# Determination of Rate Equations & the Rate Determining Step

## Determination of Rate Equations & the Rate Determining Step Revision

**Determination of Rate Equations & the Rate Determining Step**

**Rate equations** are determined experimentally by investigating the effect of **changing concentrations** and then deducing the order.

## Using Initial Rate to Calculate Reaction Variables

Rate equations have a general form:

**\text{Rate} =k[X]^n**

To work out the value of n graphically, we need to take **logarithms** of both sides of the equation to convert it to the form y = mx + c.

**\text{log Rate}=\text{log k}+\text{n log [X]}**

A graph of this equation would produce a **straight line,** where the **\text{y-intercept}** is equal to **\text{log k}** allowing us to calculate the rate constant \text{K}.

**\text{K}=10^{\text{ y-intercept}}**

The** gradient of the line** is equal to the order of reaction, **\text{n}**, with respect to X.

**The Rate Determining Step**

The slowest step in a reaction is known as the** rate-determining step **(RDS). It is rate-determining because it controls the **overall** **rate of a reaction**.

The rate equation contains species **up to** and **including** the rate-determining step. Therefore, we can use the rate equation to determine the rate-determining step. Take the reaction

\text{A} + 3\text{B} + 2\text{C} \rarr\text{D}

A suggested mechanism for this reaction is

Step 1: \text{A} + 2\text{B}\rarr\text{X}

Step 2: \text{B} + 2\text{C}\rarr\text{Y}

Step 3: \text{X} + \text{Y}\rarr\text{D}

The rate equation for the reaction is

\text{Rate} = \text{k[A][B]}^2

The rate equation tells us that there is **1 mole of A** and **2 moles of B** up to and including the rate-determining step. Step 1 contains both 1 mole of A, 2 moles of B, Step 2 contains only 1 mole of B and none of A, and Step 3 contains no moles of either species.

Step 3 cannot be the RDS as it does not contain either of the species contained in the rate equation. step 2 contains 1 mole of B, but taking both steps 1 and 2 together gives 3 moles of B in total. As the rate equation contains only 2 moles of B step 3 can’t be the RDS. Step 1 contains the correct number of both moles according to the rate equation and so must be the rate determining step.

**Determining Reaction Orders From Experimental Dat**a.

By carrying out a **series of experiments** using **different concentrations** of reactants, we can determine the rate equation of a reaction. To do this we need to compare the effects of the different concentrations on the initial rate.

Experiment | Initial Concentration of X /\text{mol dm}^{-3} | Initial Concentration of Y /\text{mol dm}^{-3} | Initial Rate /\text{mol dm}^{-3}\text{ s}^{-1} |

1 | 0.12 | 0.26 | 2.10\times10^{-4} |

2 | 0.36 | 0.26 | 1.98\times10^{-3} |

3 | 0.72 | 0.13 | 3.78\times10^{-3} |

**[5 marks]**

We can start by **comparing the rates** of experiments 1 and 2. Although **the concentration of Y remains constant**, the **initial rate changes**. That immediately tells us that **the order of X is greater than 0** and is part of the rate equation.

From experiments 1 to 2, the **concentration of X triples** while the **initial rate increases by a factor of 9**. This means that the **order of X must be 2** (you can work out the amount the initial rate increases if you divide the rates).

\text{Rate} = \text{k[X]}^2

We can now look at experiments 2 and 3 to find the effect of Y. Firstly, we have to consider the effect of X. Between experiments 2 and 3, the **concentration of X doubles**. We know that X is second order so from experiments 2 to 3, **the initial rate would be expected to quadruple** from 1.8\times10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} to 7.56\times10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}.

Since that is not the initial rate of the reaction, **Y must have an order greater than 0**. From experiments 2 to 3, the concentration of **Y halves**, meanwhile, the **rate is half the expected value** of 7.56\times10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}, so **Y must be first order**.

\text{Rate} = \text{k[X]}^2\text{[Y]}

To calculate the value for \text{k}, the rate equation needs to be rearranged to make \text{k} the subject, and then the known values can be substituted into the equation.

For example, if we use the rate equation above, \text{Rate} = k [X]^2[Y], we can find \text{k} using the equation k=\frac{\text{Rate}}{[X]^2[Y]}.

**Required Practical**

**Initial Rate Experiments – The Iodine Clock**

The initial rate of a reaction can be calculated by carrying out a clock reaction. In clock reaction, a system is designed so that it can be monitored over a set period of time or until one product has been produced in a specific concentration.

**Method**

In this common clock reaction, **hydrogen peroxide** (\text{H}_2\text{O}_2), **iodide ions** (\text{I}^-), **thiosulfate ions** (\text{S}_2\text{O}_3^{2-}) and **starch** are mixed together. The following reactions take place:

\text{H}_2\text{O}_{\text{2(l)}} + 2\text{H}^+_{\text{(aq)}} + 2\text{I}^{-}_{\text{(aq)}} \rarr \text{I}_{\text{2 (aq)}}

2\text{S}_2\text{O}_3^{2-}\text{}_{\text{(aq)}}+\text{I}_{\text{2(aq)}}\rarr2\text{I}^-_{\text{(aq)}}+\text{S}_4\text{O}_6^{2-}\text{}_{\text{(aq)}}

- Fill a
**burette**(50\text{ cm}^3) with potassium iodide solution. Rinse the 50\text{ cm}^3 burette with**potassium iodide**beforehand to improve accuracy. - Put 10.0\text{ cm}^3 of
**hydrogen peroxide**into a clean and dry measuring cylinder. - Use a measuring cylinder to add 25\text{ cm}^3 of
**sulfuric acid**to a second 250\text{ cm}^3 beaker. Then, add 20\text{ cm}^3 of**deionised**water into the beaker. - Use a plastic
**dropping pipette**to add about 1\text{ cm}^3 of starch solution to the 250\text{ cm}^3 beaker. - Use the burette to add 5.0\text{ cm}^3 of
**potassium iodide**solution to the mixture in the 250\text{ cm}^3 beaker. - Finally add 5.0\text{ cm}^3 of
**sodium thiosulfate**solution to the mixture in the 250\text{ cm}^3 beaker. - Stir the mixture in the 250\text{ cm}^3 beaker and pour the 10.0\text{ cm}^3
**hydrogen peroxide**solution from the measuring cylinder into the 250\text{ cm}^3 beaker and**immediately**start the timer. - Stop the timer when the mixture in the 250\text{ cm}^3 beaker turns
**blue-black**and**record**the time. - Repeat steps 1-8 at least two more times, changing the
**potassium iodide****concentration**each time. This is to allow the order of reaction to be determined. - Plot a graph of initial rate versus concentration to determine the order.

## Determination of Rate Equations & the Rate Determining Step Example Questions

**Question 1: **The overall equation for a reaction is as follows.

\text{NO}_{2\text{(g)}} + \text{CO}_{\text{(g)}} \rarr \text{NO}_{\text{(g)}} + \text{CO}_{2\text{(g)}}

Below is the mechanism for the reaction.

Step 1:

Step 2:

\text{NO}_3+\text{CO}\rarr\text{NO}_2+\text{CO}_2

Step 1 is the slowest step in the reaction. Determine the rate equation for the reaction. Explain your answer.

**[2 marks]**

\text{Rate} = \text{k[NO}_2\text{]}^2.

\text{NO}_2 appears twice in the rate-determining step while \text{CO} does not appear at all.

**Question 2:** When placed in a dilute alkaline solution, ethanal undergoes the following reaction.

2\text{CH}_3\text{CHO}\rarr \text{CH}_3\text{CH(OH)CH}_2\text{CHO}

Below is the rate equation of the reaction.

\text{Rate} = \text{k}\left[\text{CH}_3\text{CHO}\right]\left[\text{OH}^\text{-}\right]

The initial rate of the reaction is found to be 2.5\times10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} when the initial concentration of ethanal is 0.15\text{ mol dm}^{-3}, and the initial concentration of sodium hydroxide is 0.030\text{ mol dm}^{-3}.

Calculate the value for the rate constant at this temperature and give its units.

**[3 marks]**

\begin{aligned}\text{k}&=\frac{2.5\times10^{-3}}{0.15\times0.03}\\ &=0.555\end{aligned}

Units:\text{ mol}^{-1}\text{dm}^{3}\text{ s}^{-1}

**Question 3: **The data in the following table was obtained by carrying out a series of experiments to find the initial rate of reaction between substances X and Y.

Experiment | Initial Concentration of X /\text{mol dm}^{-3} | Initial Concentration of Y /\text{mol dm}^{-3} | Initial Rate /\text{mol dm}^{-3}\text{ s}^{-1} |

1 | 0.14 | 0.24 | 2.60\times10^{-4} |

2 | 0.42 | 0.24 | 2.34\times10^{-3} |

3 | 0.84 | 0.48 | 9.36\times10^{-3} |

Deduce the orders of the reaction with respect to X and Y.

**[2 marks]**

Order with respect to X: 2 or \text{[X]}^2

Order with respect to Y: 0 or \text{[Y]}^0