# What is the rate constant?

The rate **constant** is a proportionality constant of the Rate Law of chemical kinetics, which establishes a relationship between the molar concentration of the reactants and the rate of the reaction.

As the reagents are consumed, the reaction rates also decrease. That is why in the calculations of the velocity constant, represented by the symbol ‘k’, the initial velocities are taken into account, the starting velocities, to have a point of comparison between several experiments.

Chemical kinetics indicates that this constant depends on the reaction orders of the reactants, whose concentrations will in turn alter the speed of the reaction. On the other hand, according to the Arrhenius equation, k also depends on the temperature and the activation energy for the reaction.

That is why it is said that the rate constant is not a true constant, since it varies with temperature and also with the presence of catalysts that modify the activation energy. Therefore, there are not many tables where they are tabulated for specific reactions (and conditions).

__How to calculate the speed constant?__

__How to calculate the speed constant?__

**Experimentation**

The determination of the value of the rate constant of a reaction is done experimentally. If, for example, you want to determine the value of the rate constant for a reaction of the form:

A + B → C

To establish the value of k, three reactions can be carried out, with the following differences: in reaction 2 the initial concentration of reagent A, [A] is kept constant, but the concentration of reagent B is doubled, for example, [B].

Meanwhile, in reaction 3, [A] is doubled compared to reaction 1, keeping [B] constant.

Thus, a table of concentrations will be built that will serve to compare the initial speeds between several experiments. This in order to determine the reaction orders, and subsequently to be able to calculate the rate constant.

**Reaction orders**

**First calculation**

For the previous reaction, its speed is expressed by the equation:

Reaction rate = k · [A] ^{x} · [B] ^{y}

Where ‘x’ and ‘y’ correspond to the reaction orders for reactants A and B, respectively.

In order to determine the values of the reaction orders, a comparison between reaction 2 and 1, as well as between reaction 3 and 1, must be carried out, that is.

So: the ratio of [B] between reactions 2 and 1 is 2, as well as the ratio between the rates of the reactions. Then, the order value of the reaction with respect to B (exponent y) can be obtained, using the following approach:

The speed of experiment 2 is twice as fast as the speed of experiment 1. Thus, dividing the two equations above we will have:

2 = 2 ^{and}

Therefore, ‘y’ is equal to 1, and it can be concluded that the order of the reaction with respect to reactant B is first order.

**Second calculation**

In the same way, we proceed with reagent A to calculate ‘x’:

The speed of experiment or reaction 3 is four times that of experiment or reaction 1. Therefore:

4 = 2 ^{x}

OR

2 ^{2} = 2 ^{x}

And the value of x is 2, so the reaction is second order in relation to reactant A.

**Final clearance**

Then, the value of the rate constant can be established based on any of the reactions:

Reaction rate = k · [A] ^{2} · [B]

Suffice it to introduce the experimental values and solve for k:

k = Reaction speed / ([A] ^{2} · [B])

__Examples__

__Examples__

The following examples serve as an introduction before the exercises solved in the next section.

**Example 1**

In a chemical reaction:

a) If the initial concentration of a reagent is doubled, the reaction rate is doubled: What is the order of the reaction with respect to this reagent?

b) If increasing the concentration of the reactant causes an increase in speed by a factor of 8, what would be the order of the reaction then?

c) If the concentration of the reactant undergoes a change and the speed remains the same: What is the order of the reaction with respect to the reactant?

In a) The ratio of the reaction rates is equal to 2 and the ratio of the reactant concentrations is equal to 2 ^{x} ,

2 = 2 ^{x}

Therefore, x = 1 and the order of the reaction with respect to the reactant is 1.

In b) the ratio of the reaction rates is equal to 8 (2 ^{3} ) and the ratio of the reactant concentrations is equal to 2 ^{x} ,

2 ^{3} = 2 ^{x}

Therefore, x = 3 and the order of the reaction with respect to the reactant is 3-

While in c) the reaction rate is independent of the concentration of the reagent, so the order of the reaction with respect to the reagent is zero.

**Example 2**

The reaction:

2 NO _{2} + O _{2} → 2 NO _{2}

It has the expression of the following speed law:

Reaction rate = k · [NO _{2} ] ^{2} · O _{2} .

What is the global order of the reaction? What units would k have?

The order of the reaction with respect to O _{2} is 1, and the order of the reaction with respect to NO _{2} is 2, so the global order of the reaction is the sum of these values, that is, 2 + 1 = 3.

The speed of the reaction has units M / s, and the concentrations of the reactants are expressed in M. Doing an analysis of the units we will have:

k = Rate of reaction / ([NO _{2} ] ^{2} · [O _{2} ])

= (M / s) / (M ^{3} )

= s ^{-1} M ^{-2}

__Solved exercises__

__Solved exercises__

**Exercise 1**

The data for the following reaction were obtained at 25 ° C:

A + 2 B → C + 2 D

What is the rate law expression for this reaction? What is the value of its rate constant?

**Reaction Order Calculations**

The experimental data are:

Comparison of experiments 1 and 3 allows the following conclusions to be drawn:

Ratio between the rates of reactions 3 and 1:

3 x 10 ^{-4} M min ^{-1} / 3 x 10 ^{-4} M min ^{-1} = 1

But the ratio between the concentrations of B is:

(0.3 mol / L) / (0.1 mol / L) = 3

The speed of the reaction is independent of the concentration of B, since its increase does not affect it. Therefore, the order of the reaction with respect to reactant B is zero.

On the other hand, the comparison of reactions 4 and 1 allows the following conclusions to be drawn:

6 x 10 ^{-4} M min ^{-1} / 3 x 10 ^{-4} M min ^{-1} = 2

Meanwhile, the proportion between [A] of reactions 4 and 1 is equal to:

(0.20 mol / L) / (0.10 mol / L) = 2 ^{x}

So the relationship between the proportions will be:

2 = 2 ^{x}

x = 1

Therefore, the expression for the speed law is:

Reaction rate = k [A]

B is not taken into account because its order of the reaction is zero.

**Calculation of k**

We can calculate the rate constant from any of the kinetic data. Let’s use experiment 1:

k = Reaction speed / [A]

= 3 x 10 ^{-4} M min ^{-1} / 0.10 M

= 3 x 10 ^{-3} or 0.003 min ^{-1}

**Exercise 2**

Data on the reaction rate were obtained at a certain temperature.

2 ClO _{2} (aq) + 2 OH ^{–} (aq) → ClO _{3 }^{–} (aq) + ClO _{2} (aq) + H _{2} O (l)

**Reaction Order Calculations**

The experimental data are:

The ratio between the rates of reaction 2 and reaction 1 is:

4.14 x 10 ^{-4} M min ^{-1} / 2.07 x 10 ^{-4} M min ^{-1} = 2

And the ratio of the OH concentrations ^{–} between reactions 2 and 1 is equal to:

(0.024 mol / L) / (0.012 mol / L) = 2 ^{x}

The exponent ‘x’ represents the order of the reaction with respect to the reagent OH ^{–} . The relationship between the calculated proportions can be represented as follows:

2 = 2 ^{x}

x = 1

Therefore, the order of the reaction with respect to the reagent OH ^{–} is equal to 1.

In reaction 3, the ClO _{2} concentration of reaction 1 is doubled , while the OH ^{–} concentration remains constant.

Ratio between reaction rates 3 and 1:

(8.28 x 10 ^{-4} M min ^{-1} ) / (2.07 x 10 ^{-4} M min ^{-1} ) = 4 (2 ^{2} )

And the proportion of [ClO _{2} ] between reactions 3 and 1 is equal to:

(0.024 mol / L) / (0.012 mol / L) = 2 ^{and}

The exponent ‘y’ represents the order of the reaction with respect to the reagent ClO _{2} .

Comparing the proportions:

2 ^{2} = 2 ^{and}

4 = 2 ^{and}

y = 2

Therefore, the reaction is second to the reagent ClO _{2} .

The expression of the speed law will then be:

Reaction rate = k · [ClO _{2} ] ^{2} · [OH ^{–} ]

**Calculation of k**

Again, we can calculate k from the above expression using any of the kinetic data for the experiments. We will use experiment 3:

k = (8.28 x 10 ^{-4} M min ^{-1} ) / (0.024 M) ^{2} (0.012 M)

= 119.79 min ^{-1} · M ^{-2}