Hello, I am Mrs. Adcock and welcome to today's lesson on concentration and rate using an end-point method.

How can we perform a fair test to investigate the effect of concentration on rate using an end-point method? Today's lesson outcome is "I can carry out a fair test to investigate how the rate of reaction depends on the concentration of a reacting solution." Some of the keywords that we will be using in today's lesson include "rate of reaction," "end-point method," "precipitate," and "fair test." You can see each of those keywords written here 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 you need to.

Today's lesson on concentration and rate, a practical using end-point method is split into two parts.

First of all, we are going to look at the theory that links concentration and rate of reaction, and then we are going to move on to looking at if we are investigating how concentration affects the rate of reaction, how we can use an end-point method.

Let's get started on the first part of our lesson.

The collision theory states that for reactant particles to react, they must collide with each other and they must collide with sufficient energy.

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

Here we can see reactant particles colliding together and these particles have the activation energy.

When they collide, they react and form new products.

We can see we've got AB and that reacts with C and they react together to form new products and we have A and BC as our new products.

We can increase the frequency of collisions by increasing the pressure of reacting gases.

So at a higher pressure, there are more reactant particles per unit volume and therefore these reactant particles are closer together and therefore are more likely to collide.

So, when we increase the pressure, we increase the frequency of collisions.

Here we can see an animation showing reacting gases under low pressure, and here we've got reacting gases under a higher pressure.

And we can see those reacting gases under a higher pressure, the particles are closer together and they are more likely to collide than when they were under a low pressure.

We can also increase the frequency of collisions by increasing the concentration of a reacting solution.

When we increase the concentration of a solution, there are more reacting particles per unit volume, so this is similar to increasing the pressure of reactant gases and these particles are now more likely to collide.

Here we can see a low concentration of reactant in solution and we can compare this to this animation that shows us a high concentration of reactant in solution.

And hopefully you notice when we have the high concentration, we have got more reactant particles per unit volume and therefore they are more likely to collide.

When we increase the frequency of collisions, we increase the rate of reaction.

Time for a question.

How do the particles in a reactant solution at a high concentration compare to those in a low concentration solution? A, there are more reactant particles per unit volume.

B, there are less reactant particles per unit volume.

C, the particles are more likely to collide.

D, the particles have more energy.

We are comparing the particles in a high concentration compared to a low concentration.

The correct answers are A, at a high concentration there are more reactant particles per unit volume; and C, the particles are more likely to collide because they are closer together.

Well done if you chose A and C.

Doubling the concentration of a reactant in solution doubles the number of reactant particles in the same volume.

And we've got an illustration here that shows this.

We have particles at a low concentration and we have reactant particles when the concentration has been doubled.

And we've doubled the concentration of those atoms that are shown with a large circle.

So this is a model showing the reactant particles when the concentration of this reactant is doubled.

And hopefully you will notice that when we have doubled the concentration, then we've now got more reactant particles in that same volume.

As a result of there being double the number of reactant particles in the same volume, then the number of successful collisions per second also doubles and therefore the rate of reaction doubles.

So when we double the concentration of a reactant in solution, we double the rate of reaction.

Concentration and rate of reaction.

Just a reminder that rate of reaction is the speed at which a chemical reaction takes place.

These are directly proportional, concentration and rate of reaction.

We can see them shown here on the graph.

We have the rate of reaction on the Y axis and the concentration on the X axis.

And we can see as the concentration of a reactant in solution doubles, the rate of reaction also doubles.

The relationship between these two is directly proportional.

Let's have a go at this question.

Use the data to work out what the rate of reaction will be when the concentration of the reactant solution is 2.

5 grammes per centimetre cubed.

We have a table.

It shows us two different concentrations.

One of them the concentration is 5.

0 grammes per centimetre cubed and then the concentration is halved to 2.

5 grammes per centimetre cubed.

We can see the rate of reaction when we had five grammes per centimetre cubed concentration is 2.

0 centimetres cubed per second.

So what will the rate of reaction be when we use half the concentration of reactant solution? Will it be A, 4.

0 centimetres cubed per second? Will it be B, 2.

0 centimetres cubed per second? Or will it be C, 1.

0 centimetres cubed per second? The correct answer is C.

As the concentration of reactant solution is halved, the rate of reaction will also halve.

And we saw that on the previous slide, the relationship between concentration and rate of reaction is directly proportional.

Well done if you chose answer C.

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

Calcium carbonate reacts with hydrochloric acid to produce calcium chloride, carbon dioxide, and water.

And we can see that reaction represented there as a word equation.

What effect would increasing the concentration of hydrochloric acid have on the rate of this reaction and use the collision theory to explain your answer.

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

Pause the video now.

Answer this question in as much detail as you can and then when you come back, we'll go over the answer.

Hopefully you did well answering this question.

What effect would increasing the concentration of hydrochloric acid have on the rate of this reaction? Increasing the concentration increases the rate of reaction.

We then need to use the collision theory to explain our answer.

There will be more reactant particles per unit volume when we increase the concentration and therefore a higher frequency of successful collisions.

That's really great work if you managed to answer that question correctly.

We have had a look at the theory between concentration and the rate of reaction.

Now we are going to move on to have a look at concentration and rate and end-point method.

We can investigate the effect of concentration on the rate of reaction by changing the concentration of a reacting solution and measuring the amount of gas produced over time.

To do this, we could use a mass balance and a mass balance is used to measure the mass of product made.

If one of our products is a gas, the gas will be produced and it will escape into the surroundings.

And as our gas is produced and escapes into the surroundings, the mass of our reaction mixture will decrease.

And we can use this to calculate the amount of gas that is produced over time.

Another piece of apparatus we could use is an inverted measuring cylinder.

And we can use this to measure the volume of product made.

Again, one of our products is a gas.

Then the gas will be produced in our reaction mixture and it can travel through the delivery tube and into our inverted measuring cylinder.

It will displace the water here and we can measure the amount of gas that we have produced.

Again, if one of our products is a gas, then we can use a gas syringe to measure the volume of product made.

The gas will be produced and it will move out of the conical flask and into the gas syringe.

An end-point method can be used in a rate of reaction investigation, so we don't just need to use gas collection methods.

An end-point method measures the progress of a reaction until it has either reached completion or a desirable point.

You can see in the example we have a reaction taking place that involves a colour change.

So we could time how long it takes to reach an end-point, and in this reaction our end-point would be the colour change.

We can use an end-point method to investigate the effect of changing the concentration of sodium thiosulfate on the rate of this reaction.

And in this reaction, we have reacted sodium thiosulfate with hydrochloric acid.

And when these react together, they produce the products, sodium chloride, water, sulphur dioxide, and sulphur.

Below the word equation, you can see we have a balanced symbol equation and we have included the state symbols for these substances.

If you look carefully, you will notice that the sulphur that we produce has the state symbol "S," and this is because the sulphur that is produced is a solid, so we form a precipitate.

The sulphur that is produced is a yellow precipitate and a precipitate is an insoluble solid that is formed when we react two solutions together.

When this yellow precipitate of sulphur forms, it makes the solution that was clear and colourless turn cloudy.

As sulphur is produced, the mixture becomes cloudy.

If an "X" is placed on a card beneath the reaction mixture, we can measure the time taken for the "X" to no longer be visible.

Here is our reaction at the start, and you can see the solutions are clear and colourless.

So we can visibly see the "X" that was placed on a piece of card beneath the conical flask.

When this reaction has reached its end-point, the "X" is no longer visible.

And you can see this here in the image, we can no longer see the "X" below the conical flask, and this is because we produced that sulphur precipitate.

And as we produced the precipitate, it's made the solution cloudy.

This experiment here where we can see the "X" at the beginning and then we can't see the "X" when we've reached our end-point is sometimes referred to as the "disappearing cross experiment." If we are investigating changing the concentration of one of our reacting solutions, then we could time how long it takes for the end-point to be reached.

In this experiment, we would time how long it takes for the "X" to no longer be visible, and we could repeat this experiment using different concentrations of our reacting solution.

Let's have a go at this question.

When using a lower concentration of reacting solution, the time taken for the "X" to no longer be visible will be A, shorter.

B, the same.

C, longer.

The correct answer is C, longer.

If we use a lower concentration of reacting solution, then the time taken for the "X" to no longer be visible will be longer because using a lower concentration of reacting solution will decrease the rate of reaction and therefore the reaction will take longer to reach the end-point.

Well done if you got this question correct.

In the disappearing cross experiment, the end-point is when the "X" is no longer visible.

Here we can see the end-point using a low concentration and we can see the end-point using a high concentration.

So for both the low and the high concentration, the end-point is the same.

It's when we can no longer see that "X" beneath the conical flask.

For different concentrations of reacting solutions, each reaction will produce the same amount of product at the end-point.

It doesn't matter whether we use a low concentration or a high concentration.

The end-point is when we have produced a certain amount of product that has made us no longer able to see the "X" beneath the conical flask.

The end-point is when the "X" is no longer visible, but this is not necessarily the end of the reaction.

The chemical reaction may still not be complete.

Here we can see on the graph the mass of sulphur that's being produced in grammes on the Y axis and the time in seconds on the X axis.

And we have two lines, one representing a reactant with a high concentration and the other one representing a reactant with a low concentration.

Here is where the end-point is reached.

So the end-point is reached when we have produced a certain mass of sulphur and this mass of sulphur will make the solution cloudy.

However, the reaction has not reached completion.

If you notice, the reaction continues past this point, and it's only where we have now mapped on the graph that the reaction has ended when the mass of sulphur produced remains constant.

To ensure this is a fair test investigation, a fair test is where the variables are controlled so that only one change is affecting the dependent variable.

So we need to consider the independent variable.

This is the variable that we change or select values for.

The dependent variable, this is the variable that we measure or observe to get your results.

And control variables, these are variables that must remain the same or constant throughout an investigation.

Is this statement true or false? In the disappearing cross reaction when the end-point is reached and the "X" is no longer visible, the reaction has ended.

That statement is false.

Can you now explain why that statement is false? That statement is false because when the "X" is no longer visible, the end-point has been reached.

But this does not mean that the reaction is complete.

Well done if you got that question correct and you understand that the end-point is not always when the reaction has ended.

Time for us to have a go at another practise task.

Laura is investigating how changing the concentration of sodium thiosulfate affects the rate of reaction.

And we can see that Laura is reacting sodium thiosulfate with hydrochloric acid.

And when these react together, they produce sodium chloride, water, sulphur dioxide, and sulphur.

And in the question there we have the word equation and a balanced symbol equation for the reaction that Laura's using.

And hopefully you can notice those state symbols and remember that in this reaction we produce sulphur that is a precipitate.

Laura measures the time taken for an end-point of the reaction to be reached.

Identify the independent, dependent, and control variables in Laura's investigation.

Pause the video now, have a go at answering this question.

Then when you come back, we will go over the variables in Laura's investigation.

Let's see how you got on with identifying the independent, dependent, and control variables for Laura's investigation.

The independent variable is the concentration of the sodium thiosulfate.

That was the variable that she was changing.

The dependent variable is the time taken for the end-point to be reached because those are the measurements that Laura was going to be making.

And control variables include the concentration and volume of hydrochloric acid, and the temperature.

Hopefully you got at least a couple of those control variables.

Well done if you managed to identify all three of those control variables.

For the final part of this practise task, we are going to follow a method to collect results on the effect of concentration on the rate of reaction and we are going to use an end-point method.

You can see the steps there that you need to follow to carry out this method.

You will then repeat the experiment using different concentrations of sodium thiosulfate.

And finally, we will repeat the whole experiment two more times to obtain more reliable data.

Good luck carrying out this investigation and following this method.

I will see you when you have finished this task and collected your data.

Here we have a sample set of results for you if you were unable to carry out the investigation or you did not manage to get a full set of data.

You can see we used three different concentrations of sodium thiosulfate.

We used eight grammes per decimeter cubed, 24 grammes per decimeter cubed and 40 grammes per decimeter cubed.

And then we timed how long it took for the "X" to no longer be visible in seconds.

And we repeated this three times.

So we have data for trial one, trial two, and trial three.

We have reached the end of today's lesson on concentration and rate, a practical using an end-point method.

Let's just summarise some of the key points that we covered in today's lesson.

The reaction between sodium thiosulfate solution and hydrochloric acid produces a cloudy precipitate.

At the end-point of the disappearing cross experiment, each reaction has produced the same amount of product.

At the end point of the disappearing cross experiment, the chemical reaction may not be complete.

And increasing the concentration of a reacting solution introduces more reactant particles so they collide more frequently, and this increases the rate of reaction.

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

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