##### Reaction Kinetics Part 3: Pseudo Order Reactions

- focuschemistry
- March 14, 2018
- Reaction Kinetics

Click here to view Reaction Kinetics Part I: The Truth behind Rate Equations II

In the determination of reaction orders and rate constant, one of the common used technique is known as the method of excess. Consider a hypothetical reaction:

2A (aq) + B(aq) → 3C (aq) …………(1)

The rate equation can be written as

Rate = k[A]^{m}[B]^{n}

Assuming that the rate of the reaction as the reaction progresses is most easily followed by measuring the volume of C (g) formed at different times of the reaction and if we conduct only 1 experiment with a certain concentration of A and B, and follow the volume of C over time, we get a typical product graph with the highest rate at the beginning and rate=0 when the reaction ends.

With only 1 graph, are we able to find the orders of A and B?

Of course not!

With only 1 graph, we can only find the order of one of the reactant.

Which reactant? A or B?

There is no easy answer to this. Unless the concentrations of A and B differ by at least 20 times, with only 1 graph, neither order can be found.

However if one of the reactant’s concentration is more than 20 times the other with 1 graph, the one that has a much LOWER concentration can be found.

An example of such a question is **RJC 2002 P3**

Two substances, A and B, react together to give a new compound, A2B2, according to the following equation : 2A(g) + 2B(g) → A2B2 (aq).

To determine the rate law for the above equation, two experiments with the same initial concentration of A (aq) but different initial concentrations of B(aq) are performed.

In both experiments, the concentration of A2B2 (aq) formed is found at 5 minutes intervals.

The results of the two experiments are shown below:

**Experiment 1:**

0.120 moldm^{-3} of A(aq) is reacted with 1.2 moldm^{-3} of B(aq)

Time/min | 0 | 5 | 10 | 15 | 20 | 25 |

[A2B2(aq)]/moldm^{-3} | 0.0000 | 0.0211 | 0.0348 | 0.0436 | 0.0494 | 0.0531 |

**Experiment 2:**

0.120 moldm^{-3} of A(aq) is reacted with 1.8 moldm^{-3} of B(aq)

Time/min | 0 | 5 | 10 | 15 | 20 | 25 |

[A2B2(aq)]/moldm^{-3} | 0.0000 | 0.0374 | 0.0515 | 0.0568 | 0.0588 | 0.0595 |

- Why is B used in large excess?
- Plot the graph of [A2B2(aq)] against time for both experiments using the same axes.
- From the graph plotted,
- Determine the orders of reaction with respect to A(aq) and B(aq)
- i Determine the rate constant, k, stating its units.

**Discussion Points:**

1. Notice that A and B react in a 1:1 ratio, i.e

2A (aq) + B(aq) → [A2B2(aq)]

In experiment 1, A and B are a 1:10 ratio, and in experiment 2, A and B are in a 1: 15.

Clearly, B is not just in excess, but many more times in excess of A (in large excess).

Is there a purpose for this?

Yes, it’s a technique to find the orders of the reactants when there are 2 or more reactants. One of them is limiting and the rest made in large excess.

What happens when we do that?

Remember that we conduct kinetics experiments, our main objectives are to find

a. Orders of reaction

b. Rate constant

**Since rate constant depends on temperature, we would have to keep the temperature constant throughout the entire duration of the experiment.** That is, if it is endothermic, put it in a water bath with external heating source (like a heating coil). If it is exothermic, we need to pass cooling liquid around the vessel in which the reaction is taking place.

**If the temperature of the reaction is not controlled properly and allowed to fluctuate, the value of rate constant (k) will also fluctuate and we end up finding an average rate constant, and not THE rate constant.**

This is **very important** when we are planning for kinetics experiments!

Since the rate constant stays constant with time as the reaction progresses (through the maintenance of the reaction temperature), what happens one of the reactant is purposely made in **LARGE EXCESS?**

Its concentration stays virtually constant with time!

Thus, since the rate constant, k and the concentration of the reactant in large excess (which is reactant B in the RJC question) stays constant with time, they can be expressed in terms of a new constant, k’ (known as the apparent rate constant), where

k’ = k[B]^{n}

in the RJC question above.

Thus, the original (or theoretical rate equation) of the reaction between A and B was

Rate = k[A]^{m}[B]^{n}……….(1)

Has now apparently become

Rate’ = k’ [A]^{m}; where k’= k[B]^{n}……….(2)

Equation (2) which is the **apparent rate equation,** is also known as the **pseudo rate equation**(since pseudo means : “appears to be but is not”)

2. Since there are 2 experiments conducted in this question, we will get 2 “concentration vs time” graph, and with that we are able to find the orders with respect to A and B.

Since the pseudo rate equation is in terms of the limiting reactant (which is A in the RJC question), finding the half life of the reaction using either experiment 1 or 2 allows us to find the order of A.

How do we find the half life on a product curve? How to use half life to predict the order of A? How do we find the order of B?

Find out more by signing up for our A Levels H2 Chemistry tuition classes NOW!