CHEM_100_11                                                                                              2013               

Assignment #5

 KINETICS (chapter 14)

1.      {14}  Indicate the order ( zero, 1st or 2nd) of the rxn  Aà B  for each of the following observations

a)      A plot of the inverse of the concentration [A]  versus time yields a straight line (2nd)

b)     The unit of rate constant is Ms-1   (zero)

c)       The rxn has a half-life that is independent of the initial concentration, [A]0 (1st order)

d)     the concentration of reactant [A] fall to one-half in equal intervals of time (1st order)

e)      the half-life of the  rxn gets shorter as the initial concentration is increased (2nd order)

f)        A plot of the ln [A]  versus time yields a straight line (1st order)

g)      When the initial concentration [A] doubles, the rate doubles (1st order).

h)     When [A] is reduced by a factor of 2, the rate of the reaction is reduced by a factor of 4 (2nd order).

i)        Change in  [A] does not change the rate   (zero)

j)       the unit of the rate constant is M-1s-1

2d order

k)     the unit of the rate constant is s-1

1st order

l)        the rate law is described by the formula : rate=k[A]2

second order

m)   A plot of [A] versus time yields a straight line.  (zero)

n)     The rate law is described by the formula rate=k   (zero)

 

2.      {10} The data in the table below were obtained using the initial rate method for the reaction:

2ClO2 (aq) + 2OH- (aq) à ClO3- (aq) + ClO2-(aq) + H2O (l)

 

exp. number               [ClO2], M                    [OH-], M         Initial rate, M/s

1                                  0.060                           0.030               0.0248

2                                  0.020                           0.030               0.00276

3                                  0.020                           0.090               0.00828

 

a)      {5} analyze data and deduce the differential rate law

b)     {2} calculate the magnitude of the rate constant

c)      {1} what is the overall order of this reaction?

d)     {2} calculate the reaction rate for [ClO2]=[OH-]= 0.01M

 

Solution:   (0.060/0.020)n = (0.0248/0.00276); 3n =9, n=2

                  (0.09/0.03)m = (0.00828/0.00276);  3m = 3, m=1

a)      Reaction is first order in OH- and second order in ClO2

   rate= k [ClO2]2[OH-]

b)      k= rate/([ ClO2]2[OH-]) = 0.0248Ms-1/(0.0602M2 x 0.030 M) = 229.6= 230 M-2s-1

c)      2+1 =3

d)     rate= 230 M-2s-1[0.01]2[[0.01] = 0.00023 M/s

 

 

3.  {10} The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32 x 10-10 s-1 at 25 oC. Calculate the rate constant at 40oC if the activation energy is 116 kJ/mol.

Solution

T1 = 25+273 = 298K

T2= 40 +273= 313K,    k2 = ?

ln (k1/k2) = Ea/R (1/T2-1/T1) = (116 x 103 J/mol)/(8.314 J/K mol) [(1/313K-1/298K) = 1.395 x 104 [-0.000161] = -2.244

k1/k2 = e-2.244  = 0.106

k1/k2 =0.106; k2 = (3.32 x 10-10 s-1)/ 0.106 = 3.13 x10-9 s-1

 

4. {10} Consider the first-order reaction Aà product, with k=2.95 x 10-3 s-1. What percent of A remains after 100 s?

Solution

ln([A]t/[A]0) = -kt  = -2.95 x 10-3 s-1 x 100 s-1 = -0.295 ;   [A]t/[A]0 = e-0.295 = 0.745

[A]t = 0.745 [A]0  – the fractional part of A remaining.

0.745 x 100 = 74.5%     

 

5. {10} the reaction 2Aà B is second order with a rate constant of 52 M-1 min-1 at 25oC.

(a)   {5} Starting with [A]0 = 0.0082M, how long will it take for [A]t = 2.7 x 10-7 M?

(b)   {5} Calculate the half-life of this reaction.

Solution

t1/2 = 1/k[A]0 ;     1/[A]t = kt + 1/[A]0 ;

1/[A]t - 1/[A]0 = kt = 1/2.7x10-7M  - 1/0.0082  M  =  0.37 x 107M-1 – 122.0 M-1 = 3699878 M-1;  t = 3699878/52 =71151min=1186hr = 49.4 days

t1/2 = 1/(52x0.0082) = 2.35 min

6.   {15}  The reaction Aà 2B + C is first order. If the initial [A] =1.50 M and k=1.02 x10-3 s-1,

(a) {10} what is the value of [A] after 400 s?

(b) {5} What is the concentration of B after the 400 s? Assume that volume is constant during the rxn.

(a)    ln [A]t = -kt + ln [A]0  ;

     ln[A]400 = -1.02 x 10-3 s-1 x 400 s  + ln 1.50 = -0.408 + 0.405 = -0.003;

   [A]400 = e-0.003 = 0.997M

(b) [A] consumed = 1.50M-0.997M = 0.503M

     [B] = 2 x 0.503M = 1.01M

 

7.  {5} The reaction 2Aà B is first order in A with a rate constant of 1.5 x 10-2 s-1 at 100oC. How long (in seconds) will it take for A to decrease from 0.30M to 0.10 M?

Solution

ln [A]t = -kt + ln [A]0

  ln([A]t/[A]0) = -kt = ln(0.10/0.30) = -1.099; t= 1.099/1.5x10-2 = 73s

 

 

8. {6} For the elementary process N2O5 (g) à NO2(g) + NO3(g) the activation energy, Ea, is 154kJ/mol  and overall ΔE reaction  is  136 kJ/mol.

(a). {2} Sketch the energy profile for this reaction, and label Ea and ΔE.

(b) {2} What is the activation energy for the reverse reaction?

 

(c)    {2} Which reaction is faster, forward or reverse?

 

Solution:

 

 

Ea (reverse) = 154-136= 18 kJ

 

Reverse reaction is faster because the activation energy is lower

 

 

9. {10} Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of NO to N2 and O2 is second order with rate constant of 0.0796 M-1s-1 at 737 0C and 0.0815 M-1s-1 at 947 0C. Calculate the activation energy for the reaction.

 

Solution:

R= 8.314 J/mol K

T1 =737 + 273 = 1010K

T2= 947 + 273 =1220K

 

ln(k1/k2) =  Ea/R (1/T2 -1/T1)

 

ln(0.0796/0.0815) = Ea /8.314 J/molK(1/1220-1/1010)

Ea = 8.314(-0.023589)Jmol-1K-1/(-1.704 x 10-4)K-1 = 1.151 x 103 J/mol = 1.15 kJ/mol