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2.2 COLLISION THEORY AND RATE OF REACTION

Updated: Jan 28, 2020

IF YOU WOULD LIKE A COPY OF PAST PAPER QUESTIONS FOR THIS TOPIC OR ANY OTHER TOPIC PLEASE EMAIL ME (space is provided on the main page of my blog)

IF YOU WOULD LIKE TO GO THROUGH THE QUESTIONS WITH ME, PLEASE VISIT MY YOUTUBE CHANNEL PART 1 AND PART 2.


Hi all. Today we have an easy topic and when I say easy, this is like super easy. If you are confident with this topic, I invite you to my youtube channel for some past paper practice questions to check your knowledge. But if you are not, that's totally fine because you will in a couple of minutes.


It is all about the rate of reaction, so first of all let me tell you what rate of reaction is.

It starts with collision theory and you need to know what this is in order to understand the rate of reaction. The picture above basically explains collision theory but I will write you a short explanation.


- For the particles to react and become a product, they need to collide. But they can't just collide, they have to do this with enough energy. This is called successful collision. (If they don't have enough energy they will still collide but they won't become a product/they won't react togather.)


- Rate of reaction is the frequency of the successful collisions. It is the measure of how often the particles collide (successfully).


Because not all of the collisions will result in a product, we can try to increase the probability of the right collisions by increasing the number of collisions overall ( more collisions means more of them will have a chance of being successful).

Here's a couple of examples of how we can increase the number of collisions (rate of reaction):

1. Increase the temperature - by increasing the temperature the particles gain more kinetic energy. This means they will move faster and collide with more energy. Therefore increasing temperature increases the rate of reaction.


2. Increases the concentration - when you increase the concentration it means that we have more reactant particles in the same volume ( in the same amount of space). This means the particles are closer togather and collide more often. This increases the rate of reaction.


3. Increase the pressure - when we increase the pressure the particles are squeezed into a smaller volume. Therefore, they are again closer together and colide more often. This also increases the rate of reaction.


4. Size of particles - when you have big particles there will be just few of them, so less chance of them colliding together, compared to the same big particle being crushed (in the same volume). As I crush these big particles into smaller ones, there are much more of them ( increased surface area) and therefore there is more chance of them colliding.


5. Catalyst - last but not least, catalyst does not take part in the reaction but it increases the rate of reaction. This is because it lowers the activation level (this is the energy needed for the collisions to happen).


 

Calculating the rate of reaction:

The equation for calculating the rate of reaction is:


Rate= change in concentration/ time


It is usually calculated by taking the amount of reactants used up (the ones that successfully collide) in a given amount of time. The phrase 'amount of reactants' is the concentration. You also need to know the units of rate, concentration and time which I included in the picture above.


Let's talk about the graph. You probably realised and if not I am telling you now; The gradient of this graph is the rate of reaction. Change in y/ change in x = gradient and here we have concentration (change in y)/ time (change is x) which is equal to the rate.


The concentration against time graph is a curve. The line shows us what happens during reaction (colours relate to the picture above):


- At the beginning (brown) we have the highest concentration of reactant this means there are a lot of collisions (highest rate of reaction);


- Then (blue) as the number of reactants decrease the rate of reaction also decreases (less collisions);


- At the end (pink) there is no more reactants therefore rate of reaction is zero ( no collisions at all, reactant used up);


As always, I have examples (one easier and one slightly harder):

1. Here is our simple graph showing the rate of reaction. We are just asked to find the rate, which is gradient of this graph. The first part of the graph is a straight line (for AS-level), so you just need to find the change in y and divide by change in x.


2. This is a slightly harder example, we are asked to find the rate of reaction at 20s. So, what you need to do is go up from 20 until you touch the curve and this is where you need to draw a tangent to that point. (tangent is a straight line that goes right through the 20s point/ blue line on the graph). When you draw that line you need to find the gradient of this line (the one you just drew). This will give you the rate of raction at 20 seconds. I worked through this example in the picture above.


I want you to be be aware that you can express rate in a different way as well:

And what I mean is:

rate is proportional to 1/time


meaning rate is inversely proportional to time.

You can demonstrate this relationship on a graph and I drew two different situations for you:


1. Concentration is increasing however 1/time stays the same. This is an example of rate of reaction being independant of concentration of one of the reactants.


2. Concentration is increasing as well as 1/time. This shows us that as concentration increases rate also increases (because 1/time is proportional to rate) and rate is dependant on the concentration of this reactant.


You can check this relationship experimentally by changing the initial conentration of one of reactants at a time and see if the rate changes. And yes, you are right, I got you an example of this:

On the picture above, is a table of results where concentration of only one reactant was changed at a time and intial rate calculated every time.


In the exam you may get this type of table and be asked to see what reactant has an effect on the initial rate.


- In experiment 2 and 3 the concentration of B is increased (and the concentration of A left the same) however, this does not change the rate of reaction, so rate is independant of the concentation of reactant B.


- In experiment 1 and 2 the concentration of A s increased and concentration of B left constant and the initial rate increases. This means rate is dependant on the concentration of reactant A.


 

The last thing is, how to measure the rate of reaction:


1. Change in volume of gas - you can measure this using a gas syringe (picture of this is draw in the photo above). It is suitable for reactions where gas is produced as in the example above.


2. Change in pressure - this you measure using a Manometer. It is suitable for reactions involving gases only. If the number of moles of reactants is different to the number of moles of products there is a change in pressure. In the example above we have one mole of gases on the LHS and 2 moles of gases on the RHS.


3. Change in mass - this method can be used if a gas is produced (one of the heavier ones for example carbon dioxide) and it is relesed. This will lower the mass. You need to use an accurate balance to do this because the difference may be small.


4. Change in colour - if percipitate is formed you can use this method. You need to wait until you can't see the cross below the conical flask. (everything is drawn in the picture above).


5. Change in colour - you can also measure the rate of reaction using calorimeter. You can use this method when only one thing has a colour and the rest of them are colourless. The light passes through a sample and onto a detector where colour is recorded. An example is when iodine reacts and it has a brown colour, when its colour disappears it means there is less and less of the reactant.


PS. Please remember, I am only a student, and as anyone, I can make mistakes. If you think you can see one, don't hesitate and comment (either here on on my youtube channel) Thank you!




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