Hello, I am Mrs. Adcock, and welcome to today's lesson on temperature and rate: practical.

We are going to be investigating the effect of temperature on the rate of reaction.

Today's lesson outcome is, I can explain using the collision theory, how warming or cooling a reactant will affect the rate of reaction.

Some of the keywords we will be using in today's lesson include collision theory, activation energy, rate of reaction, and endpoint method.

Here you can see each of those keywords written in a sentence.

It'd be a good idea to pause the video now and read over those sentences.

You might even like to make some notes so that you can refer back to them later in the lesson if needed.

Today's lesson on temperature and rate: practical is split into two main parts.

First of all, we are going to be looking at the theory that links temperature and rate of reaction.

And then we are going to move on to look at the variables and method that we could use to investigate the effect of temperature on the rate of reaction.

Let's get started on the first part of our lesson, looking at the theory between temperature and the rate of reaction.

For a chemical reaction to take place, particles in the reactants must collide with each other.

If the particles collide with too little energy, then a reaction will not occur and no new products will be formed.

Here we can see in this animation the particles collide, but they do not have sufficient energy to react, and so they do not form a new product.

So the reactant particles are colliding, but they are colliding with insufficient energy, so no reaction occurs.

Our reactants are AB and they react with C, but no reaction takes place, and therefore we do not form new products.

For particles to have successful collisions, and this is where they react, they must collide with sufficient energy, and this is known as the collision theory.

The minimum energy that the particles must have in order to react is known as the activation energy.

Here we can see in this animation the particles have sufficient activation energy when they collide, and therefore they are able to react and they form a new product.

Here we can see the reactants are AB and they react with C, and a reaction does take place, because when the particles collide, they have sufficient energy, and we form the new products A and BC.

The rate of a reaction, and that is the speed with which a chemical reaction takes place, depends on the frequency of successful collisions between reactant particles.

The higher the frequency of successful collisions, the higher the rate of reaction.

There are factors that can vary the rate of reaction.

Factors that increase the frequency of successful collisions and therefore increase the rate of reaction are, increasing the concentration of reactants in solution, increasing the pressure of reacting gases, increasing the surface area of solid reactants, and increasing temperature.

These are all factors that can increase the rate of reaction.

Time for a question, which of the following will increase the rate of reaction? A, using a solid reactant as granules rather than a powder? B, using a reacting solution of a lower concentration? C, decreasing the pressure of reacting gases? Or D, increasing the temperature of a reacting solution? Choose any options that you think will increase the rate of reaction.

The correct answer is D, increasing the temperature of a reacting solution will increase the rate of reaction.

The answer is not A, because using a solid reactant as granules rather than a powder will decrease the surface area and therefore decrease the rate of reaction.

B will also decrease the rate of reaction if we use a lower concentration.

And C will also result in a decrease in the rate of reaction if we decrease the pressure of reacting gases.

Well done if you chose answer D and got that question correct.

Why does increasing the temperature increase the rate of reaction? Well, when we increase the temperature, the particles move faster, and not only do they move faster, but they move with more energy.

Here we can see in this animation we have got particles at a low temperature, and we can compare these to the particles here that are at a higher temperature, and hopefully you can notice the particles at the higher temperature are moving faster and with more energy.

Are particles at a higher temperature more or less likely to successfully collide? Why? Particles as a higher temperature are more likely to successfully collide.

And hopefully you've been able to link this back to the collision theory that we saw earlier.

And the collision theory states that in order for a reaction to take place, the particles must collide, and they must collide with sufficient energy.

Let's have a look in a bit more detail.

At a higher temperature, the particles are moving faster.

And faster moving particles collide more frequently, so this increase in the frequency of collisions will increase the rate of reaction.

There will also be more particles with energy equal to or greater than the activation energy.

So this also increases the likelihood of successful collisions.

Again, we can see the particles here at a low temperature, and when we heat them up, the particles move faster and they move with more energy.

So when we heat up our reaction mixture, we increase the likelihood of successful collisions.

Increasing the temperature increases the rate of reaction as there are more frequent successful collisions.

What about when we decrease the temperature? Well, when we cool a reactant, this means the particles move slower and less particles will have energy equal to or greater than the activation energy.

Here we can see the particles at a high temperature and then when we cool this reaction mixture down, then the particles move slower and the particles have less energy.

So we can see that here where we can see the particles at a low temperature.

Decreasing the temperature means the particles move slower and the particles have less energy, therefore this decreases the rate of reaction, as there will be less frequent successful collisions.

Time for check for understanding.

Is this statement true or false? If particles collide with energy lower than the activation energy, they will not react.

That statement is true.

Well done if you chose true and got that correct.

Now, can we all have a go at explaining why that statement is true? This statement is true.

If the particles collide with energy lower than the activation energy, they will not react.

We know this because the activation energy is the minimum energy that particles must have when they collide in order to react.

Time for our first practise task of today's lesson.

Aisha is investigating the effect of temperature on the rate of this reaction.

In the reaction she has hydrogen peroxide and that is breaking down to form water and oxygen.

And below the word equation there, we can see we have a balanced symbol equation for this reaction, and we have the state symbols included too.

Aisha predicts that decreasing the temperature of the hydrogen peroxide will increase the rate of reaction.

Is Aisha correct? You need to answer this by explaining in terms of particles and collisions.

Pause the video now, add as much detail as possible to your answer, and then come back when you're ready to go over the answer.

Let's see how you got on.

Aisha predict that decreasing the temperature of the hydrogen peroxide will increase the rate of reaction.

So is Aisha correct? And we needed to explain in terms of particles and collisions.

Aisha is incorrect, because decreasing the temperature will decrease the rate of reaction.

So hopefully we got that first part correct.

Then we need to explain why.

So this is because the particles move slower at a lower temperature and so they collide less frequently.

Also, fewer particles have the activation energy, so the frequency of successful collisions decreases.

Well done if you got that question correct, and hopefully you included lots of those details in your answer.

We have had a look at the theory that links temperature and rate of reaction.

Now we are going to move on to look at an investigation we could use to investigate how changing the temperature affects the rate of reaction.

And we are gonna focus now on the variables and method.

Andeep is considering whether an increase in temperature will affect the volume of product made.

Here's what Andeep says.

"I predict that increasing the temperature will increase the volume of product made." Do you think that Andeep is correct? Why? Andeep is not correct, because increasing the temperature will increase the rate of reaction, but it does not have an impact on the volume of product that we make.

Experimental data shows the effect of changing the temperature on the rate of reaction, and we can see here on the Y axis we've got the volume of gas produced in centimetre cubed, and the X axis we've got the time in seconds.

We've got two sets of data shown here on the graph.

We have a set of data for a lower temperature that was used for a reaction, and the same reaction repeated at a higher temperature.

When the temperature is higher, the gradient is steeper as the rate of reaction increases.

So increasing the temperature increases the rate of reaction.

We therefore have a steeper gradient on our graph.

Hopefully you can notice that for the higher temperature and the lower temperature, the final volume of gas produced in centimetre cubed is the same.

And that's because the temperature does not affect the overall volume of gas that is produced.

So temperature affects the rate of reaction, it does not affect the overall amount of product that is produced.

Andeep investigates the effect of changing the temperature of the sodium thiosulfate on the rate of this reaction.

You can see we have the sodium thiosulfate there and Andeep is changing the temperature of the sodium thiosulfate.

The sodium thiosulfate reacts with hydrochloric acid, and these react together to produce sodium chloride, and water, and sulphur dioxide, and sulphur.

Underneath that word equation, you can see we have a balanced symbol equation for this reaction, and we have got the state symbols which are included.

So the sodium thiosulfate is aqueous, and this reacts with aqueous hydrochloric acid, and this produces aqueous sodium chloride, and liquid water, and gaseous sulphur dioxide, and sulphur, which is a solid.

The sulphur that is produced is a yellow precipitate that makes the solution cloudy.

Now remember, a precipitate is an insoluble solid that is formed when we react aqueous solutions together.

And the sulphur is a coloured precipitate, it is yellow, and this makes this colourless solution cloudy.

Andeep plans on using an endpoint method where he will measure the time taken for the X to no longer be visible.

An endpoint method is where we measure the progress of a reaction until completion or until a desirable point has been reached.

So in this case, Andeep is measuring until the X is no longer visible.

Here we can see the reaction at the start and we can see the X on a piece of card placed under the conical flask.

And because the solutions are colourless, we can see that the X is visible underneath the conical flask.

Andeep is going to time until the X is no longer visible.

So we can see here that the reaction has reached the endpoint and that the solution has turned cloudy and the X is no longer visible underneath the conical flask.

Andeep will repeat the experiment using different temperatures of sodium thiosulfate solution to help him investigate the effect of changing the temperature on the rate of reaction.

Here's a question for us to have a go at.

Andeep investigates how temperature affects the rate of reaction.

What is the independent variable for this reaction? Is it A, the temperature of the reacting solution? B, the time taken for the reaction to reach completion? Or C, the concentration of the reacting solution? The correct answer is A.

So well done if you chose A.

The independent variable for and Andeep's investigation is the temperature of the reacting solution.

The independent variable is the variable that we are changing in an investigation.

Andeep is investigating how temperature affects the rate of reaction.

What is the dependent variable for this reaction? Is it A, the concentration of the reacting solution? B, the time taken for an endpoint to be reached? Or C, the volume of the reacting solution? The correct answer is B, the time taken for an endpoint to be reached.

So well done if you chose answer B.

The independent variable is the temperature, that's the thing that we are changing.

The dependent variable is the variable that we are measuring.

So in this case, Andeep is measuring the time taken for the endpoint to be reached.

We have a final check for understanding here before we move on to our next task.

So Andeep investigates how temperature affects the rate of reaction.

Which of the following are control variables for this investigation? A, the concentration of the reacting solution? B, the mass of solid produced? Or C, the volume of the reacting solution? The correct answers are A, the concentration of the reacting solution, and C, the volume of the reacting solution.

So well done if you chose both of those and identified them both as control variables.

Control variables are variables that we keep the same during an investigation.

So it's important that Andeep keeps the concentration and the volume of the reacting solutions the same, because it is the temperature that is the variable that he's changing.

Time for our final practise task of today's lesson, and this task is split into two parts.

So first of all, you need to follow the method to collect results on the effect of temperature on the rate of reaction.

Now let's just have a look at what we are going to be doing for that method.

So first of all, you're going to measure 25 centimetres cubed of room temperature sodium thiosulfate and place it in a conical flask on top of a card marked with an X.

Record the temperature of the sodium thiosulfate.

Add five centimetres cubed of hydrochloric acid to the conical flask and immediately start the timer.

Then step D, look directly through the top of the flask to observe the reaction.

Being careful here that you do not inhale directly from above the top of the flask.

Step E, record the time taken for the X to no longer be visible.

And then for step F, we are going to repeat the experiment using sodium thiosulfate cooled in an ice bath, so that will be sodium thiosulfate at a lower temperature.

And then we're going to repeat again using sodium thiosulfate heated in a water bath.

So this will give us sodium thiosulfate at a higher temperature.

You should then collect three sets of data where you've recorded the time taken for the X to no longer be visible using sodium thiosulfate at a lower temperature, at room temperature, and at a higher temperature.

Pause the video now.

Good luck completing the investigation and come back when you have finished.

If you were unable to complete the investigation or you did not manage to gather a full set of data, then there are some sample results for you here.

We can see the temperature of sodium thiosulfate and we have three different temperatures there in degrees Celsius, at eight degrees Celsius, 20 degrees Celsius, and 32 degrees Celsius.

And we can see the time taken for the X to no longer be visible in seconds at each of those different temperatures.

For the final part of this practise task, we need to look at those results.

Andeep measured how long it took for the X to no longer be visible at three different temperatures, and we can see those there recorded in the results table.

Write a conclusion for Andeep's results.

Make sure to use the collision theory to explain why we see this trend.

Pause the video now, have a go at answering this question, and when you come back we will go over the answer.

Let's see how you got on.

As the temperature increases, the time taken for the reaction to go cloudy and for the X to no longer be visible decreases.

The particles are moving faster at a higher temperature, so they collide more frequently.

More particles have energy equal to or higher than the activation energy, and therefore at the higher temperature, the frequency of successful collisions increases and therefore the rate of reaction increases.

Hopefully you have included the key trend there, that at a higher temperature, the time taken for the X to no longer be visible decreases and therefore the rate of reaction has increased, and you've used the collision theory to explain why we see that trend.

We have reached the end of today's lesson on temperature and rate: practical.

Let's just summarise a couple of key points that we have covered in today's lesson.

Increasing the temperature generally increases the rate of reaction by boosting the energy and the frequency of collisions between particles.

Cooling a reactant decreases its rate of reaction, because it reduces the collision energy and reduces the frequency of those successful collisions.

Well done for all your hard work throughout today's lesson.

I've really enjoyed the lesson, I hope you have too, and I hope you're able to join me for another lesson soon.