J2026 A LEVEL CHEMISTRY SECTION A

[00:00:00]All right, it's Niyaki. It's a Niyaki osteo zone and today we are going to have the June 2026 A level chemistry from the Zimse sec exam board as you can see on the screen. Right, so today we simply going to dissect this paper step by step and if you're doing Zimse, you need to pay attention to these instructions. Right, we're required to answer total of six questions and we simply going to revise section A. So if you haven't checked on our playlist, you need to do so. We have done number one, we have done number two and today we're simply going to focus with number number three. So always remember always remember to subscribe so that you'll be notified whenever whenever we post right. So number three says sodium theosulfate disproportionates in acidic medium according to this equation. So we are having this equation is the disproportionation reaction. So disproportionatial reaction is defined as a redux reaction in which the same element the same element is simultaneously being oxidized and reduced simultaneously oxidized and reduced forming two products two different products of different oxidation states right forming two different products of different oxidation state right so having simultaneous oxidation and reduction of the same element giving us two different products with two different oxidation oxidation states that's what you the disproportionation reaction. Right? So we are having the disproportionation reaction of the sodium theosulfate in the acidic medium.

[00:01:18]Right? So you also need to know the disproportionation reaction of chlorine with cold sodium hydroxide and also with what sodium hydroxide. You must know this one also by heart right together.

[00:01:30]And then here we're having that one of sodium theosate. Right? So you should know the oxidation states of sulfur in the ion. Right? So in sulfur we are having the average oxygen state from our normal calculation using the O level method. So we're having this one the sulfur and then we having three oxygen atoms and then two minus. So we have this one as X and then here we're having 2X and then here it's -2. So it's plus -2 by 3 to give us -2. Right? From our general procedure to calculate the oxision state. Right? So here we're having 2x being - 6 being -2. and then is 2x= 6 - 2 2x = + 4 and then x is equal to 2 + 22. So this plus 2 is the average oxidation state of sulfur according to our normal calculation of the oxidation state. Right? So I want you to take note on that in sulfa we are not having this one is the the oxidation state of sulfur right? So in sulfa we are simply going to have a different approach. Why? Because of the difference in the geometry the difference in the bonding right we are simply going to have this one is the the theosoph is the theosoph theosophit ion right? So this one is the dot cross of the the iron, right? So as you can see we're having this one. This one is the terminal sulfur, right? Terminal terminal sulfur and this one is our central our central sulfur, right? So the terminal sulfur has got an oxidation state of minus1. And then this one the central has got an oxidation state of + 5 to give us the overall average here which of four. So it's minus1 + 5 and then we simply going to have this +4 all together. So this is what you must you must bear this in mind that in sulfur we're having the terminal and the central sulfur having different oxygen states. So the one is being reduced here. So this one the minus one is being oxidized to give this sulfur and then the plus 5 is being reduced to give this one with the plus with the plus4 together. So this is what you call the disproportionation within the theosulfate theosulfate ion. We are having the terminal sulfur or the peripheral sulfur being oxidized to this sulfur zero oxygen state and then this one reduced to sulfur in the sulfur dioxide with the plus plus oxidation state. So oxidation is simply defined as the as the increment in the oxidation number right and then reduction is being divided the decrement in the oxidation oxidation number. You must know these two by hearts are are you together? So let us simply proceed at the next part.

[00:03:41]So here we have explained how the disproportionation within the theosate ion is simply going we are simply going to have it right. So here we go to the first part the table shows the results of the kinetics of the reaction right so we having experiment number one experiment number two number three and then the concentration of hydrochloric acid the acidic medium and then we having that one of the theosulfate and then the time the time taken away together. So here we are simply going to be required to state any two methods of determining the rate of this of this reaction right. So this reaction is of importance when it comes to reaction kinetics. You must know this reaction by heart. Right? So here we're going to have the production of sulfur the yellow precipitate the yellow precipitate of sulfate. So we're still going to have a cross beneath the container or the conical flask in which we have to carry out this this reaction. So here we are still going to produce this solid sulfate in the yellow precipitate. So to measure the rate we simply going to measure the time taken to obscure to cover this the sulfur which is being produced all together. So this is the best method we are simply going to to use the disappearing cross method. Are you together? So you must know this one.

[00:04:44]You must know this experiment by heart.

[00:04:45]So it is called the disappearing disappearing cross disappearing cross method. Are you together? We simply going to have the cross beneath the conical flask. And then we're simply going to have the sulfur which is being produced covering this this cross. Are you together? So it is called the disappearing cross method. And then also we simply going to have the production of this gas. So we're simply going to measure the time taken to produce to reach a certain volume of the gas. Are you together? So simply going to have time against the gas in the syringe, right? So we're simply going to have time against gas in the in the syringe. Are we together? So this is another method we simply going to use to measure the the rate of the reaction. Are we together?

[00:05:22]So let us proceed to the next part. So the next part says determine the rate the rate equation. Right? So we're required to determine the rate the rate equation. Right? So rate equation also known as the rate the rate law is easily defined as the mathematical equation that shows how the rate how the rate of a reaction depends on the concentration of the reactant right so how the rate of a reaction or of a chemical reaction depends on the concentration of the reactants are together of the of the reactants right so this is what you call the rate equation or the rate law and then we also have another term which needs to be defined the Order of the reaction. So the order of the reaction is defined as the exponents of the concentration in the rate law. Right? So here we're simply going to have the exponents the powers right the exponents of the concentration or the reactants in the rate law right or in the rate equation in the rate law rate equation. Right? So here we're simply going to have let us say this one is our rate equation rate is equal to K and then we're having the concentration of A to the M and then concentration of B to the N. So m and b the orders of the reaction the exponents to which the reactants are being raised in the in the rate law right and then we also have what we call the rate constant. So this one k is our rate constant. Right? So rate conant right?

[00:06:39]So rate constant. So rate constant is simply defined as the proportionality constant in the rate law that relates the rate to the reactant concentration.

[00:06:47]Right? So it is a proportionality proportionality constantity constant in the rate law in the rate the rate law that relates reaction the rate of a reaction to the reactant concentration the reactant concentration are together. So you must know the definition of the rate constant, the order of reaction and then also the rate equation. You must know these definitions, these three definitions by heart, right? You must know these three by by heart together. So let us now go back to the question. We are being examined on the rate the rate equation.

[00:07:28]Right? So we're given this table and then we simply going to be examined on the rate the rate equation. Right? So here we're simply going to assume that in both in all these three experiments we're simply going to have the same amount of sulfur which is being produced. Right? So that's the first assumption we are simply going to to have right obviously the rate of the reaction is now simply going to be affected by the by these two right by these two concentrations right so here we are going to have the first and the second experiment whereby we're simply going to assess on this one hydrochloric the hydrochloric acid all together so here we having 0.1 and then here we're having 0.2 2 meaning say we've doubled the concentration and then in terms of the time we're simply going to have the time so it's 63 right over over 30 32 right so here we go straight to the calculator we go straight to the calculator so 63 / 32 we're going to have 1.96 875 is approximately equal to approximately equal to two right if you round one number it is equal to to two together so this would mean to say doubling the conant concentration doubling the concentration of this hydrochloric acid doubles of the reaction, right? Doubles the the rate of the reaction. So the rate the concentration is directly proportional to the rate. So this one it implies for the first order reaction whereby we're simply going to have the rate directly proportional directly proportional to the concentration of the reactant. So here we're having direct proportionality. I'm going to say this one is a first order first order reaction. Right? And then we go on to that one of the theosophion. We are simply going to have this one and this one. Right? And then we take this one again and in this one. Right? So it's simply going to here is 0.1.01 and then here is 0.05. So this one has got increased five five times. Right?

[00:09:13]And then simply going to have 63 again divided by 13. Right? We're simply going to have 63 divided by 13. So here it's 63 divided by 13. And then we are going to have um 4.8 8 4 6 1 5 3 8 4 6 right so here this one is approximately equal to five if you round to the nearest one number so here we have increased five times and then also here we are simply going to have the rate increasing fivefold right so the rate here is increased fivefold so we simply going to have again direct proportionality we to say this one also is in first order first order reaction are we together so the rate law the rate equation we are simply going to have the rate being equal to the rate constant the concentration of the proton to the one and then we're simply going to clear here and then we're going to have the concentration of the the sulfate ion this one to the two right raised to the power to the power one so here these are the first order first order reaction so you should know the properties of a first order reaction you should know the properties of a first order reaction so in terms of the first order reaction we are simply going to have direct proportionality right that's first in the major principle the major effect you must know direct proportionality right so if you to double the concentration.

[00:10:25]We're simply going to double the rate.

[00:10:27]Triple the concentration, triple the rate. Halfing the concentration, halfing the rate. Right? Increasing five times here, increasing fivefolds the rate. Are we together? You must know that we're having direct proportionality from the first order first order reaction. Are we together? And then we go on to the next where we are going to what you call the second the second order. So if you have the second order reaction, we are simply going to have the rate being proportional to the square of the concentration. So the rate in the second order is proportional proportional to the square of the concentration to the square of of the concentration. Right? So this one is what you call the second order reaction. Are you together? So if you want to double the concentration, we are going to square the double to four four times. Right? So if to double double is increasing two times. So we're simply going to square. And then if we are to um triple, we're simply going to increase nine times away together. So you must know the second order by by heart. You must know the second the second order by by heart. Right? So this one it quadruples. Right? Meaning to say increase four times, right? And then we now move on to the next to the next part. So we have done the rate equation.

[00:11:31]And then we are now moving on to the next the next part. Are we together? So the next part says a volume of 25 cm of saturated solution of gypsum. This one was run through an ion exchange resin.

[00:11:45]Right? A reasonable quantity of water was added to the washings and then the wash requires about 35 cm of 0.1 moles per cubic decimeter sodium hydroxide.

[00:11:56]The chemical reactions that occur as shown in the equations. So this one is the exchange reaction and this one is the wash reaction. Now together you must know these two by heart. Right? So here we are simply going to have the ion exchange system. Right? So you must know the ion exchange system using these reasons. Right? So the first thing we should be able to define what is meant by the term the ion exchange ion exchange process right. So ion exchange process must be able to define the ion exchange process. So it is simply a process in which ions in a solution are being exchanged with ions attached to an insoluble resin. Right? So ions in a solution in a solution are being exchanged being exchanged to ions to ions attached to an insoluble resin right to an insoluble insoluble resin. So this one is the simple definition of the ion exchange process. So you want to remove specific ion from the solution. So here we want to remove the the calcium ions right. So we want to remove these calcium ions from the solution. So we're simply going to have the rein containing two hydrogen atoms, two hydrogen ions, the protons so that we can easily maintain the electron usually. So the calcium ion is two plus. So we're simply going to have the rein attached to two hydrogen atoms here. So if we to have aluminium 3+, we simply going to have H3 and then we have the rein. And then if we to have sodium, we were simply going to have one because it's plus one. So we need to replace this + one with a proton of plus one. And then this plus three with three protons. this plus two with these two protons. Are we together to maintain electrical neutrality? Are we together? So this is the the ion exchange process. Right? So you must know these ones. You must know this one by by heart. Are we together? So here we're simply going to remove this calcium the calcium ions. Right? And then we're simply going to replace it with the two the two protons. Right? Are we together? So here from calcium sulfate we are simply going to have this calcium ion the sulfate ion. So the sulfate ion will then combine with these two protons to give the sulfuric acid.

[00:13:47]Are we together? and then sulfic acid will then react with the sodium hydroxide and then we can easily quantify the amount of the calcium sulfate all together. So this is how the ion exchange is simply going to to occur. We're simply going to have the resin catching the calcium ions like we have alluded here and then we going to have the the sulfate ions being left in the solution and then they will combine with this two protons and then we're going to have the sulfuric acid which will then react with the sodium hydroxide all together. So this is how we going to have the ion exchange. So we want to remove these calcium ions from the solution together we want to remove them from the solution right in exchange of the ions attached to this insoluble resin right so this is what you call ion exchange right so let me clear here so that we can easily go back to the to the questions right so let me clear so that we can easily go to the the questions right so the first part says calculate the number of moles of sodium hydroxide so we are given the volume then we also given the concentration so it's N is equal to CF so here we're having C the concentration being equal to 0.1 by V which is 35 over 1,000. Here we're having the volume in cubic centimeters.

[00:14:49]So we want it to be in cubic decime.

[00:14:51]Right? So here we're simply going to straight to our calculator and then we are going to have 35 / 1,00 and then we multiply by 0.1 and then we are simply going to have 0.0035 moles. Right? So here we're having 0.0035 moles is the number of moles of sodium hydroxide. Are we together? So in standard form we can have it as 3.5 by 10 -3 moles right the number of moles of sodium hydroxide and then we go to the next part which says calculate the number of moles of the calcium sulfate in 25 of the saturated solution right so we want to calculate the number of moles of the sulfate ion. So here as you can see we having one mo of calcium sulfate and then one more of the sulfuric acid.

[00:15:34]So we want to have the number of moles of sulfuric acid that reacted with this one the sodium hydroxide. So here we've calculated that we're going to produce 0.0 035 moles of sodium hydroxide. So two according to this stochometrical ratios two moles here you react with with one mole. So what about 0.35?

[00:15:52]We're simply going to have it reacting with with less. So we're simply going to have 0.35 / 2 by 1 for us to calculate the number of moles of sulfuric acid that reacted.

[00:16:01]Right? So here we're having this one divided by by two. Right? And then we are simply going to have 0.00175 moles from the calculator. These are the number of moles of sulfuric acid that reacted. So the number of moles of sulfic acid that reacted are the number of moles of sulfic acid that were produced here. So these number of moles of sulfuric acid are being produced in the ratio one one is to one with calcium calcium sulfate. So the number of moles are the same. So the number of moles of calcium sulfate are the same as to those number of moles of thisic acid. And then we're going to have the answer being equal to again 0.001.75 moles. Right? So these are the number of moles or we can have it as 1.75 by 10 to the -3 right moles right so these are the number of moles of calcium calcium sulfate right and then we now move on to the next we now move on to the next part let me clear here so that you can have the next part more clear right so the next part says table 3.2 two shows the latis enthalpy of some ionic compounds.

[00:17:01]So latice enthalpy so it is the heat evolved when one more of a solid ionic compound is formed from from it separate gaseous ions. Right? So having this one the cation x in the gas space plus this ion the anion y in the gas space to give the solid xy the solid. So here you should take note of these physical states. This one is a gas. This one is a gas. This one is a solid together. One mole of a solid ionic compound. Right?

[00:17:25]This that's the definition of the latis latis energy right and then we're having sodium chloride we're having sodium chloride we're having magnesium oxide right so first we need to understand the factors which affect the latis energy right so for the factors we have two major factor we have two major factors so the first one is ionic the ionic charge right so it's ionic charge so this one is the dominant right so the higher the charge the stronger the attraction so the higher the charge the stronger the attraction the stronger the attraction Right? And then the second one, we're simply going to have the ionic size. So the ionic size, right? So in terms of the ionic size, we're simply going to have smaller ions having stronger attractions. Right?

[00:18:03]Smaller and then stronger. So S for smaller, S for stronger, right? Smaller ions are simply going to have stronger attractions. Right? So we're simply going to have the last one which is ion packaging. Right? So it's ion packaging, the structure, the latice structure.

[00:18:17]Right? Is the charge and then the size and then the packaging. So these are the three factors which affect the latis the latis structure. So here we say ionic ionic charge. So the higher the charge the stronger the attraction ionic size.

[00:18:29]The smaller ones you've got the stronger attractions. Right? So here we are required to explain the difference in the enthaly changes between sodium fluoride and sodium chloride. So here we're having the same sodium metal sodium metal. So what is changing? We still going to have the fluoride and the chloride ion. Right? So going to the the principles we said the smaller the smaller ones they've got the stronger attraction. So between florine and chlorine which one is smaller. So in the group seven the hallogens we are starting with florine we go to chlorine.

[00:18:56]So obviously we're simply going to have this one having a smaller grade compared to this one. This one is one sh smaller than the chlorine atom. Right? So simply going to have this one smaller and then we say the smaller ones they've got the stronger attractions. Right? So we're simply going to have the stronger electrostatic of attraction between the sodium and the fluoride ion than the sodium and the chloride ion together. So here we need to understand that the issue of charge doesn't contribute here.

[00:19:19]Why? Because the fluoride ion is in group seven and then the chloride ion is also in group seven. They both have the negative one here. So it doesn't contribute the issue of the of the charge. So the issue contributing here it is the ionic the ionic size. So the smaller the ion the stronger the attraction. they still going to have the stronger electrostatic force of attraction in the sodium fluoride than in the sodium chloride. That's why you're simply going to have a lot of energy being involved when these two are simply going to combine right to form one more of a solid ionic compound right and then we go to the next we go to the next one where we simply going to have sodium chloride and magnesium oxide. So here in terms of magnesium oxide we are simply going to have plus2 for magnesium and then minus2 for the oxygen. So here the issue of the charge contributes higher charges stronger attraction right so here is plus2 here it's minus2 and then here for the sodium is + one and then for the chloride is -1 right and also we're having this one the oxygen in the period 2 and then this chlorine in the period 3 meaning to say the oxygen is smaller than the than the chlorine one so here we're having oxygen being smaller than the chlorine one right so chlorine is larger oxygen is is smaller so this one is larger chlorine is is smaller right so the issue of size contributes smaller strong attractions Right? We're having this one smaller in higher charge mean going to have strong electrostatic force of attraction and also having the magnesium ion magnesium 2 plus being smaller than the sodium the sodium ion. So we're simply going also to have the shorter inter ionic distance in magnesium oxide compared to magnesium chloride we're having the smaller inter ionic distance right in magnesium oxide compared to this one magnesium sodium chloride. So here we going to have in lat structure. Therefore we going to have the strong electrostatic forces of attraction in magnesium oxide compared to the to the magnesium chloride. Are we together? So you must know these three factors by heart. You must know ionic charge. You must know ionic size and then the packaging. Right? So we are we are done. Right. So always remember always remember to subscribe so that you'll be notified whenever whenever we post. Right. All right. So this one is our Nyaki online tutotoring as you can see on the screen. So we specialize in sciences. We specialize in sciences. So we have pyramids, chemistry, physics, biology, combined science, mechanics and statistics for both the O levels and the A levels. Right? So we have our standard package, we have our premium package, we have our premium pro max package, right?

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