This example follows the kinetics for a dye called Royal Violet.

We write the rate law for the bleaching reaction as:    Rate  =    k [ RV⁺ ]^m  [ OH^- ]ⁿ
where:     m    = order of the reaction with respect to RV⁺, and
               n     = order of the reaction with respect to OH^-

We use a large excess of OH^-.  Therefore, while [ RV⁺ ] will decrease as the reaction proceeds, [ OH^- ] will not change significantly during the course of the reaction and [ OH^- ]ⁿ  will be constant in the rate law..

If we let q designate the product of constants, k [ OH^- ]ⁿ, we can write a simplified rate law as:    Rate  =    q [ RV⁺ ]^m

This permits determination of m, the order of the reaction with respect to RV⁺  by  following only the concentration of RV⁺ as a function of time.

The order of the reaction with respect to RV⁺, m, is determined by establishing which function of time best represents the concentration of RV⁺.

• if the reaction is first order ( m = 1) in RV⁺, we expect      ln( [ RV⁺ ] )   to provide a linear plot against time
• if the reaction is second order ( m = 2) in RV⁺, we expect 1 / [ RV⁺ ]      to provide a linear plot against time

1. Preliminaries:

Concentration of NaOH stock solution                       0.1134 M
Concentration of Royal Violet stock solution             1.0 X 10^-5 M
Beer's Law slope for RV⁺                                         5.2 X 10⁴ L/mol                                  

2. Prepare Initial Solutions

4.0 mL of NaOH solution in 50 mL of water  
10 mL of RV⁺ solution in 50 mL of water

3. Mix Solutions

Total volume of reaction solution (50 mL + 50 mL)                     = 100 mL
Concentration of NaOH  (4.0 X 0.1134 / 100)                           =  4.5 X 10^-3 M
Initial concentration of RV⁺ (10.0 X 1.0 X 10^-5 / 100)                =  1.0 X 10^-6 M

[Note that the concentration of NaOH is 4,500 times as large as that of RV⁺]

4. Collect Data - Absorbance vs Elapsed Time

5. Enter Data into Spreadsheet - 2 Possible Outcomes

5A. Log plot hs the larger RSQ and                   q₁ = 7.0 X 10^-4

5B. Reciprocal plot has the larger RSQ and       q₁ = 2.9 X 10^-3

5A. ln [ RV⁺ ] is linear when plotted against time

[Note that the quantity we actually plot is the absorbance of RV⁺, not its concentration.  This is without consequence in the case of a first order kinetic plot. The integrated form of the first order rate law is

                                    ln [ RV⁺ ]    =    ln [ RV⁺ ]₀ + k t

If the quantity we actually measure is proportional to the concentration, e.g., the absorbance, Abs = e [ RV⁺ ], with e = absorptivity, the rate law becomes

                                    ln( Abs / e )        =    ln( Abs₀ / e )         +  k t            ,or
                                    ln( Abs ) - ln( e ) =    ln( Abs₀ ) - ln( e ) +  k t            , or
                                    ln( Abs )             =    ln( Abs₀)               +  k t

In this case, the slope of the straight line does not depend on the proportionality constant between Abs and concentration! ]

The reaction is now known to be first order in RV⁺, and the rate law is known to be:   

Rate  =    q^₁ [ RV⁺ ]                where q₁  = k [ OH^- ]ⁿ and is equal to 7.0 X 10^-4

How do we determine n, the order of the reaction with respect to OH^-?

We conduct a second run with a different concentration of OH^-. Using the same stock solutions as in the first run, we prepare Initial solutions as follows:

8.0 mL of NaOH solution in 50 mL of water  
10 mL of RV⁺ solution in 50 mL of water

We mix the solutions to obtain:

Total volume of reaction solution (50 mL + 50 mL)                     = 100 mL
Concentration of NaOH  (8.0 X 0.1134 / 100)                           =  9.0 X 10^-3 M
Initial concentration of RV⁺ (10.0 X 1.0 X 10^-5 / 100)                =  1.0 X 10^-6 M

collect the absorbance as a function of elapsed time, plot the data (using the logarithm function, since we now know that the reaction is first order in RV⁺) and obtain the slope of the plot, q₂. Suppose the value of the slope in the second run is q₂ = 3.0 X 10^-3

We have recorded the variation of [ RV⁺ ] for two different, known, values of [ OH^- ], as follows:

We can write:
            q₂           k [ OH^- ]₂ⁿ                 [ OH^- ]₂ⁿ
            ----    =    ------------    =    ----------
            q₁            k [ OH^- ]₁ⁿ                [ OH^- ]₁ⁿ

or, taking logarithms of both sides of the equation,

                            log ( q₂ / q₁ )    =    log( [ OH^- ]₂ / [ OH^- ]₁ )ⁿ   =    nlog( [ OH^- ]₂ / [ OH^- ]₁ )
and, solving for n,

                           n    =    log ( q₂ / q₁ ) / log( [ OH^- ]₂ / [ OH^- ]₁ )

Substituting the values from TABLE 1 above, we get:

                           n    =    log ( 3.0 X 10^-3 / 7.0 X 10^-4 ) / log ( 9.0 X 10^-3 / 4.5 X 10^-3  )    =    log (4.3) / log (2.0) = 0.63/0.30 = 2.1. Since we seek an integer solution in this case, a reasonable solution is n = 2

I.e., for the above data, n, the order of the reaction with respect to OH^- is now known to be 2 and the rate law is of the form:

                              Rate  =     k [ RV⁺ ] [ OH^- ]²

What remains is to determine the specific rate constant, k.

Remembering that the quantities q₁ and q₂ involved both the (constant) concentration of OH^- and k, we can go back to TABLE 1 and determine k for each of the kinetic runs, namely, for the data shown in TABLE 1,

q₁    =    7.0 X 10^-4    =    k [ OH^- ]₁²    =    k X (4.5 X 10^-3)², from which we get    k    =    34 L² / mol² sec
and from the second run
q₁    =    3.0 X 10^-3    =    k [ OH^- ]₁²    =    k X (9.0 X 10^-3)², from which we get    k    =    37 L² / mol² sec

Our best estimate will be the average of these two values, or k = 35 L² / mol² sec

We have now determined all of the quantities that determine the kinetics of the reaction, namely, m, n and k

If the rate law had been found to be first order in [ OH^- ], we would have proceeded in a similar fashion to determine the specific rate from the supporting data.  Note, however, that the units of the specific rate would be different.
 

 We have now determined all of the quantities that determine the kinetics of the reaction in this case, namely, m, n and k and our rate law is

Rate = 35 [RV⁺]² [OH^-]

5B. 1 / [ RV⁺ ] is linear when plotted against time - Reaction is 2nd order in RV+ with q₁ = 2.9 X 10^-3

[Note that the quantity we actually plot is the reciprocal of the absorbance of RV⁺, not its concentration.  Unlike in the first order case, we cannot ignore this. The integrated form of the second order rate law is

                                   1 / [ RV⁺ ]    =    1 / [ RV⁺ ]₀ + k t

If the quantity we actually measure is proportional to the concentration, e.g., the absorbance, Abs = e [ RV+ ], with e = absorptivity, the rate law becomes

                                   e / [ RV⁺ ]    =    e / [ RV⁺ ]₀ + k t

or,                                 

                                   1 / [ RV⁺ ]    =    1 / [ RV⁺ ]₀ + k t / e

In this case, the slope of the straight line does depend on the proportionality constant between Abs and concentration! ]

The reaction is now known to be second order in RV⁺, and the rate law is known to be:   

Rate  =    q^₁ [ RV⁺ ]²                where q₁  = k [ OH^- ]ⁿ and is equal to 2.9 X 10^-3

We determine n, the order of the reaction with respect to OH^- just as we did in the first order case.

We conduct a second run with a different concentration of OH^-. Using the same stock solutions as in the first run, we prepare Initial solutions using 8.0 mL of NaOH solution in 50 mL of water  and 10 mL of RV⁺ solution in 50 mL of water as before, mix the solutions to obtain:

Total volume of reaction solution (50 mL + 50 mL)                     = 100 mL
Concentration of NaOH  (8.0 X 0.1134 / 100)                           =  9.0 X 10^-3 M
Initial concentration of RV⁺ (10.0 X 1.0 X 10^-5 / 100)                =  1.0 X 10^-6 M

collect the absorbance as a function of elapsed time, plot the data (using the reciprocal function, since we now know that the reaction is second order in RV⁺) and obtain the slope of the plot, q₂. Suppose the value of the slope in the second run is q₂ = 6.0 X 10^-3

We have recorded the variation of [ RV⁺ ] for two different, known, values of [ OH^- ], as follows:

In this case, we write:
            q₂           k [ OH^- ]₂ⁿ  / e            [ OH^- ]₂ⁿ
            ----    =    --------------    =    ----------          (Note that the Beer's Law slope cancels here.)
            q₁            k [ OH^- ]₁ⁿ / e            [ OH^- ]₁ⁿ

again, solving for n as above

                           n    =    log ( q₂ / q₁ ) / log( [ OH^- ]₂ / [ OH^- ]₁ )

Substituting the values from TABLE 2 above, we get:

                           n    =    log ( 6.0 X 10^-3 / 2.9 X 10^-3 ) / log ( 9.0 X 10^-3 / 4.5 X 10^-3  )    =    log (2.1) / log (2.0) = 0.32/0.30 = 1.1. Since, again, we seek an integer solution, a reasonable solution is n = 1

I.e., for the above data, n, the order of the reaction with respect to OH^- is now known to be 1 and the rate law is of the form:

                              Rate  =     k [ RV⁺ ]² [ OH^- ]

We again determine the specific rate constant, k.

In this case, the quantities q₁ and q₂ involve the (constant) concentration of OH^-,k and the Beer's Law slope, e whose value is given in Section 1 as 5.2 X 10⁴ L / mol.        

q₁    =    2.9 X 10^-3   =   k [ OH^- ]₁/ e  =   k X 4.5 X 10^-3 / 5.2 X 10⁴ which gives    k  = 3.3 X 10⁵ L² / mol² sec
and from the second run
q₁    =    6.0 X 10^-3   =   k [ OH^- ]₁/ e  =   k X 9.0 X 10^-3 / 5.2 X 10⁴ which gives    k  = 3.5 X 10⁵ L² / mol² sec

Our best estimate will be the average of these two values, or k = 3.4 X 10⁵ L² / mol² sec

We have now determined all of the quantities that determine the kinetics of the reaction in this case, namely, m, n and k and our rate law is

Rate = 3.4 X 10⁵ [RV⁺]² [OH^-]

If the rate law had been found to be first order in [ OH^- ], we would have proceeded in a similar fashion to determine the specific rate from the supporting data.  Note, however, that the units of the specific rate would be different.