Rate equations and the rate constant

In a nutshell

Rate equations show how the rate is impacted by the concentration of each reactant. The initial rates method is a useful way of calculating the orders of the reaction, for each compound. Once the rate and orders of the reaction are known, the rate constant can be calculated.

Rate equation

Rate equations tell us how the concentration of the reactants affects the rate of reaction. For the reaction, $A\ +\ B\ \rightarrow\ C\ +\ D$ , the rate equation would be:

$rate\ =\ k[A]^m[B]^n$

Note: The units of the rate constant are typically $mol\ dm^{-3}\, s^{-1}$ .

$k$ is the rate constant. The larger the rate constant, the faster the reaction.

$m$ and $n$ are the orders of reaction, with respect to reactants $A$ and $B$ . This shows how the concentration of A and B affects the rate.

The overall order of the reaction is $m\ +\ n$ .

Example

The chemical equation below is the equation for the formation of pure nitrogen gas which is used to make fertilisers.

$2NO\ +\ 2H_2\ \rightarrow\ N_2\ +\ 2H_2O$

The rate equation is:

$rate\ =\ k[NO]^2[H_2]$

If a compound has a $0^{th}$ order it is not included in the rate equation (anything to the power of zero is equal to one).

Even though catalysts are not used up in the equations, they can still be part of rate equations, and often are, as they help speed up the rate of reaction. Spectator ions are not usually included in the rate equations. Additionally, even though two molecules of hydrogen are involved in the overall reaction, only one molecule is involved in the rate-determining step. Hence, why hydrogen has a reaction order of one, not two.

Initial rates

The initial rate method can be used to calculate the reaction order of a reactant. By comparing the initial rate of reaction with varying concentrations of the reactants, the order of reaction can be calculated for reactants. Once the equation and orders of a reaction are known, the rate equation can be written.

Example

Bromate ions are reacted with bromide and acid to form bromine and water.

The table shows the results of an experiment where the initial rate was calculated for the reaction:

$BrO_3^-\ +\ 5Br^-\ + \ 6H^+\ \rightarrow\ 3Br_2\ +\ 3H_2O$

experiment	$Concentration\ (mol\ dm^{-3})$			$Initial\ rate\ (mol\ dm^{-3}\ s^{-1})$
	$[BrO_3^-]$	$[Br^-]$	$[H^+]$
$1$	$0.10$	$0.10$	$0.10$	$8.00 \times 10^{-4}$
$2$	$0.20$	$0.10$	$0.10$	$1.60 \times 10^{-3}$
$3$	$0.20$	$0.15$	$0.10$	$2.40 \times 10^{-3}$
$4$	$0.10$	$0.10$	$0.25$	$5.00 \times 10^{-3}$

Comparing experiments $1$ and $2$ , as the concentration of $[BrO_3^-]$ is doubled, the rate also doubles. Therefore the reaction is $1^{st}$ order with respect to $BrO_3^-$ .

Comparing experiments $2$ and $3$ the concentration of $[Br^-]$ is increased by $\times 1.5$ and the initial rate increases by the same amount. Therefore, the reaction is $1^{st}$ order with respect to $Br^-$ .

Comparing experiments $1$ and $4$ the concentration of $[H^+]$ is increased by $\times 2.5$ and the initial rate increases by $\times5$ . Therefore the reaction is $2^{nd}$ order with respect to $H^+$ .

Now that the order with respect to each reactant is known, the rate equation can be written:

$rate\ =\ k[H^+]^2[BrO_3^-][Br^-]$

Calculating the rate constant

Once the orders of reaction and the rates are known, the rate constant, $k$ can be calculated. The rate constant changes with temperature. If the temperature is increased, the rate constant will also increase. Therefore, reactions which occur at a higher temperature, happen faster. The units for the rate constant vary, so they will also have to be worked out.

Example

Using the data above for the formation of bromine, the rate constant and its units can be calculated.

$BrO_3^-\ +\ 5Br^-\ + \ 6H^+\ \rightarrow\ 3Br_2\ +\ 3H_2O$

$rate\ =\ k[H^+]^2[BrO_3^-][Br^-]$

Now the equation can be re-arranged and the concentrations and rate can be inserted to calculate the rate constant and its units.

$rate\ =\ k[H^+]^2[BrO_3^-][Br^-]$

	$Concentration\ (mol\ dm^{-3})$			$Initial\ rate \ (mol\ dm^{-3}s^{-1})$
Compound	$[BrO_3^-]$	$Br^-$	$[H^+]$
Value used	$0.10$	$0.10$	$0.10$	$8.00\ \times\ 10^{-4}$

$k\ =\ \frac{rate}{[H^+]^2[BrO_3^-][Br^-]}\ =\ \frac{8.00\ \times 10^{-4}}{0.10^2\ \times\ 0.10\times\ 0.10}\ =\ 8$

Find the units for $k$ using the units for the rate and concentration:

$\frac{mol\ dm^{-3}s^{-1}}{(mol\ dm^{-3})^2\ \times\ mol\ dm^{-3}\ \times\ mol\ dm^{-3}}\ \newline = \frac{\sout{mol\ dm^{-3}}s^{-1}}{mol\ dm^{-3}\ \times\ mol\ dm^{-3}\ \times\ mol\ dm^{-3}}\$

$=dm^9mol^{-3}s^{-1}$ 

The final answer is: $k\ =\ \underline{8\ dm^9mol^{-3}s^{-1}}$