In the previous post, we talked about the **integrated rate law **and its use for determining the concentration of a reactant at a given time when the reaction order was provided.

Now, there are questions where the **reaction order needs to be determined** using the data on how the concentration changes with time.

**For example,**

Given the following kinetics data, determine **the order **of the reaction, the **rate constant**, and predict the **concentration of A at 450. s**.

A → B + C

To solve these types of problems, you first need to remember which plot of concentration vs time gives a straight line for the zero-, first-, and second-order reactions:

So, the strategy is to plot the data given in the problem and see whether the [A], ln[A], or 1/[A] gives a straight line:

A straight line is obtained when 1/[A] is plotted vs time, indicating a second-order reaction. I used the Origin program to obtain the graphs above, however, this can be done in excel:

\[\frac{{\rm{1}}}{{{{\left[ {\rm{A}} \right]}_t}}}\; = \;kt\; + \;\frac{{\rm{1}}}{{{{\left[ {\rm{A}} \right]}_0}}}\]

y = mx + b

The **rate constant** is the slope of the graph, which can also be obtained **in excel**. For this, *create the plot 1/[A] vs time, and once it is ready, double click on the graph line > choose “linear” under the format templates on the right > check the box for “Display equation on chart”. *

According to excel, *k* = 0.0038, and because it is a second-order reaction, the units of k are *M*^{-1} ⋅ *s*^{-1}.

Alternatively, the rate constant can be calculated by Δy/Δx (rise/run). So, let’s see how the values of k compare when calculated using these two methods. We can pick any two y and two x values from the graph. Let’s do this for the time interval 200-600 s, since it is relatively easy to see the y values for these points. When x = 200, y = 1.1, when x = 600, y = 2.6. The rate constant then is:

k = (2.6-1.1)/(600-200) = 0.00375, which is very close to the value in excel.

So far, we have answered two questions: a) The reaction is **second order**, b) *k* = 0.0038 M^{-1} **⋅**** s ^{-1}**

Now that we know the value of rate constant, we can calculate the **concentration of A at 450. s**., by using integrated rate law for second-order reactions:

\[\frac{{\rm{1}}}{{{{\left[ {\rm{A}} \right]}_t}}}\; = \;kt\; + \;\frac{{\rm{1}}}{{{{\left[ {\rm{A}} \right]}_0}}}\]

\[\frac{{\rm{1}}}{{{{\left[ {\rm{A}} \right]}_t}}}\; = \;{\rm{0}}{\rm{.0038}}\;{{\rm{M}}^{{\rm{ – 1}}}}{{\rm{s}}^{{\rm{ – 1}}}}\;{\rm{ \times }}\;{\rm{450}}{\rm{.}}\;{\rm{s}}\;{\rm{ + }}\;\frac{{\rm{1}}}{{{\rm{2}}{\rm{.85}}\;{\rm{M}}}}\]

**[A] _{t} = 0.485 M**

This number makes sense, because according to the table given in the problem, the concentration of A at 400 s is 0.54 *M*, and at 500 s, it is 0.447 *M*. So, at 450 s, the concentration must be between 0.54 *M* and 0.447 *M*.

**Here is a 77-question, Multiple-Choice Quiz on Chemical Kinetics:**

**Check Also**

- Reaction Rate
- Rate Law and Reaction Order
- How to Determine the Reaction Order
- Integrated Rate Law
- The Half-Life of a Reaction
- Units of Rate Constant k
- How Are Integrated Rate Laws Obtained
- Activation Energy
- The Arrhenius Equation
**Chemical Kinetics Practice Problems**