CHM 2046C   Module Nine Sample Exam Answers    

Part D: Rate Laws from Experimental Data: Initial Rates Method      10 points

 

In the reaction:      2 NO (g)   +      Cl 2 (g)    à       2 NOCl (g)  

 

Was studied at -10 oC.  The following results were obtained:

 

[NO] (mol/L)                  [Cl 2] (mol/L)               Initial Rate    (Mol/L min)

     0.10                                  0.10                           0.18

     0.10                                  0.20                           0.35

     0.20                                  0.20                             1.45

 

a.   What is the Rate Law?

 

We can a general write the rate expression for the overall reaction, but we can not be specific with the rate order of each reactant. They must be determined from the experimental data or the rate determining step of the accepted mechanism of the reaction. The general expression is:

       Rateoverall = koverall [NO]x[Cl2]y

 

b. What is the rate order of each reactant?

We may determine the order by substitution data from the experimental data above:

 

From Experiment 1:                                        From Experiment 2:

Rate1= k1[NO]x[Cl2]y                           Rate2= k2[NO]x[Cl2]y

 

Isolating k1:                                            isolating k2

                                                                                                             

     Rate1

 

 

    

 

    Rate2

---------------

=

k1

                   k2

=

---------------

[NO]x[Cl2]y

 

 

[NO]x[Cl2]y

 

[NO]x[Cl2]y

    

At constant temperature:    

k1   =  k2   =    k3   =    koverall

                                                       

Therefore set k1= k2  and substitute values for k1 and k2 from the rate expressions:                                                                                                            

     Rate1

 

    Rate2

 

  Rate3 

 

---------------

=

---------------

Also =

---------------

When necessary

[NO]x[Cl2]y

(values exp #1)

 

[NO]x[Cl2]y

(values exp #2)

 

[NO]x[Cl2]y

(values exp #3)

 

 

 

Rearrange the equation to isolate Rates on one side and the concentrations from the experiments on the other side:

 

    

      Rate1

 

(values exp #1)

[NO]x [Cl2]y

---------------

=

---------------

     Rate2

 

[NO]x [Cl2]y

(values exp #2)

 

 

 

Now put the actual values into the equation and solve for x and y from experiment #1 & #2:

 

    

0.18 Mol/L min

 

(values exp #1)

[0.10 M]x [0.10 M]y

---------------

=

-------------------------

 0.36 Mol/L min

 

[0.10 M]x [0.20 M]y

(values exp #2)

 

Canceling the units:

 

     0.18 Mol/L min

 

[0.10 M]x [0.10 M]y

     ---------------

=

------------------------

       0.36 Mol/L min

 

[0.10 M]x [0.20 M]y

 

 

Canceling numerator with dominator:

 

     0.18

 

[0.10]x [0.10]y

     -------

=

-------------------

       0.36

 

[0.10]x [0.20]y

 

 

Simply the fractions:

 

½  = (½) y

 

Therefore y = 1

 

 

 

 

Now to find the value of x look at the experiments and find two experiments where the concentrations of the Cl2 are constant. This would be Experiment #2 and Experiment #3. Substitute the data into the following rearranged expression after setting k21 = k3:

 

    

      Rate2

 

(values exp #2)

[NO]x [Cl2]y

---------------

=

---------------

      Rate3

 

[NO]x [Cl2]y

(values exp #3)

 

 

 

Now put the actual values into the equation and solve for x and y from experiment #2 & #3:

 

    

0.35 Mol/L min

 

(values exp #2)

[0.10 M]x [0.20 M]y

-----------------

=

-------------------------

 1.45 Mol/L min

 

[0.20 M]x [0.20 M]y

(values exp #3)

 

Canceling the units:

 

     0.35 Mol/L min

 

[0.10 M]x [0.20 M]y

     ---------------

=

------------------------

       1.45 Mol/L min

 

[0.20 M]x [0.20 M]y

 

 

Canceling numerator with dominator:

 

     0.35

 

[0.10]x [0.20]y

     -------

=

-------------------

       1.45

 

[0.20]x [0.20]y

 

 

Simply the fractions (Since we are working for a one significant answer change the 1.45 to 1.40 to make an even fraction:

 

¼  = (½) x

 

Therefore x = 2

 

 

c. What is the overall rate order of the reaction?

 

Since x = 2  and y = 1

We say the reaction is second order with respect to the concentration of NO and first order with respect to the concentration of Cl2.

 

Then:

2 + 1 = 3 or the reaction is overall third order, or a termolecular reaction. This means that in the reaction mechanism two NO molecules must simultaneously strike a Cl2 molecule, which is very rare to have a three body collision for a reaction to take place (by coincidence the stoichiometric coefficients in the overall reaction happen to be the rate order of each reactant):

 

Rateoverall = koverall [NO]2[Cl2]1

 

 

d. What is the value of the rate constant k?

 

Now we have determined the order of each reactant, we may take any set of experimental data and calculate the value of k. In lab we would calculate k for each experiment and average the values. But for a textbook/test problem one set is sufficient.

 

From Experiment 1:

Rate1= k1[NO]2[Cl2]1

                                                                                                            

     Rate1

 

 

---------------

=

k1

[NO]2[Cl2]1

 

 

 

Substituting the values:

 

                                                                                                            

     0.18 Mol/L min

 

 

---------------

=

k1

[0.10 M]2[0.10 M]1

 

 

 

Simplify:

                                                                                                             

     0.18 Mol/L min

 

 

     ---------------

=

k1

      (0.10 M)3

 

 

 

                                                                                                            

     0.18 Mol/L min

 

 

     ---------------

=

k1

      0.0010 mol3/L3

 

 

 

 

k1 = 180 L2/mol2 ∙min

 

 

e. What are the units of the rate constant?

 

 

L2/mol2 ∙min

 

Read section 15.3 Page 706-712.

Work all problems #7-16 p 744-745