Determination of Mechanism in Chemistry

Mathematical Tools in Kinetics

MK17. Solutions 

Problem MK9.1.

 

Problem MK9.2.

 

Problem MK9.3.

 

 

Problem MK.9.4

 

 

Problem MK.9.5

 

Problem MK10.1.

a) ΔG = -RTlnK = -(1.987 cal K-1mol-1)(273.15 + 35 K) ln(0.93)

= -(1.987)(308.15)(-0.0726)

= 44.4 cal mol-1 or 0.0444 kcal mol-1 (or 0.186 kJ mol-1)

 

b) ΔG = -RTlnK = -(1.987 cal K-1mol-1)(273.15 + 50 K) ln(7.6)

= -(1.987)(323.15)(2.028)

= -1,302 cal mol-1 or 1.30 kcal mol-1 (or 5.45 kJ mol-1)

 

c) ΔG = -RTlnK = -(1.987 cal K-1mol-1)(273.15 + 70 K) ln(3.8)

= -(1.987)(343.15)(1.335)

= -910 cal mol-1 or -0.910 kcal mol-1 (or 3.81 kJ mol-1)

 

Problem MK10.2.

a)  ΔG = RT{ln(kBT/h) - lnk}

= (1.987)(273.15 + 45 K){ln[(1.38 x 10-23 J K-1)(273.15 + 45 K)/(6.626 x 10-34 J.s-1)] - ln[3.7 x 10-5 s-1]}

= (1.987)(318.15){ln[6.626 x 1012] + 10.2046}

= (632.16)(29.552 + 10.2046)

= 25,132 cal mol-1  = 25.1 kcal mol-1 = 105 kJ mol-1

 

b)  ΔG = RT{ln(kBT/h) - lnk}

= (1.987)(273.15 + 65 K){ln[(1.38 x 10-23 J K-1)(273.15 + 65 K)/(6.626 x 10-34 J.s-1)] - ln[1.9 x 10-6 s-1]}

= (1.987)(338.15){ln[7.0427 x 1012] + 13.1737}

= (671.904)(29.583 + 13.1737)

= 28,7282 cal mol-1  = 28.7 kcal mol-1 = 120 kJ mol-1

 

c)  ΔG = RT{ln(kBT/h) - lnk}

= (1.987)(273.15 + 85 K){ln[(1.38 x 10-23 J K-1)(273.15 + 85 K)/(6.626 x 10-34 J.s-1)] - ln[ 4.8 x 10-7  s-1]}

= (1.987)(358.15){ln[7.459 x 1012] + 14.5495}

= (711.644)(29.6405 + 14.5495)

= 31,448 cal mol-1  = 31.5 kcal mol-1 = 132 kJ mol-1

 

Problem MK10.3.

a)

b)

c)

 

Problem MK11.1.

a)

b)  I is a global minimum; II is a local minimum; III is a maximum.

 

Problem MK11.2.

a)

b) I is a local minimum; II is a global minimum; III is a maximum.

 

Problem MK11.3.

a)

b)  I is a maximum; II is a local minimum; III is a global minimum.

 

Problem MK12.1.

BDE = 2.86 cal/cm-1(19,898 - 7,598 cm-1) = 35.178 kcal/mol.

Problem MK12.2.

In both cases, a resonance-stabilized radical leads to easier bond cleavage, hence a weaker bond strength.

Problem MK12.3.

a) ΔG = - nFΔE = -(1)(96,485 J/V)(1 cal/4.184J)(1 kcal/1,000 cal)ΔE = -23.06 kcal/V

ΔG = - RTlnK = - 2.303RTlogK = 2.303RTpKa = (2.303)(1.987 cal/K mol)(298 K) pKa  = 1.36 pKa kcal/mol

b)  BDE (kcal/mol) = 1.37pKa + 23.06Eoox(PhO-) - 23.06Eored(H+) + 46 kcal/mol

= (1.37)(18) + (23.06)(0.55) + (23.06)(-0.4) +46 kcal/mol

= 24.66 + 12.68 - 9.22 + 46 kcal/mol = 74.12 kcal/mol

Problem MK13.1.

Approximating with whole number values for atomic mass:

a)  a 12C-1H bond

μ = m1m2 / (m1 + m2)

= 12 x 1 / (12 + 1)

= 12 / 13

= 0.923

 

vs. a 12C-2H bond

μ = 12 x 2 / (12 + 2)

= 24 / 14

= 1.71

 

b)  a 12C-1H bond

μ = 12 x 1 / (12 + 1)

= 0.923

 

vs. a 13C-1H bond

μ = 13 x 1 / (13 + 1)

= 13 / 14

= 0.929

 

c)  a 35Cl-35Cl bond

μ = 35 x 35 / (35 + 35)

= 1,225 / 70

= 17.5

vs. a 35Cl-37Cl bond

μ = 35 x 37 / (35 + 37)

= 1,295 / 72

= 18.0

 

Problem MK13.2.

The value of μ almost doubles for a C-D vs. a C-H bond.  In cases of heavier atoms, the change in reduced mass is only a small percentage.  That leads to a very small isotope effect when heavier atoms are concerned.

 

Problem MK13.3.

This is a primary isotope effect in which a Pt-H or Pt-D bond is being broken.  The relatively low value results from the non-linear geometry, since the methyl and hydride are cis- to each other on the platinum.

 

Problem MK14.1.

The phenolate anion delocalizes negative charge into the aromatic ring, whereas the benzoate anion does not.  Thus, π effects show up more strongly in σ- than in σ.

 

Problem MK14.2.

The platinum is changing from Pt(II) to Pt(0).  The more electron-donating the phosphine is, the more it stabilizes the higher oxidation state of the platinum, retarding the reductive elimination reaction.

 

Problem MK15.1.

a) approximate slope = -0.012 M / 25 s = -4.8 x 10-4 mol L-1 s-1

b) approximate slope = -0.002 M / 50 s = -4 x 10-5 mol L-1 s-1

Note that these slopes correspond to reaction rates.

 

Problem MK15.2.

a) dy/dx  = -6e-3x           b) dy/dx = 3e5x

 

Problem MK15.3.

a) dy/dx  =  3 - 4x          b) y =  0.9 - 10x + 24x2        

 

 

This site was written by Chris P. Schaller, Ph.D., retired, College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

 

Support for this project was provided by the Opens Textbooks Pilot Program of the U.S. Department of Education through a collaboration with the Libre Texts project at University of California, Davis.

 

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