Organic Chemistry, Ebook And Single-course Homework Access
Organic Chemistry, Ebook And Single-course Homework Access
6th Edition
ISBN: 9781319085841
Author: LOUDON
Publisher: MAC HIGHER
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Chapter 9, Problem 9.7P
Interpretation Introduction

(a)

Interpretation:

The ratio of rate constants kA/kB at 25°C for two reactions A and B is to be stated.

Concept introduction:

Relative rate is defined as the ratio of rate of a reaction and rate of a standard reaction. The rate constant is numerically equal to the rate under standard condition under which concentration is 1molL1 for all reactants.

The relationship between relative rate of two reactions A and B under standard condition is shown below.

log(kAkB)=ΔG°B+ΔG°A+2.303RT

Expert Solution
Check Mark

Answer to Problem 9.7P

The ratio of rate constants kA/kB at 25°C for two reactions A and B is 288.4.

Explanation of Solution

The ratio of rate constants kA/kB at 25°C for two reactions A and B is calculated by the formula given below.

log(kAkB)=ΔG°B+ΔG°A+2.303RT…(1)

Where,

ΔG°B+ΔG°A+ is difference in standard free energies of activation for reaction B and reaction A.

R is the gas constant (8.314Jmol1K1).

T is temperature (25°C).

The values of ΔG°B+ΔG°A+, R and T are 14kJmol1, (8.314Jmol1K1) and 298K respectively.

Substitute the values of ΔG°B+ΔG°A+, R and T in equation (1).

log(kAkB)=ΔG°B+ΔG°A+2.303RT=14kJmol12.303×8.314Jmol1K1×298K=14kJmol15705.84Jmol1

The conversion of Jmol1 to kJmol1 is shown below.

1Jmol1=103kJmol1

Therefore, the conversion of 5705.84Jmol1 to kJmol1 is shown below.

5705.84Jmol1=5705.84×103kJmol1=5.70kJmol1

Solve the above equation.

log(kAkB)=14kJmol15.70kJmol1log(kAkB)=2.46kAkB=antilog(2.46)=288.4

Therefore, the ratio of rate constants kA/kB at 25°C for two reactions A and B is 288.4.

Conclusion

The ratio of rate constants kA/kB at 25°C for two reactions A and B is 288.4.

Interpretation Introduction

(b)

Interpretation:

The difference in the standard free energies of activation at 25°C of two reactions A and B if reaction B is 450 times faster than reaction, is to be calculated.

Concept introduction:

Relative rate is defined as the ratio of rate of a reaction and rate of a standard reaction. The rate constant is numerically equal to the rate under standard condition under which concentration is 1molL1 for all reactants.

The relationship between relative rate of two reactions A and B under standard condition is shown below.

log(kAkB)=ΔG°B+ΔG°A+2.303RT

Expert Solution
Check Mark

Answer to Problem 9.7P

The difference in the standard free energies of activation at 25°C of two reactions A and B is 15.16kJmol1 and ΔG°A+ is greater than ΔG°B+.

Explanation of Solution

The formula to calculate relative rate is written below.

log(kAkB)=ΔG°B+ΔG°A+2.303RT…(1)

Rearrange the equation (1) to calculate the value of difference in the standard free energies of activation as shown below.

ΔG°B+ΔG°A+=log(kAkB)×2.303RT…(2)

Where,

kB and kA is rate constant of B and A.

R is the gas constant (8.314Jmol1K1).

T is temperature (25°C).

The reaction B is 450 times faster than reaction A. Therefore the value of kAkB is 1450.

The values of R and T are (8.314Jmol1K1) and 298K respectively.

Substitute the value of R and T in equation (2).

ΔG°B+ΔG°A+=log(kAkB)×2.303RT=log(1450)×2.303×(8.314Jmol1K1)×298K=log(0.0022)×5705.84Jmol1=(2.657)×5705.84Jmol1

ΔG°B+ΔG°A+=(15160.42Jmol1)

The conversion of Jmol1 to kJmol1 is shown below.

1Jmol1=103kJmol1

Therefore, the conversion of 15160.42Jmol1 to kJmol1 is shown below.

15160.42Jmol1=15160.42×103kJmol1=15.16kJmol1

Solve the above equation.

ΔG°B+ΔG°A+=(15160.42Jmol1)=15.16kJmol1

The difference in the standard free energies of activation at 25°C of two reactions A and B is 15.16kJmol1. The negative sign indicates that ΔG°A+ is greater than ΔG°B+.

Conclusion

The difference in the standard free energies of activation at 25°C of two reactions A and B is 15.16kJmol1 and ΔG°A+ is greater than ΔG°B+.

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Chapter 9 Solutions

Organic Chemistry, Ebook And Single-course Homework Access

Ch. 9 - Prob. 9.11PCh. 9 - Prob. 9.12PCh. 9 - Prob. 9.13PCh. 9 - Prob. 9.14PCh. 9 - Prob. 9.15PCh. 9 - Prob. 9.16PCh. 9 - Prob. 9.17PCh. 9 - Prob. 9.18PCh. 9 - Prob. 9.19PCh. 9 - Prob. 9.20PCh. 9 - Prob. 9.21PCh. 9 - Prob. 9.22PCh. 9 - Prob. 9.23PCh. 9 - Prob. 9.24PCh. 9 - Prob. 9.25PCh. 9 - Prob. 9.26PCh. 9 - Prob. 9.27PCh. 9 - Prob. 9.28PCh. 9 - Prob. 9.29PCh. 9 - Prob. 9.30PCh. 9 - Prob. 9.31PCh. 9 - Prob. 9.32PCh. 9 - Prob. 9.33PCh. 9 - Prob. 9.34PCh. 9 - Prob. 9.35PCh. 9 - Prob. 9.36PCh. 9 - Prob. 9.37PCh. 9 - Prob. 9.38PCh. 9 - Prob. 9.39PCh. 9 - Prob. 9.40PCh. 9 - Prob. 9.41PCh. 9 - Prob. 9.42PCh. 9 - Prob. 9.43PCh. 9 - Prob. 9.44APCh. 9 - Prob. 9.45APCh. 9 - Prob. 9.46APCh. 9 - Prob. 9.47APCh. 9 - Prob. 9.48APCh. 9 - Prob. 9.49APCh. 9 - Prob. 9.50APCh. 9 - Prob. 9.51APCh. 9 - Prob. 9.52APCh. 9 - Prob. 9.53APCh. 9 - Prob. 9.54APCh. 9 - Prob. 9.55APCh. 9 - Prob. 9.56APCh. 9 - Prob. 9.57APCh. 9 - Prob. 9.58APCh. 9 - Prob. 9.59APCh. 9 - Prob. 9.60APCh. 9 - Prob. 9.61APCh. 9 - Prob. 9.62APCh. 9 - Prob. 9.63APCh. 9 - Prob. 9.64APCh. 9 - Prob. 9.65APCh. 9 - Prob. 9.66APCh. 9 - Prob. 9.67APCh. 9 - Prob. 9.68APCh. 9 - Prob. 9.69APCh. 9 - Prob. 9.70APCh. 9 - Prob. 9.71APCh. 9 - Prob. 9.72APCh. 9 - Prob. 9.73APCh. 9 - Prob. 9.74APCh. 9 - Prob. 9.75APCh. 9 - Prob. 9.76APCh. 9 - Prob. 9.77APCh. 9 - Prob. 9.78APCh. 9 - Prob. 9.79APCh. 9 - Prob. 9.80APCh. 9 - Prob. 9.81APCh. 9 - Prob. 9.82APCh. 9 - Prob. 9.83APCh. 9 - Prob. 9.84APCh. 9 - Prob. 9.85APCh. 9 - Prob. 9.86APCh. 9 - Prob. 9.87AP
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