Lesson video

Hello, my name is Mrs. Collins and I'm going to be taking you through the learning today.

This lesson forms part of the Unit Industrial Chemistry and is called Factors affecting equilibrium.

During today's lesson, you will explain how changes in pressure, temperature and concentration, as well as addition of catalysts, affect the equilibrium position of a reversible reaction.

Here are the keywords for today's lesson.

Le Chatelier's principle, dynamic equilibrium, equilibrium position, and reaction rate.

Here are those keywords in context.

Now, I suggest you pause the video here, read through those explanations and make notes of anything that you are unsure of.

We will emphasise those during the session, but you do need to make sure that you understand what those words mean before we start.

Today's lesson is divided into two parts.

Le Chatelier's principle and concentration, and then pressure, temperature and catalysts.

Now, some elements of today's lesson will be challenging, but I'll help support you through the process.

So let's begin with Le Chatelier's principle and concentration.

Some reactions can go forwards and backwards, and we can see that in this equation here.

So the forward reaction, A and B are reacting together to form C and D.

And in the backward reaction, C and D are reacting together to form A and B.

And we can see it's a reversible reaction because it has the double-headed arrow in the centre.

So these are called reversible reactions and under the right conditions and in a closed system, remember they can reach a state where the rates of the forward and backward reactions are equal.

This is called dynamic equilibrium.

When a reaction has reached dynamic equilibrium, the concentrations of reactants and products remain constant.

Remember, that doesn't mean they're the same, they're just remaining at a constant level.

They're not the same as each other.

The reactions have not stopped.

They simply occur at the same rate.

So the forward reaction is happening and the reverse backward reaction is also happening, and they're happening at a constant rate.

Here's a question based on that learning.

When a reaction reaches dynamic equilibrium, which of the following are true? So pause the video here, answer the question, and I'll see you when you're finished.

Welcome back.

So hopefully you've recognised that the rates of the forward and the backward reaction are equal to each other, so well done if you've got that correct.

Henry Louis Le Chatelier was a French chemist in the late 19th and early 20th centuries.

He's best known for his contributions to understanding chemical equilibrium.

His work helps chemists predict how a changing conditions affects the equilibrium position.

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the equilibrium position shifts to counteract the change.

So here's an example.

We've got a forward reaction and we've got a backward reaction.

So if we influence something that increases the rate of the forward reaction, the equilibrium position will shift to counteract that change.

So in this case, if we increase the forward reaction, the equilibrium position shifts to counteract the change and increases the backward reaction in response.

When the equilibrium position shifts, we are referring to the concentrations of the reactants and the products at the new equilibrium relative to the original concentrations, IE before the change in conditions.

So here's a question based on that learning.

According to Le Chatelier's principle, what happens to a dynamic equilibrium if it's disturbed by a change in conditions? So pause the video here, answer the question, and I'll see you when you're finished.

Welcome back.

So the answer to the question is the equilibrium position shifts to counteract the change.

So well done if you've got that correct.

Le Chatelier's principle states that if you increase the concentration of reactants in a system equilibrium, the equilibrium shifts to produce more products.

So here's an example.

So we've got our A, B, C, and D.

So increase A and B, equilibrium will move forward to increase the amount of products, so we reach a new equilibrium.

So the size of the reactants in this image represents their concentration so increase the concentration of A and B, and you can see the concentration of C and D will increase as a result of that.

So the equilibrium position shifts to the right, so it's shifting towards the products.

This produces more products and reduces the concentration of the reactants.

So if we've increased the concentration of the reactants, the equilibrium shifts to reduce the concentration of those reactants, and that in turn increases the products.

More products will be formed until equilibrium is reached again.

Le Chatelier's principle states if you decrease the concentration of reactants in a system at equilibrium, the equilibrium shifts to produce more reactants.

So here we are reducing the reactants and you can see that the equilibrium is shifting to increase those reactants again.

So the equilibrium position shifts to the left this time to produce more reactants, reducing the concentration of the products and more reactants will be formed until equilibrium is reached again.

So it depends what you want in a reaction.

If you want more A and B, then you need to remove A and B from the reaction, and then the reaction will shift to the left producing more A and B.

So we're going to consider in context now.

So let's consider it in terms of hydrogen and nitrogen and ammonia.

So this is the Haber process, remember? So this is the Haber process.

Industry uses this reaction to produce ammonia, which is used for a lot of things including fertilisers.

Here is the ammonia here.

So we want the reaction to shift in that direction.

We want to produce as much ammonia as we possibly can.

So in order to produce more ammonia, the equilibrium position must be shifted to the right so we want it to produce more of that.

We want to shift it in that direction.

We can do this either by increasing the concentration of nitrogen and hydrogen.

So that will push the equilibrium position towards ammonia, towards the right, or we could decrease the ammonia concentration by removing it, and that again would shift the reaction towards the right.

So here's a question based on that learning.

Nitrogen dioxide and dinitrogen tetroxide reach a dynamic equilibrium in a sealed container made of glass.

The equation for the reaction is shown.

What happens to the equilibrium position of this reaction if the concentration of the reactance is reduced? And why? So consider which way it will be shifted, first of all, and then consider why that's the case.

So take your time on this, really read through the information.

Pause the video here, I'll see you when you're finished.

Welcome back.

So hopefully you've realised it will shift towards the left if we reduce the concentration of reactants, so we're removing the reactants.

So the equation will shift, the equilibrium will shift towards those reactants to try and replace them.

So the equilibrium shifts to increase the concentration of reactants.

If we decrease the concentration of reactants, then the reaction will shift, the equilibrium will shift to reverse that, to increase the concentration of the reactants.

So well done if you've got that correct.

Increasing the concentration of reactants increases the reaction rate of the forward reaction.

More reactant particles lead to more frequent successful collisions, increasing the likelihood of a reaction taking place.

So remember this from your rates of reaction work.

If we increase the concentration of reactants, there will be more frequent successful collisions, okay.

So here's a question based on that learning.

Nitrogen dioxide and dinitrogen tetroxide reach a dynamic equilibrium in a sealed container made of glass.

The equation for the reaction is shown.

what immediately happens to the rate of the forward and backward reactions if the concentration of reactants is reduced.

So we are looking at what happens immediately, not what happens in terms of equilibrium.

So we've reduced the concentration of reactants.

What happens to the rate of the forward reaction? What happens to the rate of the backward reaction? So pause the video here, consider your answer and I'll see you when you're finished.

Welcome back.

So the answer to this question is B.

So the rate of the forward reaction increases and the rate of the back reaction remains the same.

So well done if you've got that correct.

We are now moving on to Task A.

A revision website uses an analogy to help students understand dynamic equilibrium.

The analogy describes a person who runs up a down escalator, so they're going in the reverse direction of the escalator.

The person remains in the same position.

A, what aspect of dynamic equilibrium does this analogy help to explain? And B, what aspect of dynamic equilibrium does this analogy not help to explain? So think about basically the advantages and disadvantages of this as a model.

Pause the video here, answer the question.

I'll see you when you're finished.

Welcome back.

So let's have a look at the answer to that question.

So A, this analogy helps to explain that when the forward and reverse reaction rates are equal, there is no overall change.

This results in no observable change.

So we see the person at the same position on the escalator.

And then B, this analogy does not help to explain what happens at the submicroscopic level.

So the dynamic aspect of equilibrium, so individual molecules are changing.

There are a mixture of reactants and products at dynamic equilibrium.

So it's not showing what's happening on a particle level, it's not showing what's happening to the molecules.

It's looking at the overall reaction.

So well done if you got that correct.

Moving on to question two.

So Alex adds hydrochloric acid to an aqueous solution of compound A.

The word equation for the reaction is shown.

He notices that the reaction mixture turns and stays green after a while.

He concludes that dynamic equilibrium has been reached, and both compound A and B are present in the solution.

So you're going to need to use information from the equation to answer this question.

So A, the concentrations of the compounds in this reaction do not change a dynamic equilibrium.

Explain why.

And B, how can the equilibrium position be shifted to make the reaction mixture more yellow? So pause the video here, really consider your answers and I'll see you when you're finished.

Welcome back.

So 2A, the forward and reverse reactions are taking place at the same rate.

And then for B, add more compound A or hydrochloric acid, that will shift the equilibrium to the right.

Or removing water shifts equilibrium to the right.

As the system will try and replace the removed product by producing more compound B and water.

Remember, we don't want to remove compound B because that's what's yellow.

We need that to stay in the solution.

So take care, you need to really think carefully when answering questions like this.

So removing one of the products, but not both of the products.

So well done if you've got that correct.

Now we're moving on to the second part of the lesson.

Pressure, temperature and catalysts.

So one mole of any gas occupies the same volume under the same temperature and pressure conditions.

So one mole of anything, remember, contains 6.

02 times 10 to the 23 things.

So one mole of carbon atoms will contain 6.

02 times 10 to the 23 carbon atoms. One mole of carbon dioxide molecules will contain 6.

02 times 10 to the 23 carbon dioxide molecules.

And we've got an example here of nitrogen, water and argon.

So changing the number of moles of gas in a fixed volume will change the pressure.

Now, by fixed volume we mean the volume remains the same.

So we're gonna show that using a diagram.

So here we've got one mole of gas in a container and that's exerting a pressure on the container.

So as the particles hit the surface of the container, they exert a pressure.

If we add another mole of gas, but we don't change the volume, it will double the pressure as there are twice as many particles colliding with the walls of the container.

And you can see that in these two animations.

So if you need to, just pause the video to see what's happening.

Changing pressure affects only equilibrium where each side that's the reactants in the products, has an unequal number of moles of gas.

So firstly, it only works where each of the reactants in the products are gaseous or gaseous, and it only works if there's an unequal number of moles of gas in that equation.

So let's have a look at an example.

So here, remember, one mole of any gas has the same volume at the same temperature and pressure.

So we've got carbon monoxide reacting with hydrogen to produce methanol.

We've got one mole of carbon monoxide reacting with two moles of hydrogen to form one mole of methanol.

And this is a reversible reaction.

We can see that because of the symbol in the centre.

So here are the reactants, here are the products.

And we can see there are three moles of gaseous reactants and there's one mole of gaseous products.

So it's a definitely an unequal number of moles of gas.

And you need to think carefully about what would happen if we increased or decreased the pressure.

Which direction would it push the equilibrium? So the reactants would have a pressure three times greater than the products because we've got three moles of gas on the reactant side, but only one mole of gas on the product side.

Remember back to those diagrams showing pressure of gas inside the same volume.

And remember, we're in a fixed volume here.

So according to Le Chatelier's principle, the equilibrium position shifts to minimise the effect of pressure change.

So increasing pressure favours the side with fewer gas molecules.

Decreasing the pressure favours the side with more gas molecules.

So here's an example.

We've got nitrogen reacting with hydrogen to form ammonia.

So one mole of nitrogen reacting with three moles of hydrogen to produce two moles of ammonia.

In order to produce more ammonia, the equilibrium position must be shifted to the right 'cause we want more ammonia.

So high pressure favours the production of ammonia because there are fewer gas molecules on the right-hand side, so it forces or shifts the equilibrium position to the right.

Here's a question based on that learning.

So nitrogen dioxide and dinitrogen tetroxide reach a dynamic equilibrium in a sealed container made of glass.

And here's the equation for the reaction.

What happens to the equilibrium position of this reaction if the pressure is increased? And why? So what I'd like you to do is pause the video here, really consider your answer, really think about how many moles are on each side of the equation, and then I'll see you when you're ready.

Welcome back.

So the equilibrium will be shifted to the right, and that's because there are fewer gas molecules on the right-hand side.

So well done if you've got that correct.

As one direction of a reversible reaction is exothermic and one is endothermic, changes in temperature affect the equilibrium position.

So here we've got the endothermic direction for these two reactions, and we've got the exothermic direction.

So remember, if one direction is exothermic for reaction, the other direction must be endothermic.

Le Chatelier's Principle states that equilibrium position shifts to counteract the temperature change.

So increasing temperature favours the endothermic reaction and it absorbs heat from the surroundings.

The temperature of the surroundings will decrease.

Decreasing the temperature favours the exothermic reaction because it releases heat into the surroundings and the temperature of the surroundings will increase.

So you can see it's counteracting the change.

So if we increase the temperature, the endothermic reaction will decrease the temperature in response.

If we decrease the temperature, the exothermic reaction will increase the temperature of the surroundings in response.

Consider the reaction of sulphur dioxide and oxygen to make sulphur trioxide.

So we've got the equation there, it's a gaseous equation.

This is the contact process, remember, industry uses this reaction as a way to produce sulfuric acid.

The forward reaction is exothermic.

So lower temperatures favour the production of the product, SO3, as the equilibrium position is shifted to the right.

Remember the forward reaction is exothermic.

So the reverse reaction must be endothermic.

We want that forward reaction.

We want it to go in that direction.

So we need to lower the temperature to do so.

Here's a question based on that learning and we are using that same equation.

You'll notice all the way through.

So nitrogen dioxide and dinitrogen tetroxide reach a dynamic equilibrium in a sealed container made of glass.

And there's the reaction.

The forward reaction is exothermic.

What happens to the equilibrium position of this reaction if the temperature is increased? And why? So pause the video here, answer the question, and I'll see you when you're finished.

Welcome back.

So this shifts the equilibrium position to the left because increasing the temperature favours the endothermic reaction.

And the endothermic reaction, remember is the reverse reaction.

Catalysts speed up the rate at which equilibrium is achieved.

So we've got examples of catalysts there in the contact process, which is the one that produces sulphur dioxide and eventually hydrochloric acid.

And the catalyst iron in the Haber process reaction, which produces ammonia.

Both the forward and the backward reaction rates speed up equally.

So remember, a catalyst speeds up a chemical reaction without being used up in the reaction.

It provides an alternative pathway with a lower activation energy, and it's speeding up the forward reaction and the reverse reaction equally.

So they do not change the equilibrium position.

And this is often a question on examination papers.

So bear that in mind.

So they do increase the rate of reaction, but they don't change the equilibrium position.

Catalysts are crucial for increasing production efficiencies.

They make it much more efficient because the reaction rate is higher and you need less energy to start the reaction.

Here's a question based on that piece of work.

So nitrogen dioxide and dinitrogen tetroxide reach a dynamic equilibrium in a sealed container made of glass.

There's the equation for the reaction.

What happens to the equilibrium position of this reaction if a catalyst is used? And why? So pause the video here and answer the question.

I'll see you when you're finished.

Welcome back.

So the answer is stays the same because catalysts only affect the reaction rate, not the equilibrium position.

So well done if you got that correct.

Here's a follow-up question.

What happens to the rates of the forward and backward reactions if a catalyst is used? This time, we are thinking about the rate of the reaction.

So choose the correct line in that table, pause the video here, answer the question, and I'll see you when you're finished.

Welcome back.

So the answer to that question is A.

Both reaction rates are increased and remember they're increased by the same amount, so well done.

Changing pressure and temperature also affect the reaction rate.

So if you increase the temperature of a reaction, the reaction rate will increase.

Remember, 'cause we are having more successful collisions happening.

So increasing the temperature increases the reaction rate because it increases the energy of the particles and there are more frequent successful collisions.

So remember that back from your rate of reaction work.

Here's a diagram showing pressure.

So we're increasing the pressure, this time we're using the units kilopascal, and we've got the reaction rate there.

So increasing the pressure of the gases increases the reaction rate by forcing gas particles closer together so there are more frequent successful collisions.

Here's a question based on that learning.

Nitrogen dioxide and dinitrogen tetroxide reach a dynamic equilibrium in a sealed container made of glass.

There's the equation for the reaction.

What happens to the rates of the forward and backward reactions if the temperature is increased? So think very carefully about.

We're talking about the rate here, not the equilibrium position.

So pause the video here and answer the question.

I'll see you when you're finished.

Welcome back.

So the answer is C, increases.

So it increases the rate of the forward and the rate of the reverse reaction, as well as having an impact on the equilibrium.

So well done if you've got that correct.

Here's a follow-up question.

What happens to the rates of the forward and backward reactions if the pressure is increased.

So think carefully about this.

We are thinking about the rate of reaction.

Remember not the equilibrium position.

So pause the video here, answer the question and I'll see you when you're finished.

Welcome back.

So the answer is C.

It increases both the forward and the backward reactions, the rate of those reactions, and that's because the particles are closer together so there are more frequent successful collisions.

So well done if you got that correct.

We're now moving on to Task B.

A class is discussing the decomposition of hydrogen iodide gas at high temperatures.

The equation for the reaction is shown.

Jun writes the following passage, "Increasing the pressure will shift equilibrium position to the right favouring the production of H2 and I2 because increasing pressure always shifts the equilibrium position to the side with the fewer gas molecules." So is Jun correct? If not, rewrite his answer making any corrections.

So pause the video here, answer the question, and I'll see you when you're finished.

Welcome back.

So Jun is not correct.

Increasing the pressure has no effect on the equilibrium position because there are equal moles of gas molecules on each side.

So we've got two moles of gas on one side and two moles of gas on the other.

Therefore, the equilibrium position will stay the same.

So well done if you've got that correct.

We're now going to have a look at questions two and three.

Sophia reads Jun's passage and adds the following.

"Adding a catalyst will increase the rate of the forward reaction, meaning more products will be formed." So is Sophia's statement correct? If not, rewrite the answer making any corrections.

And then three, their teacher later reveals that the forward reaction is exothermic.

How could the equilibrium position be manipulated to produce more product? What might be the problem with your suggestion? So pause the video here and answer the question.

Welcome back.

So let's have a look at the answers.

So Sophia is not correct.

Adding a catalyst increases the rate of both the forward and the backward reactions equally.

A catalyst does not change the equilibrium position.

The equilibrium position remains the same.

So well done if you've got that correct.

And then three, what they could do is lower the temperature, this would shift the equilibrium position to the right and more products will be formed favouring the exothermic reaction.

But do note that decreasing the temperature lowers the rate of reaction.

That means it will take longer to reach equilibrium and produce the product.

So we want to lower the temperature because we want it to go in the exothermic direction.

But the problem is, is that also reduces the rate of the reaction.

So well done if you got that correct.

Here is a summary of today's lesson.

Catalysts, pressure, temperature, and concentration of solutions can affect equilibrium position of a reversible reaction.

If the conditions of a reversible reaction are changed, the equilibrium will move to counteract that change.

A catalyst increases the rate at which equilibrium is achieved.

Increasing temperature increases the rate of both reactions.

Equilibrium shifts in favour of the endothermic reaction.

Increasing the concentration or pressure of a substance increases the rate of reaction, and equilibrium shifts in response.

So well done, that was quite a challenging lesson, but I hope you've learned a lot from it.

Thank you for joining me.