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

We are going to look how we can conduct a fair test to investigate the effect of surface area on the rate of reaction.

Today's lesson outcome is, I can carry out a fair test to investigate how the rate of reaction depends on the surface area of a solid reactant.

Some of the keywords we will be using in today's lesson include rate of reaction, surface area to volume ratio, independent variable, dependent variable, and control variable.

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

It would be a good idea to pause the video now and read through 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 surface area and rate practical is split into two parts.

First of all, we are going to be looking at surface area and rate the variables.

Then we are going to move on to look at surface area and rate and we will look at the equipment and method.

Let's get started on the first part of our lesson, surface area and rate variables.

The rate of reaction is the speed with which a chemical reaction occurs and the rate of reaction can be affected by changing the surface area of solid reactants.

To increase the surface area of a solid, we can break it into smaller pieces.

Here we can see in the image we have metal strips and these have a smaller surface area than if we broke them up into smaller pieces.

If we break them up into smaller pieces such as a powder, then the metal powder has a much larger surface area.

If we break a solid into smaller pieces, the volume stays the same, but the surface area increases.

If we took those metal strips, if we had five grammes of metal strip and we broke that into smaller pieces so that we had five grammes of metal powder, then we would still have the same volume, but the surface area will have increased.

Let's have a look at another example.

Here we have a large cube and our cube has sides of two centimetres.

We can work out the total surface area of our cube.

Well, the surface area of one side will be two centimetres multiplied by two centimetres, which is four centimetres squared.

If we want to know the total surface area of all six sides, we would then do four times by six because there's six sides on our cube and we get a total surface area of 24 centimetres squared.

We could now work out the volume for our cube.

Our cube has sides of two centimetres, and to work out the volume, we do width times height, times depth, so we would do two times two times two, and this gives us a volume of eight centimetres cubed.

We could then work out the surface area to volume ratio, and this tells us the surface area relative to the volume.

We have a surface area of 24 and a volume of eight, so our ratio would be 24 to eight and we can simplify this as three to one.

What happens then if we break our large cube into smaller pieces? Well, we can have a look here.

We have broken our cube into eight smaller cubes, and these cubes each have a side of one centimetre.

Let's work out the total surface area of our eight smaller cubes.

Well, the surface area of one side will be one multiplied by one, which is one centimetre squared.

We have six sides on a cube and we have eight cubes in total.

So if we multiply six by eight, we know that we have a total of 48 sides.

So the total surface area of these 48 sides will be one multiplied by 48, which is 48 centimetres squared.

We can see that the surface area of the smaller cubes is larger than the surface area of the one large cube.

What about the volume? To work out the volume of all our eight cubes, we will do one times one times one, which is one centimetre cubed.

That is the volume of one cube.

If we times that by eight, we'll get the volume for the eight cubes and this will be one times by eight, which is eight centimetres cubed.

Now you should notice that the volume for the one large cube and the eight smaller cubes is the same.

They both have a volume of eight centimetres cubed, so the surface area increased, but the volume stayed the same.

Let's look at the surface area to volume ratio.

This time for the smaller cubes, it's 48 to eight, which we can simplify as six to one.

So you should notice that the smaller cubes have a larger surface area to volume ratio.

Time for a question.

Which solid has the largest surface area to volume ratio? Is it A, the sugar cubes, B, granulated sugar, or C, powdered sugar? The correct answer is C, the powdered sugar.

If we break those sugar cubes into granules and then break the granules further into powdered sugar, then we are increasing the surface area to volume ratio.

Here's another question to have a go at.

Is this statement true or false? Five grammes of small marble chips have a smaller surface area than five grammes of large marble chips, and we've got an image there of the marble chips so you can see what they look like.

Marble chips are actually calcium carbonate, but we often refer to them as marble chips.

Is that statement true or false? That statement is false, well done if you got that correct.

Now can we all have a go explaining why this statement is false? Smaller marble chips have a larger surface area than larger chips of the same mass.

Jacob wants to investigate how changing the surface area of a solid reactant affects the rate of reaction.

Jacob says, "I predict that increasing the surface area will increase the rate of reaction." Do you think that Jacob's prediction is correct? Why? Jacob's prediction is correct.

As we increase the surface area, the rate of reaction will increase.

This is because when we have a larger surface area, there are more particles exposed, therefore we can have more collisions per second, and this increases the frequency of successful collisions.

So we can link it back to the collision theory to explain why an increase in surface area will increase the rate of reaction.

To ensure this is a fair test investigation, Jacob needs to consider three main variables.

The independent variable, this is the variable that we change or select values for.

Sometimes it helps to remember it starts with I and it's the variable that I change.

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

So this starts with D and it's D for the data that you are going to collect, and the control variables, these are variables that must remain the same throughout an investigation.

Control starts with C-O-N, so it's like constant.

What do we keep constant in the investigation? Time for question.

Jacob wants to investigate how changing the surface area of a solid reactant affects the rate of reaction.

Which of the following is a control variable for this investigation? Is it A, time taken for the reaction, B, mass of solid reactant used, or C, surface area of solid reactant used? The correct answer is B, the mass of solid reactant used.

This is a control variable in this investigation.

The time taken for the reaction is the dependent variable and the surface area of the solid reactant used in this investigation, this is the independent variable.

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

Jacob is investigating how changing the surface area of calcium carbonate affects the rate of reaction.

Jacob reacts calcium carbonate with hydrochloric acid and you can see the reaction written out as a word equation for you there.

Jacob uses a mass balance to measure the decrease in mass of the reaction mixture as carbon dioxide is made and lost to the surroundings.

You can see an image at the top of the slide which shows you how the equipment is set up.

Your task is to identify the independent, dependent, and control variables in Jacob's investigation.

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

Let's see how you got on.

For this investigation, identify the independent, dependent, and control variables.

The dependent variable is the mass of the reaction mixture.

The independent variable is the surface area of the calcium carbonate, and control variables include the concentration and volume of hydrochloric acid, the mass of calcium carbonate and the temperature.

Well done if you correctly identified the independent and dependent variables, and hopefully you listed many of those control variables too.

We have had a look at the variables.

Now we are going to move on to the second part of our lesson where we look at surface area and rate equipment and method.

To determine the rate of reaction, we need to know either the amount of reactant used in a given time and we can use this equation here to work out the rate of reaction where we do the amount of reactant used divided by the time taken, or we need to know the amount of product that has been made in a given time and we could then work out the rate of reaction by doing the amount of product made divided by the time taken.

The rate of reaction can be determined using the following equipment.

We've got a gas syringe, a conical flask containing our reaction mixture and a delivery tube.

We could use this equipment to measure the volume of product if it's a gas made in a given time.

So we would time the reaction mixture and at set intervals we would measure the volume of gas that has escaped through the delivery tube and into the gas syringe.

Another way we could measure the rate of reaction is using the following equipment.

Here we have a mass balance at the bottom.

We've got our reaction mixture containing our marble chips and acid, and there's some cotton wool at the top.

We could use this equipment to measure the decrease in mass as a gas product is made and leaves the reaction vessel in a given time.

As a reaction took place and a gas is produced, then the mass of the reaction mixture would decrease as the gas escapes to the surroundings.

Time for a question.

At the start of a reaction, the reaction vessel has a mass of 124.

75 grammes.

What mass of gas is produced in the reaction in 10 seconds? In the image there we can see we've got the mass balance and this shows us the reading after 10 seconds.

So what mass of gas is produced? The correct answer is C, 0.

10 grammes because the initial mass of the reaction vessel was 124.

75 grammes, and now we can see that the mass is 124.

65 grammes.

So the mass has decreased by 0.

10 grammes and this is because we have produced this mass of gas and it has escaped into the surroundings.

Well done if you got that question correct.

The results from measuring a decrease in mass of a reaction mixture in a given time may look similar to this graph.

We've got the mass of reaction mixture in grammes and we've got the time in seconds.

We can see that the mass decreases rapidly per second at the beginning of the reaction, so we have a high rate of reaction at the start of the reaction.

The gradient then becomes shallower, and this shows us that the decrease in mass per second is less so we have a lower rate of reaction.

So the rate of reaction decreases as the reaction progresses, and then finally the mass stays the same and this indicates that the reaction has ended.

Over the course of a reaction, the mass of the reaction mixture decreases as a gas is produced and escapes to the surrounding, and then eventually the mass becomes constant and this is because the reaction has ended.

Why will the mass of the reaction mixture eventually reach a constant mass? Is it because A, the rate of reaction is increasing? B, the rate of reaction is decreasing, or C, the reaction has ended? The correct answer is C.

The reaction has ended, and this explains why the reaction mixture stays at a constant mass.

The data that is collected should be presented in a table of results.

The headings for a table of results are always the independent and dependent variables.

The independent variable is the variable we change and are investigating, and the dependent variable is the variable that we measure in our investigation.

We have the independent variable and the units and the dependent variable and the units in our table of results.

Control variables, these need to remain constant in an investigation are not recorded in a table of results because they do not change.

Jacob has produced this results table for measuring the effect of surface area on the rate of reaction.

What mistake has he made? A, the independent variable is missing.

B, the units are incorrect.

C, a control variable is included.

Have a look at what Jacob was investigating to help you work out what are the independent and dependent variables that Jacob should have included in his table.

The mistake that Jacob has made is that he has included a control variable in his table.

Jacob has got the concentration of acid in the first column and this is a control variable and should remain constant throughout the investigation.

In the other column, Jacob has got the mass of reaction mixture in grammes.

This is the dependent variable and should be in Jacob's table of results.

We have reached the final practise task of today's lesson.

Number one, Jacob wants to measure the effect of changing the surface area of calcium carbonate on the rate of reaction.

In the images there you can see we have calcium carbonate chips and we've got calcium carbonate powder.

And remember, if we break something up into smaller pieces such as breaking that calcium carbonate chips into a powder, we increase the surface area.

Jacob reacted two grammes of calcium carbonate chips with 20 centimetres cubed of hydrochloric acid.

He repeated the experiment using powdered calcium carbonate.

Create a table of results that Jacob could use to record the decrease in mass during these reactions.

Pause the video now, draw up your results table, then when you come back, we'll go over the answer.

Let's see how you got on.

Your results table may look like this.

We have the time in the first row and at set intervals, every 30 seconds, we are going to record the mass of the reaction mixture.

You can see in the next row we have the mass of the reaction mixture using calcium carbonate chips and we've included the units which are grammes, and then in the final row we have the mass of the reaction mixture using calcium carbonate powder, which is also measured in grammes.

So we have the time and we've got the units included, and then we've got the mass for both our chips and our powder.

Hopefully your results table looks similar to this.

Make sure you do have information about the dependent and independent variable in your results table and you have the units.

The dependent variable is the mass of the reaction mixture and the independent variable is the surface area, so we are using chips and powder.

All this information is present in this results table here.

For the second part of this task, you need to follow the method to collect results on the effect of surface area on the rate of reaction.

So step A, you'll measure 20 centimetres cubed of hydrochloric acid and place it in a conical flask on top of a mass balance.

B, measure two grammes of large pieces of calcium carbonate.

C, add the calcium carbonate chips to the acid and start the timer.

D, record the initial mass.

Then step E, record the mass every 30 seconds until the mass remains constant.

At this point here, we know the reaction has ended.

F is to repeat the experiment using powdered calcium carbonate, which has a larger surface area.

If you were unable to collect a set of results, then here is a sample set of results that shows what happens to the mass of the reaction mixture over time when we use the calcium carbonate chips and when we use the calcium carbonate powder.

Thank you for all your hard work during today's lesson on surface area and rate practical.

We have reached the end of today's lesson.

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

The changing rate of a reaction can be found by measuring the change in mass as the reaction progresses.

Smaller marble chips have a larger surface area than larger chips of the same mass, and the larger the surface area, the faster the rate of reaction.

I hope that you've enjoyed today's lesson and I hope that you're able to join me for another lesson soon.