12.3 Rate laws  (Page 3/10)

Determining rate laws from initial rates

Solution

As in [link] , we can approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k . In this example, however, we will use a different approach to determine the values of m and n :

1. Determine the value of m from the data in which [NO] varies and [Cl ₂ ] is constant. We can write the ratios with the subscripts x and y to indicate data from two different trials:
   

   $\frac{{\text{rate}}_x}{{\text{rate}}_y}=\frac{k{\text{[}\text{NO}\text{]}}_x^m{\text{[}{\text{Cl}}_2\text{]}}_x^n}{k{\text{[}\text{NO}\text{]}}_y^m{\text{[}{\text{Cl}}_2\text{]}}_y^n}$

   Using the third trial and the first trial, in which [Cl ₂ ] does not vary, gives:
   

   $\frac{\text{rate 3}}{\text{rate 1}}=\frac{0.00675}{0.00300}=\frac{k(0.15{)}^m(0.10{)}^n}{k{\text{(0.10)}}^m(0.10{)}^n}$

   After canceling equivalent terms in the numerator and denominator, we are left with:
   

   $\frac{0.00675}{0.00300}=\frac{{(0.15)}^m}{{(0.10)}^m}$

   which simplifies to:
   

   $2.25={\left(1.5\right)}^m$

   We can use natural logs to determine the value of the exponent m :
   

   $\begin{array}{ccc}\hfill \text{ln}\left(2.25\right)& =& m\text{ln}\left(1.5\right)\hfill \\ \hfill \frac{\text{ln}\left(2.25\right)}{\text{ln}\left(1.5\right)}& =& m\hfill \\ \hfill 2& =& m\hfill \end{array}$

   We can confirm the result easily, since:
   

   ${1.5}^2=2.25$

2. Determine the value of n from data in which [Cl ₂ ] varies and [NO]is constant.
   

   $\frac{\text{rate 2}}{\text{rate 1}}=\frac{0.00450}{0.00300}=\frac{k(0.10{)}^m(0.15{)}^n}{k(0.10{)}^m(0.10{)}^n}$

   Cancelation gives:
   

   $\frac{0.0045}{0.0030}=\frac{{\left(0.15\right)}^n}{{\left(0.10\right)}^n}$

   which simplifies to:

   $1.5={(1.5)}^n$

   Thus n must be 1, and the form of the rate law is:
   

   $\text{Rate}=k{\left[\text{NO}\right]}^m{\left[{\text{Cl}}_2\right]}^n=k{\left[\text{NO}\right]}^2\left[{\text{Cl}}_2\right]$

3. Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol ³ /L ³ . The units for k should be mol ⁻² L ² /s so that the rate is in terms of mol/L/s.

   To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k :
   

   $\begin{array}{ccc}\hfill 0.00300\text{mol}{\text{L}}^{-1}{\text{s}}^{\mathrm{-1}}& =& k{\left(0.10\text{mol}{\text{L}}^{\mathrm{-1}}\right)}^2{\left(0.10\text{mol}{\text{L}}^{\mathrm{-1}}\right)}^1\hfill \\ \hfill k& =& 3.0{\text{mol}}^{\mathrm{-2}}{\text{L}}^2{\text{s}}^{\mathrm{-1}}\hfill \end{array}$

Check your learning

Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.

Reaction order and rate constant units

In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case.