CHEMISTRY 113

HOMEWORK 6

NAME___________________________

Due Feb 21, 2001

 

Computer Homework: 16. Kinetics: all
  1. The following is the energy profile for certain reaction that proceeds by two step mechanism. On the energy profile indicate
  1. The position of reactants and products.

    2. The reactants correspond to the horizontal line at the left; the products to the horizontal line at the right

  2. Activation energy for the overall reaction

    3. The activation energy for the overall reaction corresponds to the most energy intensive step or the second hump

  3. Enthalpy change for the reaction

    4. The enthalpy change for the reaction corresponds to the difference between the energy the products and reactants or the difference in energy represented by the horizontal lines representing them.

  4. Which point on the plot represents the energy of the intermediate

    5. The intermediate is represented by the well between the two activation energy humps.

  5. Which step in the mechanism of this reaction is rate determining?

The second step because it has the higher energy requirement to overcome.

 

 2. The rate constants were measured at various temperatures for the reaction HI(g) + CH3I(g) -> CH4(g) + I2(g) The following data was obtained

Rate constant(Lmol-1s-1 Temp oC

1.91 x 10 –2 205

2.74 x 10 –2 210

3.09 x 10 –2 215

5.51 x 10-2 220

7.73 x 10-2 225

1.08 x 10 –1 230

a)Determine the activation energy the activation energy equation in kJ/mol .

 The activation energy could be obtained by the slope of the best straight line in a graph of ln rate constant vs 1/Tabs. This would have the advantage of giving an average value for Ea. The Arrhenius equation could also be used for any two points to calculate the Ea represented by their data. Because the next problem is to calculate a higher temperature yet, the last two data points seem a good choice.

ln(k1/k2) =(Ea/R)(1/T2-1/T1) = ln(1.08 x 10-1/7.73 x 10-2) = ln 1.397 =0.3344 = (Ea/8.314 J/mol deg)(1/498 - 1/503)

Ea = (0.33441.393 x 105 J/mol * 8.314 J/mol deg)/ 1.996 x 10-5 deg-1 = 1.393 x 105 J/mol

 

Ea = |_1.4 x 105 J/ mol_____

b) The rate constant at 240oC.

ln(k1/1.08 x 10-1) = (1.393 x 105 J/mol/8.314 J/mol deg) (1/503 - 1/513) =

1.675 x 104deg-1 (3.875 x 10-5 deg) = 0.6493

k1/1.08 x 10-1 = inv ln 0.6493 = 1.914

k1 = 1.914 x 0.108 = 0.206

k = |_2.1 x 10-1________

3. The experimental rate law for the reaction NO2 + CO à CO2 + NO, is Rate = k[NO 2]2 . If the mechanism is

2NO2 à NO3 + NO (slow) Rate determining step

NO3 + CO à NO2 + CO2 (fast)

Show that the predicted rate equation is the same as the experimental.

Rate is proportional to the concentrations of the reactants in the rate-determining step.

Rate = k[NO2]2

 Pred Rate eq.= _ Rate = k[NO2]2

 

4. The following mechanism (a type of chain reaction) has been proposed for the gas phase reaction of chlorine with carbon monoxide to form phosgene COCl2

Cl2 = 2 Cl fast

Cl + CO = COCl fast

COCl + Cl2 = COCl2 + Cl slow Rate-determining step

Determine the rate law for this mechanism (and the reaction).

a) Identify the intermediates in the reaction, if any Cl and COCl

b) Write the equation for the rate limiting step. Rate = k5[COCl][Cl2]

c) Write the complete reaction for the formation of phosgene Removing intermediates,

Cl2 + CO = COCl2

d) Determine the rate law for the reaction.

 Using the second equation to determine the intermediate COCl,

k3[Cl][CO] = k4[COCl]

 [COCl] =(k3/k4)[Cl][CO]

 Use the first equation to determine the intermediate Cl, k1[Cl2] = k2[Cl]2

[Cl] = {(k1/k2)[Cl2]}^0.5

[COCl] = k3/k4[Cl][CO] = (k1/k2)^0.5 (k3/k4)[Cl2])^0.5 ][CO]

Rate = k5[COCl][Cl2] = k5 (k1/k2)^0.5 (k3/k4)[Cl2]^3/2[CO]

 

5. Write the equilibrium law for the following reactions in terms of molar concentrations.

Br2 + 5 F2 ó 2BrF5

K = [Products]x/[Reactants]y = [BrF5]2 / [Br2][F2]5

Kc = [BrF5]2 / [Br2][F2]5

 

b. If the equilibrium concentration of Br2 is 0.050 mol/L , F2 is 0.25mol/L and BrF5 is 0.15 mol/L

calculate the value for Kc

K = [BrF5]2 / [Br2][F2]5 = [0.15 M]2 / [0.050 M][0.25M]5 = 460.8

 

Kc = __4.6 x 102__________

6. Use the following equilibria

 

2CH4 ó C2H6 + H2

Kc = 9.5 x 10 –33

 

CH4 + H2O ó CH3OH + H2

Kc = 2.8 x 10-21

to calculate Kc for 2CH3OH + H2 ó C2H6 + 2H2O

Need to reverse and double the second equation so K = (1/2.8 x 10-21)^2 = [CH4]2 [H2O]2/ [CH3OH]2 [H2]2

Use first equation as written so K = 9.5 x 10-33 = [C2H6][H2] / [CH4]2

K = (3.57 x 1020 )2 x 9.5 x 10-33 = 1.28 x 1041 x 9.5 x 10-33 = 1.21 x 109

K = [CH4]2 [H2O]2/ [CH3OH]2 [H2]2 x [C2H6][H2] / [CH4]2 = [C2H6][H2O]2 / [CH3OH]2[H2]

Kc = _1.2 x 109 __________