1.7 CHEMICAL EQUILIBRIA, EQUILIBRIUM CONSTANT & ACID-BASE EQUILIBRIA
- Chemistry at glance
- Jun 10, 2020
- 10 min read
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Hello everyone, how are you? Today, we are going to work on chemical equilibria. As always, I am going to explain it all in the easiest way possible with no difficult words. The aim is to teach you everything there is to know for your as-level exam without worrying about difficult words that are used in books. I am not a talented writer myself but I won't let this stop me from sharing my chemistry knowledge with others.
Ok, so you've been told before that reactants go to products and that all reactants are used up to form products. However, in reality, we usually have products and reactants being together because the reaction also goes in reverse from products to reactants. This reaction is called a reversible reaction, which means that it can go either way depending on the conditions used. We also use reversible arrows instead of just a single arrow for this reaction. I included an example on the photo above.
We have calcium carbonate which goes to calcium oxide and carbon dioxide. This is a reversible reaction (we know this by the arrow used), so both reactants go to products and products go to reactants.
When the temperature is kept constant, the products and reactants will reach a stage of equilibrium, when we will no longer see any changes happening in the reaction. This is called a:
Dynamic equilibrium - continuous reaction in both directions and the rate of the forward reaction is equal to the rate of the reverse reaction. (We can't see by eye that anything happens there, but actually, the products constantly change to reactants and reactants to products).
On the graph above, you can see that we keep using the reactants and forming products until they reach equilibrium; until they are balanced.
In some cases, we are going to have more products than reactants in the equilibrium mixture and that's when we say the reaction is complete (for as-level, we just assume it is a one-way reaction) or we might have more reactants than products (that's where the reaction doesn't happen and we might need to change the conditions used).
Remember that equilibrium changes when conditions change. To observe a dynamic equilibrium, all of the conditions need to be kept constant.
There is a relationship between the concentration of reactants and products at equilibrium and this is called the position of equilibrium.
As we mentioned before, when you change conditions, the reaction is no longer at equilibrium and here comes the:
Le Chatelier's principle- if an equilibrium is subjected to a change, then the position of equilibrium will shift to minimise that change.
What this means is, whenever you change the conditions, the position of equilibrium will move in such a way to restore the equilibrium ( to restore the balance between products and reactants). Every reaction has its own equilibrium, which is the balancing point between products and reactants.
And now we are going to have a look at the conditions that could be changed and how the position of equilibrium will change:
-->Change in concentration:
Here is an equation (photo above) of hydrogen and iodine reacting together in a reversible reaction to form hydrogen iodide. This reversible reaction is at equilibrium, however when we increase the concentration of, for example, hydrogen gas the position of equilibrium has to shift in such a way to minimise that change and restore equilibrium.
In this example, if we have more hydrogen gas, we need to make more HI, to use up this extra hydrogen gas. Therefore, the position of equilibrium moves to the right-hand side.
-->Change in temperature:
Here, the position of equilibrium (when we change the temperature) will depend on the reaction; whether it is exothermic or endothermic.
Endothermic: on the photo above I included an example reaction of carbon dioxide and carbon, making carbon monoxide. It is an endothermic reaction (positive enthalpy change) and when we increase the temperature, the reaction will want to lower that temperature to get back to equilibrium. During an endothermic reaction, we use up the heat (lower the temperature of surroundings), so that's what we want to do to restore equilibrium. Our endothermic reaction is to the right, therefore, the position of equilibrium shifts to the right.
Exothermic: this time we have a different reaction but the enthalpy change is negative now, suggesting it is an exothermic reaction. When we increase the temperature, the reaction wants to lower that temperature, and as we said before it can do it by using up the heat in an endothermic reaction. The forward reaction is exothermic therefore endothermic it to the left and that's also where the position of equilibrium will shift (to LHS).
--> Change in pressure:
Partial pressure: for example if we increase the pressure of iodine, the position of equilibrium will shift to the right-hand side, to increase the partial pressure of HI and restore equilibrium.
Total pressure: when we change the total pressure, the situation is different. You need to count the number of gas molecules on each side of the equation and the equilibrium will shift to the side with fewer gas molecules. In this example, to the right-hand side.
--> Catalyst:
Catalysts do not affect the position of equilibrium but the equilibrium is reached faster. In a reaction, both forward and reverse reactions are sped up.
Now, I've got an example for you, to practice the shifting of the position of equilibrium.
We have nitrogen and hydrogen gas forming ammonia.
Increase the concentration of a reactant- when we increase the concentration of any reactant, the reaction will want to use up this reactant and more ammonia will be produced. So, the equilibrium shifts to RHS.
Increase in total pressure- we have 4 molecules on the LHS and 2 molecules on the RHS. There are fewer gas molecules on RHS, and that's where the equilibrium will shift.
The decrease in temperature - this is an exothermic reaction (negative enthalpy change) and we want to increase the temperature to come back to equilibrium. During exothermic reaction we release heat and therefore we need to perform exothermic reaction to restore the equilibrium. This is to the RHS.
Equilibrium constant:
For your AS-level course, you need to be able to write an equation for equilibrium constant and you can see how it looks on the photo above. It is quite difficult to explain but when you have an equation, we divide the concentration of products at equilibrium over the concentration of reactants at equilibrium. Obviously, we have more than one product and more than one reactant and their concentrations need to be multiplied and raised to the mole ratio, just like in the photo above.
The equilibrium constant will be a constant number if the temperature is kept the same.
The units of equilibrium change depending on the mole ratio and we'll have a look at this when working on the examples.
Examples:
1. We have N2O4 as a product with mole ration 1 (don't need to include this mole ratio in the equation as it won't change anything) and NO2 as a reactant with mole ratio 2 (this is the power to which NO2 will be raised). Next are the units. The unit of concentration moldm-3, and now, however many you have powers, that many moldm-3 you need to write. So, at the top, I only have one and at the bottom, we have two (power is raised to 2). Cancel out some of the moldm-3 and you're left with moldm-3 but it is at the bottom, so when you take it up the mol becomes mol-1 and dm-3 becomes dm3 giving us mol-1dm3.
2. Again, we do exactly the same thing, put all the products at the top and all reactants at the bottom. See if they have any mole ratio (coefficients) in front of them. In this example, they don't. And now the units, we have two concentrations at the top and one at the bottom. Cancel out the ones you can and you're left with moldm-3 at the top, so you don't have to d anything else with it.
3. Lastly, you have products with a coefficient of 2 at the top and two reactants at the bottom. 2 x moldm-3 at the top and 2x moldm-3 at the bottom, they cancel out leaving us with no units.
--> If Kc is big- this means more products than reactants are present in the equilibrium mixture (the position of equilibrium is to the right). For example, when you have an endothermic reaction and we increase the temperature, we more the position of equilibrium to the right to lower the temperature and we produce more products than reactants. In this situation Kc is high.
--> If Kc is small - more reactants are present at equilibrium than products and the position of equilibrium is to the left. An example of this is when we increase the temperature in an exothermic reaction, the equilibrium moves to the left (in endothermic direction to use up this extra heat) and the value of Kc is small.
e.g.
On the picture above, there is also an example for you. From the text, you can find all initial moles and the final mol of ethanoic acid. Whatever the mol of ethanoic acid (product) you need to take away that much from each reactant's initial mol value. Like this, you get final moles ( initial - change = final) for all of them.
We're asked to find the Kc value, therefore, we need concentrations and not mol, but we are given volume, so we just need to use our well-known equation n=cv and rearrange to find the concentration. Our Kc equation will look like the one above with all products at the top and all reactants at the bottom. Substitute in the concentrations and put it all into your calculator to find the final answer. Lastly, find the units (just like we did it before).
--> Conclusion:
1. Take out information from the text - initial moles of reactants and final mol of product.
2. Whatever, the mol of product, take away that much from each reactant's moles.
3. Work out the final moles.
4. Convert the moles to concentrations using the volume and n=cv.
5. Write the Kc equation.
6. Calculate the Kc.
7. Find the units.
Acids and bases:
Acids - proton donors ( they donate H+ ions)
Bases- proton acceptors ( they accept H+ ions)
This donating and accepting happens in aqueous solutions and is called dissociation.
e.g. When hydrochloric acid is added to water, it dissociates into H+ and Cl- ions. When in water, HCl gives away H+ ions (donates H+ ions) therefore it is an acid.
e.g. When NaOH is added to water, it also dissociates into ions, Na+ and OH-. When a base is dissociated in water, it is called an alkali and OH- is a common ion of this dissociation.
--> Strong acids - completely dissociates into ions. If acid is strong, it is easier for it to donate H+ ions.
The overall equation looks just like the one with HCl. You have acid and then it dissociates into H+ (which is donated) and a negative ion.
e.g. HCl --> H+ + Cl-
Hydrochloric acid is a strong acid, so it completely dissociates into ions. What that means, is that there is no more HCl in the solution, just the ions. It is also worth to mention that this is no longer a reversible reaction (it goes to completion, for AS-level it is a one-way reaction). Concluding, in this reaction the concentration of HCl is equal to the concentration of H+ ions.
--> Weak acids- partially dissociate into ions. It is not easy for a weak acid to dissociate into ions, therefore, we have some acid and some ions in the solution. Because it is harder for it to dissociate, it will also not be able to donate H+ ions as much as a strong acid would.
e.g. CH3CO2H --> <-- CH3CO2- + H+
This time it is reversible because the reaction is not complete, the acid didn't dissociate fully. Here, the concentration of the H+ ions is smaller than the concentration of the acid.
Conclusion:
Strong acid - fully dissociates in aqueous solution;
Weak acid - partically dissociates in aquesous solution;
Concentrated acid - consists of small quantity of water and high concentration of the acid;
Dilute acid - consists of large quantity of water and low concentration of the aicd;
Strength - depends on how many H+ ions it can take or donate;
Concentration - depends on how many H+ there are in the solution;
pH scale:
pH scale- is used to see how acidic or basic a solution is using the concentration of the solution ( more exactly the concentration of H+ ions).
This scale was found by a Danish chemist as the concentration itself may be present over a wide range even to 1x10^-14. To avoid using such big numbers, he proposed a pH scale.
It goes from 1 (being the most acidic) to 14 (being the most basic). pH scale 7 is the neutral point and this is the pH of water.
Now, how do we relate concentration to pH?
pH= log[H+] where [H+] is the concentration of H+ ions in the solution.
If we increase the concentration of H+ (the solution is more acidic) the pH decreases and lower pH, means a stronger acid.
However, you might be given the pH of a given solution and asked to find the concentration, we use this equation:
[H+]= 10^-pH
Let's try some examples:
1. Calculate the pH using the following concentration.
We get the concentration of H+ ions and we just put it in the equation and then calculator. We get pH=1, which is a very strong acid.
2.Calculate the concentration of a solution with pH=6.5
Now, we use the other equation to find the concentration. Again put the number into the equation and then calculator and you get the answer.
3. Calculate the pH of 2x10^-3 moldm-3 HCl.
This time we are not given the concentration of H+ ions but of the acid but remembering what we said before about strong acids, we know that Hcl dissociates fully into ions. Therefore, the concentration of acid is the concentration of H+ ions.
Now, just put it into the equation and the pH is 2.7.
Neutralisation:
This is the last part of equilibria. A neutralisation reaction is when acids react with bases and they become neutralised by each other.
For example, when we react NaOH with HCl. They are both a strong acid and strong base and they will neutralise each other to form salt and water.
During a neutralisation reaction, we always produce salt and water. However, if we use a carbonate instead of the base, apart from the salt and water, we will also get carbon dioxide from the reaction.
Ok, this was a really long topic but if you're reading this, you made it. If you think you're ready for past paper questions go-ahead to my youtube channel and practice some with me (the link is at the top of this page). See you next time.
PS. Please remember, I am only a student, and as anyone, I can make mistakes. If you think you can see one, don't hesitate and comment (either here on on my youtube channel) Thank you!
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