Reactivity in Chemistry

Reaction Kinetics

MK16. Solutions to Selected Problems

 

Problem MK1.1.

A plot of ln(k) vs. 1/T gives a stright line fitting the equation: y = mx + b.

m = -Ea/R or Ea = -mR = -m * 8.314 J K-1 mol2-1

b = lnA or A = eb

 

Problem MK1.2.

A plot of ln(k/T) vs. 1/T gives a stright line with the equation: y = -4606.6x + 10.2.

ΔH‡ = mR = 4606.6 K * 8.314 J K-1 mol2-1 = 38.3 kJ mol-1

ΔS‡ = 8.314 * [b - ln(kB/h)] = 8.314(10.2 - 23.759) = -112.7 J K-1 mol-1

 

Problem MK1.3.

A plot of ln(k/T) vs. 1/T gives a stright line with the equation: y = -5228.6x + 12.739.

ΔH‡ = mR = 5228.6 K * 8.314 J K-1 mol2-1 = 43.5 kJ mol-1

ΔS‡ = 8.314 * [b - ln(kB/h)] = 8.314(12.739 - 23.759) = -91.6 J K-1 mol-1

 

Problem MK1.4.

A plot of ln(k/T) vs. 1/T gives a stright line with the equation: y = -11308x + 20.9.

ΔH‡ = mR = 11308 K * 8.314 J K-1 mol2-1 = 94.0 kJ mol-1

ΔS‡ = 8.314 * [b - ln(kB/h)] = 8.314(20.927 - 23.759) = -23.5 J K-1 mol-1

 

Problem MK1.5.

A plot of ln(k/T) vs. 1/T gives a stright line with the equation: y = -6262.5x + 5.2411.

ΔH‡ = mR = 6262.5 K * 8.314 J K-1 mol2-1 = 52.1 kJ mol-1

ΔS‡ = 8.314 * [b - ln(kB/h)] = 8.314(5.2411 - 23.759) = -154.0 J K-1 mol-1

 

Problem MK3.1.

a)  slope =  (0.107-.125)/140 =  -1.29 x 10-4 mol L-1 sec-1

b)  slope =  (0.00440-.00499)/51 = -1.16 x 10-5 mol L-1 min-1

c)  slope =  (0.0063-.007)/10 = -7.0 x 10-5 mol L-1 min-1

 

Problem MK3.2.

a)   t1/2 = 0.693/k

  t1/2 = 0.693 / 2 x 10-2 min-1  = 35 minutes

number of half lives = 175 minutes / 35 minutes = 5 half lives

 

b)   t1/2 = 0.693/k

  t1/2 = 0.693 / 1.5 x 10-2 min-1  = 46 minutes

number of half lives = 90 minutes / 46 minutes = 1.96 half lives

 

c)   t1/2 = 0.693/k

  t1/2 = 0.693 / 0.127 hr-1  = 5.46 hours

number of half lives = 18 hours / 5.46 hours = 3.3 half lives

 

Problem MK3.3.

a) k = rate / concentration = slope = (.005 - 0.0015)/(0.08 - 0.02) = 5.8 x 10-2 s-1

 

b) k = rate / concentration = slope = (.0028 - 0.001)/(0.6 - 0.2) = 4.5 x 10-3 s-1

 

c) k = rate / concentration = slope = (.012 - 0.003)/(0.04 - 0.01) = 0.3 s-1

 

Problem MK3.4.

a)  rate1/rate2 = k[Reactant]1m / k[Reactant]2m = ([Reactant]1 / [Reactant]2)m

0.189 / 0.0944 = (0.0194 / 0.0097)m

log(2.00) = m log(2.00)

m = .301 / 0.301 = 1

first order

 

b)  rate1/rate2 = k[Reactant]1m / k[Reactant]2m = ([Reactant]1 / [Reactant]2)m

0.806 / 0.101 = (050 / 0.25)m

log(7.98) = m log(2.00)

m = 0.902 / 0.301 = 3

third order

 

c)  rate1/rate2 = k[Reactant]1m / k[Reactant]2m = ([Reactant]1 / [Reactant]2)m

0.000508 / 0.000127 = (3.00 / 1.50)m

log(4.00) = m log(2.00)

m = 0.602 / 0.302 = 2

second order

 

Problem MK3.5.

a) subject/standard = 0.79; 1.24; 1.42  increasing

b) subject/standard = 1.50; 1.50; 1.50 constant

c) subject/standard = 1.17; 1.08; 0.913 decreasing

 

Problem MK4.1.

a)

d[Co']/dt = 0 = k1[Co] - k-1[dmb][Co'] - k2[nor][Co']

k1[Co] = k-1[dmb][Co'] + k2[nor][Co']

[Co'] = k1[Co] / ( k-1[dmb] + k2[nor])

d[Product]/dt = k2[nor][Co']

= k1k2[nor][Co] / ( k-1[dmb] + k2[nor]) = kobs[Co]

b)

1/kobs = ( k-1[dmb] + k2[nor]) / (k1k2[nor])

1/kobs = (k-1 / k1k2)([dmb) / [nor]) + 1 / k1

A plot of 1/kobs vs. [dmb]/[nor] is linear with slope = (k-1 / k1k2) and y intercept = 1 / k1.

 

Problem MK5.1.

a) non-competitive

b)  competitive

c)  competitive

d) non-competitive

 

Problem MK6.1.

a)  Vmax = 1.8 x 10-5 mol L-1 s-1

     Vmax / 2 = 9 x 10-6 mol L-1 s-1  so Km = 6 mol L-1   

b) Vmax = 6.5 x 10-7 mol L-1 s-1

     Vmax / 2 = 3.25 x 10-7 mol L-1 s-1  so Km = 7 mol L-1   

c)  Vmax = 2.6 x 10-5 mol L-1 s-1

     Vmax / 2 = 1.3 x 10-5 mol L-1 s-1  so Km = 6 mol L-1   

d)  Vmax = 1.2 x 10-5 mol L-1 s-1

     Vmax / 2 = 6 x 10-6 mol L-1 s-1  so Km = 6 mol L-1   

e)  Vmax = 6.0 x 10-7 mol L-1 s-1

     Vmax / 2 = 3 x 10-7 mol L-1 s-1  so Km = 13 mol L-1   

   

Problem MK6.2.

a) Kdissoc = k-1/k1 = 0.00765/0.0101 = 0.757

b) Kdissoc = k-1/k1 = 0.321/0.197 = 1.63

c) Kdissoc = k-1/k1 = 0.542/0.775 = 0.699

d) Kdissoc = k-1/k1 = 0.499/0.0228 = 0.0457

 

Problem MK6.3.

a) Km = k-1/(k1 + k2) = 0.00765/(0.0101 + 0.00654) = 0.460

a) Km = k-1/(k1 + k2) = 0.321/(0.197 + 0.0000654) = 1.63

a) Km = k-1/(k1 + k2) = 0.542/(0.775 + 0.234) = 0.537

a) Km = k-1/(k1 + k2) = 0.0228/(0.499 + 0.0087) = 0.0449

 

Problem MK6.4.

In all cases, Kdissoc = k-1/k1 = 0.195/0.326 = 0.598

a) Km = k-1/(k1 + k2) = 0.195/(0.326 + 0.00038) = 0.597

a) Km = k-1/(k1 + k2) = 0.195/(0.326 + 0.723) = 0.186

a) Km = k-1/(k1 + k2) = 0.195/(0.326 + 0.00239) = 0.594

a) Km = k-1/(k1 + k2) = 0.195/(0.326 + 0.0345) = 0.0541

Km is quite similar to Kdissoc provided k2 is much smaller than k-1.

 

Problem MK6.5.

a) kcat = vmax/[E] = (125.3 µM min-1*106pM/µM)/31.5 pM = 1,250 min-1

b) kcat = vmax/[E] = (1.587 mM min-1*109pM/mM)/7.04 pM = 15,900 min-1

c) kcat = vmax/[E] = (3.21 nM min-1*103pM/nM)/0.0129 pM = 465 min-1

d) kcat = vmax/[E] = (0.129 µM min-1*106pM/µM)/2.27 pM = 3.93 min-1

 

Problem MK6.6.

a) Efficiency = kcat/Km = 409 min-1 / 31.5 nM = 39.6 L nmol-1 min-1

b) Efficiency = kcat/Km = 22.6 min-1 / 1.56 mM = 1.26 L mmol-1 min-1

c) Efficiency = kcat/Km = 1.76 min-1 / 17.4 mM = 1,760 L mmol-1 min-1

 

 

Problem MK6.7.

a) Vmax = 1/b = 1/0.129 L s nmol-1= 7.75 nmol L-1 s-1

Km = -1/x int = m/b = 0.00326 s-1/0.129 L s nmol-1= 0.0253 nmol L-1

b) Vmax = 1/b = 1/0.0957 L s nmol-1= 10.5 nmol L-1 s-1

Km = -1/x int = m/b = 0.0759 s-1/0.0957 L s nmol-1= 0.793 nmol L-1

c) Vmax = 1/b = 1/0.129 L s nmol-1= 1.31 nmol L-1 s-1

Km = -1/x int = m/b = 0.000434 s-1/0.765 L s nmol-1= 0.000567 nmol L-1

 

Problem MK7.1.

a)  1/Vmax = 30 L s mol-1 so Vmax = 3.3 x 10-2 mol L-1 s-1     

     -1/Km = -40 L mmol-1 so Km =  2.5 x 10-2 M L-1            

b)  1/Vmax = 50 L s mol-1 so Vmax = 2.0 x 10-2 mol L-1 s-1     

     -1/Km = -70 L mmol-1 so Km =  1.4 x 10-2 M L-1           

c)  1/Vmax = 60 L s mol-1 so Vmax = 1.7 x 10-2 mol L-1 s-1     

     -1/Km = -70 L mmol-1 so Km =  1.4 x 10-2 M L-1           

d)  1/Vmax = 50 L s mol-1 so Vmax = 2.0 x 10-2 mol L-1 s-1     

     -1/Km = -100 L mmol-1 so Km =  1.0 x 10-2 M L-1           

e)  1/Vmax = 30 L s mol-1 so Vmax = 3.3 x 10-2 mol L-1 s-1     

     -1/Km = -100 L mmol-1 so Km =  1.0 x 10-2 M L-1           

a)  1/Vmax = 30 L s mol-1 so Vmax = 3.3 x 10-2 mol L-1 s-1     

     -1/Km = -60 L mmol-1 so Km =  1.7 x 10-2 M L-1           

 

 

Problem MK7.2.

a)  uncompetitive

b)  mixed

c)  noncompetitive

d) competitive

e)  uncompetitive

f)  noncompetitive

g) competitive

 

 

 

 

This site was written by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (retired) with other authors as noted on individual pages.  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. 

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This material is based upon work supported by the National Science Foundation under Grant No. 1043566.

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