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I can represent graphically and explain how the rate of reaction depends on the concentration of one of the reactants.

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This content is © Oak National Academy Limited (2023), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

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Key learning points

The changing rate of a chemical reaction can be represented by a graph of volume of gas produced against time.
When reacting excess acid, the same volume of gas is produced for equal amounts of magnesium.
Doubling the concentration doubles the rate of reaction.
The gradient of a rate of reaction graph can be calculated from a tangent drawn at a point on the curve.

Common misconception

Thinking that as the concentration of a solution in excess changes, so does the volume of gas produced.

The slide deck uses graphical data to show that the same volume of gas is produced when different concentrations are used, so long as the reacting solution is in excess.

Keywords

Gradient - The gradient on a graph represents the rate at which one quantity changes relative to another.
Rate of reaction - Rate of reaction is the speed with which a chemical reaction takes place.
Excess - A reactant in excess is present in a greater amount than is necessary to completely react with the limiting reactant.
Mean rate of reaction - The mean rate of reaction is the average rate at which reactants are converted into products in a given time.
Tangent - A tangent is a straight line that touches a curve at exactly one point without crossing it.

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Hello, I am Mrs. Adcock, and welcome to today's lesson.

Today's lesson is on concentration and rate analysis.

We are going to be looking at how we can analyse data to determine the relationship between concentration and rate of reaction.

Today's lesson outcome is: I can represent graphically and explain how the rate of reaction depends on the concentration of one of the reactants.

Some of the keywords we will be using in today's lesson include gradient, rate of reaction, excess, mean rate of reaction, and tangent.

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 concentration and rate analysis is split into three main parts.

First of all, we are going to be looking at analysing graphs.

Then we are going to move on to calculate the mean and instantaneous rate of reaction.

And then finally, we'll finish the lesson by looking at the relationship between concentration and the rate of reaction.

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

Sofia is investigating the effect of changing the concentration of acid on the rate of this reaction.

In this reaction, we have magnesium reacting with hydrochloric acid, and these react together to produce magnesium chloride and hydrogen.

Here you can see we have a balanced symbol equation for this reaction and we've included the state symbols too.

We have Mg, which is magnesium, and that's a solid, and that reacts with HCL, which is hydrochloric acid, and that's aqueous, and this produces aqueous magnesium chloride and hydrogen gas.

How can we measure the volume of gas produced in a reaction? You may have thought to use an upturned measuring cylinder that's filled with water, and as a gas is produced, it will displace the water and we can measure the volume of gas produced.

You may have thought about using a gas syringe to collect and measure the volume of gas produced.

Or you could use a mass balance in which the decrease in mass is measured as the hydrogen gas is produced and escapes to the surroundings.

Sofia uses an upturned measuring cylinder to measure the volume of hydrogen gas produced at specific time intervals.

Here we can see Sofia's equipment set up.

We have the reaction mixture, and you can see the strip of magnesium there and that's reacting with hydrochloric acid.

Hydrogen gas is being produced.

We can see the bubbles in the reaction mixture.

And the hydrogen gas will leave the conical flask and travel down through the delivery tube.

We then have water in the water trough.

An upturned measuring cylinder, which would at first be filled with water.

However, as the gas is produced and bubbles up through into the upturned measuring cylinder, the water will be displaced.

And here we will collect our volume of gas.

Sofia plots a graph of volume of hydrogen gas produced versus time for a concentration of acid that she used, and she is trying to show how the rate changes throughout the reaction.

So we can see on the y-axis we have the volume of hydrogen gas made in centimetres cubed, and on the x-axis we have the time in seconds.

The gradient of Sofia's graph, or of a volume versus time graph, represents the rate of reaction, and the steeper the slope, the greater the rate of reaction.

We can see from studying Sofia's graph that the gradient changes throughout the reaction and therefore we know that the rate of reaction is changing throughout the reaction.

Time for a question.

We have a graph there with three sets of data.

We have A, B, and C.

These are three different reactions that are shown on the one graph.

So which reaction has the highest rate of reaction? Is it A, B, or C? The correct answer is A, so well done if you chose answer A.

Remember, the gradient of this volume versus time graph tells us the rate of reaction, and the steeper the gradient, the higher the rate of reaction.

A has the steepest gradient there and therefore has the highest rate of reaction.

Typically, the gradient becomes shallower as the rate of reaction decreases over the course of a reaction.

If we look at the graph, we can see the rate of reaction is highest at the start of the reaction.

This is where our gradient is steepest.

Then the rate of reaction slows down as the reaction progresses, and we can see this because the gradient has become shallower.

And finally, the rate of reaction is zero.

And if we look at the y-axis, we can see that the volume of hydrogen being made remains constant at this point, and that is because the reaction has ended.

When reacting equal amounts of magnesium with excess hydrochloric acid, the same volume of gas is produced regardless of the concentration.

Let's have a look at this on the graph.

On the graph we can see we've got two sets of data.

We have a set of data where a high concentration of hydrochloric acid has been used and a set of data where we've used a low concentration of hydrochloric acid.

And for both of these different concentrations, we can see that the final amount of hydrogen produced for the high and low concentration of acid is the same, and this is because the acid is in excess.

So changing the concentration of the acid does not affect the overall volume of products that we make.

It is the magnesium that is the limiting reactant, and this will control the volume of hydrogen that is produced.

Time for another check for understanding.

We have a statement here and you need to decide whether it's true or false.

When reacting equal amounts of magnesium with excess acid, a higher concentration of acid will produce a larger volume of gas compared to a lower concentration of acid.

Is that statement true or false? That statement is false, so well done if you identified that that statement was false.

Can you now explain why this statement is false? This statement is false because the acid is in excess and therefore does not control the volume of product that is produced.

A high and a low concentration of acid will produce the same volume of gas.

The starting mass of magnesium is the thing that limits the reaction.

So well done if you got that correct and you identified that both the high and low concentration of acid will produce the same volume of gas.

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

For this task, you need to plot a graph of volume of hydrogen gas produced in centimetres cubed against time in seconds.

And you're doing this for reactions using a 1.

0 mole per decimeter cubed concentration of hydrochloric acid and using 1.

5 mole per decimeter cubed concentration of hydrochloric acid.

All the data that you need is there in that table.

You are going to plot the volume of gas that is made using the lower concentration and the higher concentration on the same graph.

Pause the video now, plot your graph, and then come back when you finished your graph.

Let's see how you got on with plotting your graph.

It should look similar to this where you have the volume of hydrogen produced in centimetres cubed on the y-axis.

You should have the time in seconds along the x-axis, and then you will have both sets of data, the 1.

0 moles per decimeter cubed and the 1.

5 moles per decimeter cubed concentration, both plotted on the same graph.

Well done if you've drawn your graph correctly.

Just make sure that you've included a key or that you've labelled your two different curves to show which is for the lower concentration and which one represents the higher concentration.

We have looked at analysing graphs.

We are now going to move on to see how we can use our graphs to calculate the rate of reaction.

Sofia calculates the mean rate of reaction for the reaction shown on the graph below.

Now remember, the mean rate of reaction is the average rate at which reactants are converted into products in a given time.

Let's have a look how Sofia calculates the mean rate of reaction.

We've got our labels there on our graph.

So we have the volume of hydrogen gas made in centimetres cubed on the y-axis and we have the time in seconds along the x-axis.

So first of all, Sofia looks at the total volume of product that was made, and this is 74 centimetres cubed for this data here.

Then Sofia looks at how long did the reaction take to reach completion, and the reaction takes 120 seconds to reach completion.

Sofia can then calculate the mean rate of reaction by using the equation amount of product made divided by time.

So the amount of product made was 74 centimetres cubed, divide that by the time, which was 120 seconds, and we get the answer of 0.

62 centimetres cubed per second.

So the mean rate of reaction is 0.

62 centimetres cubed per second.

The instantaneous rate of reaction is the rate of reaction at a specific time in the reaction.

And the instantaneous rate of reaction can be determined from the gradient of the curve at a specific time.

So we're not looking at the average rate of the reaction.

We want to know at a specific time what is the rate of reaction.

So to calculate the instantaneous rate of reaction from a graph at a specific time, we first of all need to draw a tangent to the curve.

A tangent is a straight line that touches the curve at a point.

And then we are going to calculate the gradient of the tangent, and it reminds us there how we calculate the gradient, and that's change in y divided by change in x.

Which of the following statements are true? A, the instantaneous rate is the rate at a specific time.

B, the mean rate is the rate at a specific time.

C, the instantaneous rate can be calculated from a tangent drawn at a specific point on the curve.

And D, the mean rate can be calculated from a tangent drawn at a specific point on the curve.

Choose any statements that you think are true.

Let's have a look.

Statement A is true, the instantaneous rate is the rate at a specific time.

And C is also true, the instantaneous rate can be calculated from a tangent drawn at a specific point on the curve, and we then work out the gradient of the tangent.

Well done if you chose answers A and C and got that question correct.

Sofia calculates the instantaneous rate of reaction at 20 seconds for the reaction shown on the graph below.

So first of all, Sofia identifies where the time is 20 seconds and marks that with a cross on the graph.

She then draws a tangent at time = 20 seconds.

And we can see that straight line and it's touching the curve at 20 seconds.

We then calculate the gradient of the tangent line, and to work out the gradient, we do change in y divided by change in x.

The change in y is from 55 down to 2.

5, so 55 - 2.

5 gives us an answer of 52.

5 centimetres cubed.

And the change in x there we can see is 47.

5 - 0, and that's 47.

5 seconds.

We can then work out the gradient by doing 52.

5 divided by 47.

5, and that gives us our instantaneous rate of reaction at 20 seconds, which is 1.

11 centimetres cubed per second.

Sofia then calculates the instantaneous rate of reaction at 80 seconds for the same reaction as shown on the graph below.

Sofia identifies where 80 seconds is and draws a tangent where time = 80 seconds.

Sofia then calculates the gradient of the tangent line, and we can see the change in y is 75 - 60, which is 15 centimetres cubed.

And the change in x is 100 - 60, so the change in x is 40 seconds.

So to work out the gradient, we do 15 divided by 40, and the instantaneous rate of reaction at 80 seconds is 0.

38 centimetres cubed per second.

A higher rate of reaction means that more reactant is used or more product is made in a given time.

Sofia's results show that the rate of reaction decreases during the reaction.

We have here in the table the time at 20 seconds and the time at 80 seconds.

And for both of these, we worked out the instantaneous rate of reaction.

So at 20 seconds, the instantaneous rate of reaction was 1.

11 centimetres cubed per second, and at 80 seconds the rate of reaction was 0.

38 centimetres cubed per second.

We can see that as the reaction progressed, the rate of reaction decreased, and a higher volume of hydrogen is produced per second at 20 seconds compared to our 80 seconds.

If we just go back and have a look at the data there, we can see at 20 seconds we are making 1.

11 centimetres cubed of hydrogen gas every second, whereas at 80 seconds the rate has decreased and we're only making 0.

38 centimetres cubed of hydrogen gas per second.

We have Sofia's results table here, and it shows us the rate of reaction at 20 seconds and at 80 seconds.

Which statement or statements correctly describe Sofia's results? A, as the reaction progresses, more product is made per second.

B, as the reaction progresses, less product is made per second.

C, as products are formed, there are less frequent successful collisions between reactant particles.

Or D, as products are formed, there are more frequent successful collisions between reactant particles.

The correct answers are B, as the reaction progresses less product is made per second, and we can see this because the rate of reaction has decreased.

And also C, as products are formed, there are less frequent successful collisions between reactant particles.

And again, we know this because we can see that as the reaction progresses, the rate of reaction decreases.

Well done if you chose answers B and C.

Time for another practise task.

First of all, you need to calculate the mean rate of reaction for the reaction using 1.

0 moles per decimeter cubed concentration of hydrochloric acid.

Let's see if you've done that correctly.

So you should have identified the total volume of product made is 74 centimetres cubed.

The time taken for the reaction to reach completion is 120 seconds.

And then to work out the mean rate of reaction, we do the amount of product made divided by time, so this will be 74 divided by 120, and the answer there is 0.

62 centimetres cubed per second.

Well done if you calculated that mean rate of reaction correctly.

For question two, you need to calculate the mean rate of reaction for the reaction using 1.

5 moles per decimeter cubed of hydrochloric acid.

Have a go at answering this question.

Then when you come back, we will go over the answer.

When Sofia used the higher concentration of acid, the total volume of product made was still the same, so that was still 74 centimetres cubed.

However, the reaction was completed in a shorter amount of time, so the reaction only took 70 seconds to reach completion.

Then working out the mean rate of reaction, we do the amount of product made divided by the time, so you should have 74 divided by 70, and that gives us a mean rate of reaction of 1.

06 centimetres cubed per second.

You may have a slightly different time if you've used your own graph or when you were reading from this graph, but you should hopefully still have an answer that's very close to that one there.

For question three, we need to calculate the instantaneous rate of reaction at 10 seconds for the reaction using 1.

0 moles per decimeter cubed of hydrochloric acid.

Pause the video now, work out the instantaneous rate of reaction at 10 seconds for this reaction.

Let's go over the answer.

We draw a tangent at t = 10 seconds.

It's a good idea when you're drawing your tangent line to extend your tangent line to the grid lines.

This helps you to then calculate the change in y and change in x more easily.

Once we've drawn our tangent at time = 10 seconds, you should have then calculated the gradient of your tangent line.

For the tangent line that I drew here, we've got the change in y is 47.

5 - 2.

5, which equals 45, and the change in x is 40 - 0, which equals 40.

So the change in y is 45 divided by the change in x, which is 40, and that gives us an answer of 1.

13 centimetres cubed per second.

Now depending on how you've drawn your tangent line, you may have a different value for the change in y and for the change in x.

Overall, you should still get a very similar value for the instantaneous rate of reaction at 10 seconds for the one that you can see shown there.

Question four is the final question as part of this task, and you need to work out the instantaneous rate of reaction at 10 seconds for the reaction using that higher concentration of 1.

5 moles per decimeter cubed of hydrochloric acid.

Pause the video now, work out this instantaneous rate of reaction.

Then when you come back, we'll go over the answer.

To calculate the instantaneous rate of reaction, you should have first of all drawn a tangent at time = 10 seconds.

Then calculated the gradient of the tangent line.

And remember, gradient is change in y divided by change in x.

The change in y here is 45 - 2.

5, which equals 42.

And the change in x is 25 - 0, which equals 25.

So the gradient will be 42.

5 divided by 25, which gives us an instantaneous rate of reaction at 10 seconds of 1.

70 centimetres cubed per second.

Your values for change in y and change in x might be slightly different based on the length of your tangent line, but overall, you should still get a very similar value for the instantaneous rate of reaction.

We have analysed graphs and then moved on to calculate the mean and instantaneous rate of reaction.

We are going to finish the lesson by explaining how the concentration affects the rate of reaction.

Sofia compared the values of mean rate of reaction for the high concentration, which was 1.

5 moles per decimeter cubed, and the low concentration, which was 1.

0 moles per decimeter cubed.

And these are the two different concentrations of acid that Sofia used.

Here we can see them written in a table.

We have the mean rate of reaction for those two different concentrations of acid.

We can see that when we had 1.

0 mole per decimeter cubed of acid, the mean rate of reaction was 0.

62 centimetres cubed per second.

And when the concentration was increased to 1.

5 mole per decimeter cubed, then the mean rate of reaction also increased, and that increased to 0.

98 centimetres cubed per second.

Sofia concluded that the higher the concentration of acid, the higher the rate of reaction.

Why does the rate increase when the concentration is increased? Hopefully, you mentioned the collision theory in your answer.

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

When we increase the concentration of a solution, there are more reactant particles per unit volume, and these particles are therefore more likely to collide.

If we look here at this model, we can see a low concentration of reactant in solution, and we can compare this to this animation which shows a high concentration of reactant in solution.

And hopefully when you compare these, you notice that when we have a high concentration of reactant in solution, we have more reactant particles per unit volume, and therefore these particles are more likely to collide.

Therefore, increasing the concentration increases the rate of reaction as there are more frequent successful collisions.

What happens when we double the concentration? Well, doubling the concentration of a reactant in solution doubles the number of reactant particles in the same volume.

And we can look at this model to help us understand.

We have an image showing the particles in a low concentration of reacting solution, and then we can see the particles in that solution when the concentration has been doubled.

If we focus on those larger green particles, then we can see that this model shows the reactant particles when the concentration of this reactant is doubled.

And when we double the concentration, then the number of that reactant particle in the same volume has also doubled.

The number of successful collisions per second then also doubles, and therefore the rate of reaction doubles.

Here we can see the relationship between rate of reaction and concentration shown on a graph.

We have the rate of reaction on the y-axis and the concentration on the x-axis, and we can see that concentration and rate of reaction are directly proportional.

So as the concentration of a reactant in solution doubles, the rate of reaction also doubles.

Time for a question.

What effect does halving the concentration of a reactant solution in excess have on the rate of reaction? A, the rate of reaction doubles.

B, the rate of reaction stays the same.

Or C, the rate of reaction halves.

The correct answer is C.

So well done if you chose answer C.

When we halve the concentration of a reactant solution, then we will have half the number of reactant particles in the same volume.

Therefore, the number of successful collisions per second will halve and the rate of reaction will halve.

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

Sofia is investigating the effect of concentration on rate of reaction.

She calculated the instantaneous rate of reaction at 10 seconds for reactions using different concentrations of reactant solution.

We can see in the table there we have two different concentrations of reactant solution, 1.

0 and 1.

5 moles per decimeter cubed.

And we have the instantaneous rate of reaction at 10 seconds for each of those different concentrations.

Predict the instantaneous rate of reaction at 10 seconds when using a 2.

0 mole per decimeter cubed concentration of reactant solution.

Use the collision theory to explain your answer.

If you pause the video now, have a go at answering this question in as much detail as possible and then come back when you're ready to go over the answer.

Doubling the concentration from 1.

0 to 2.

0 moles per decimeter cubed will double the rate of reaction, so it'll go from 1.

13 to 2.

26 centimetres cubed per second.

Well done if you got that first part of the question correct.

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

There will be twice as many reactant particles per unit volume, and therefore the frequency of successful collisions will double, which means the rate of reaction will also have doubled.

We have reached the end of today's lesson on concentration and rate analysis.

Before we go, let's just summarise some of the key points that we have covered in today's lesson.

The changing rate of a chemical reaction can be represented by a graph of volume of gas produced against time.

When reacting excess acid, it doesn't matter the concentration of that acid because the same volume of gas will be produced for equal amounts of magnesium.

Doubling the concentration doubles the rate of reaction.

And the gradient of a rate of reaction graph can be calculated from a tangent drawn at a point on the curve.

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.

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"Question", "question": { "text": "In the collision theory, what does \"successful collision\" mean?" }, "questionType": "Multiple choice", "correct": [ { "text": "a collision that results in a reaction" } ], "choices": [ { "text": "any collision between reactant particles" }, { "text": "a collision that results in a reaction" }, { "text": "a collision that increases the energy of particles" }, { "text": "a collision with no change in particles" } ], "id": "3294eb0f-e7a8-5483-5600-f084407bcf69", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" }, { "type": "Question", "question": { "text": "What is the purpose of drawing a tangent to the curve on a rate of reaction graph?" }, "questionType": "Multiple choice", "correct": [ { "text": "to find the instantaneous rate of reaction at a specific time" } ], "choices": [ { "text": "to find the total volume of gas produced" }, { "text": "to determine the average rate of reaction" }, { "text": "to find the instantaneous rate of reaction at a specific time" }, { "text": "to measure the concentration of reactants" } ], "id": "64515c51-4619-912e-a759-f4e236c8307e", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" }, { "type": "Question", "question": { "text": "What does a flat horizontal line on a graph of volume against time indicate?" }, "questionType": "Multiple choice", "correct": [ { "text": "The reaction has stopped." } ], "choices": [ { "text": "The reaction is occurring rapidly." }, { "text": "The reaction has stopped." }, { "text": "The reaction is occurring slowly." }, { "text": "The reaction rate is increasing." } ], "id": "bf30dacb-2aff-df1c-4bd5-94e906ff646c", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" }, { "type": "Question", "question": { "text": "For most reactions, when is the rate of reaction highest?" }, "questionType": "Multiple choice", "correct": [ { "text": "at the start" } ], "choices": [ { "text": "at the start" }, { "text": "in the middle" }, { "text": "towards the end" }, { "text": "when the reaction has reached completion" } ], "id": "765e43dd-bda2-f9c7-c950-16518cdade2a", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" } ], "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84", "meta": { "gradeStageId": "ks4", "subject": "Chemistry", "tier": "Foundation", "examBoard": "AQA", "grade": "Year 10", "topic": null, "unit": "Rate of reaction", "lessonCount": "11", "age": [ 14, 15 ], "gradeId": "age_14_15", "gradeStage": "GCSE", "lesson": "Concentration and rate: analysis" } }, { "id": "82766930-56b3-8310-6a31-21eb4fc0daf5", "type": "Resource", "resourceType": "Quiz", "title": "Concentration and rate: analysis", "quizType": "Exit quiz", "items": [ { "type": "Question", "question": { "text": "What does the gradient of the line on a graph of volume of gas produced against time represent?" }, "questionType": "Multiple choice", "correct": [ { "text": "rate of reaction" } ], "choices": [ { "text": "total volume of gas produced" }, { "text": "rate of reaction" }, { "text": "time taken for reaction" }, { "text": "concentration of reactants" } ], "id": "bd58908e-949e-62d6-9b87-f7c8e5401860", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" }, { "type": "Question", "question": { "text": "When plotting a graph of the volume of gas produced against time, how would the graph change if the concentration of the reacting solution is increased?" }, "questionType": "Multiple choice", "correct": [ { "text": "The graph would be steeper." } ], "choices": [ { "text": "The graph would be less steep." }, { "text": "The graph would be steeper." }, { "text": "The graph would not change." }, { "text": "The graph would be a constant horizontal line." } ], "id": "5b2fc329-ec31-2e0b-d575-d9a2cc223f39", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" }, { "type": "Question", "question": { "text": "What happens to the rate of reaction when the concentration of reactants is doubled?" }, "questionType": "Multiple choice", "correct": [ { "text": "It doubles." } ], "choices": [ { "text": "It stays the same." }, { "text": "It halves." }, { "text": "It doubles." }, { "text": "It quadruples." } ], "id": "ff854ca0-dd90-9eb0-a5ae-5f26a8f88964", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" }, { "type": "Question", "question": { "text": "How can the gradient of the curved section on a rate of reaction graph be determined at a specific point?" }, "questionType": "Multiple choice", "correct": [ { "text": "by drawing a tangent to the curve at that point" } ], "choices": [ { "text": "by calculating the area under the curve" }, { "text": "by drawing a horizontal line" }, { "text": "by drawing a tangent to the curve at that point" }, { "text": "by measuring the distance between two points on the curve" } ], "id": "62442f8c-b0fc-7dac-53f7-9f4d49146459", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" }, { "type": "Question", "question": { "text": "If you use the same mass of magnesium and react it with different concentrations of excess acid, how will this affect the volume of gas produced?" }, "questionType": "Multiple choice", "correct": [ { "text": "The same volume of gas will be produced for different concentrations of acid." } ], "choices": [ { "text": "Different volumes of gas will be produced for different concentrations of acid." }, { "text": "The same volume of gas will be produced for different concentrations of acid." }, { "text": "No gas will be produced for high concentrations of acid." }, { "text": "Higher concentrations will produce higher volumes of gas." } ], "id": "97712eb8-360e-61bf-a395-c778959fa022", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" }, { "type": "Question", "question": { "text": "If two experiments produce the same volume of gas but one reaction takes less time to complete, what can you say about the concentration of reactants?" }, "questionType": "Multiple choice", "correct": [ { "text": "The reaction that took less time had a higher concentration of reactants." } ], "choices": [ { "text": "The reaction that took less time had a higher concentration of reactants." }, { "text": "Both reactions had the same concentration." }, { "text": "The reaction that took more time had a higher concentration of reactants." }, { "text": "Concentration does not affect reaction rate." } ], "id": "1feba638-b1ba-3194-3a76-7c148035c240", "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84" } ], "lessonId": "b6d64f9f-a9eb-30d9-3de8-5721f6b2bd84", "meta": { "gradeStageId": "ks4", "subject": "Chemistry", "tier": "Foundation", "examBoard": "AQA", "grade": "Year 10", "topic": null, "unit": "Rate of reaction", "lessonCount": "11", "age": [ 14, 15 ], "gradeId": "age_14_15", "gradeStage": "GCSE", "lesson": "Concentration and rate: analysis" } } ] }