 4.9P Ch. 4 - Prob. 4.10P

Concept explainers

Videos

Answer to Problem 4.28P

Explanation of Solution

Answer to Problem 4.28P

Explanation of Solution

Want to see more full solutions like this?

Chapter 4 Solutions