Unit 2 October 2025 - AS Chemistry Edexcel - Dr Hanaa Assil

Añadido: 2026-05-12

[00:00:01]Hello, this is Dr. Hannah Oil and this is unit two of October 2025 uh chemistry fur at Excel international a level. So let's take a look at the questions and discuss the answers. The first question here was what color is the solution that forms when iodine dissolves in a hydrocarbon solvent? You should remember that when iodine dissolves in water, it's yellow. But when it dissolves in a hydrocarbon solvent such as hexane, it has a purple color.

[00:00:51]Which row shows the correct trends down group two?

[00:00:57]Well, let's look reactivity of group two. You should remember in group two, the reactivity increases going down the group. So, barium is much more reactive than calcium or magnesium.

[00:01:14]Solubility of hydroxides and solubility of sulfates.

[00:01:20]We should remember that hydroxides are small annions.

[00:01:26]A small annion with a small cation would be very strongly attracted. It would be less soluble.

[00:01:36]While with a big cation, it would be more soluble. So going down the group the solubility of hydroxides increases.

[00:01:47]But when we talk about carbonates or sulfates or nitrates going down the group the solubility decreases.

[00:01:58]The big cations will be very strongly attracted to the big annions like carbon sulfate or nitrate and uh less soluble in water.

[00:02:16]The gas nitrogen dioxide exists in equilibrium with another gas diit.

[00:02:24]What happens to the color of the equilibrium mixture when a catalyst is added? Remember that when we add a catalyst to a reversible reaction, it speeds up both forward and backward reactions. So actually when we add a catalyst there is no change to the color because it doesn't favor forward or backward reactions. And also remember it will have no effect on the yield of the uh forward reaction. So it will not give more N204 or less N204.

[00:03:08]Which isomer has the lowest boiling temperature? Well, let's see what determines the boiling temperature. You should be familiar with the fact that a boiling temperature is higher if we have more carbons. So the first thing we do the one with more carbons has higher boiling temperature. What if they have the same number of carbons? Then the one with less branching has higher boiling temperature. If it is something that can do hydrogen bonding, the one with more hydrogen bonding will be higher boiling temperature. And you should remember that the fact that a long chain has higher boiling temperature. That is because this three chain will have stronger interactions between the molecules. So more energies need to separate them. While if it is more branched then the branching will prevent the molecules from coming too close to each other. So less attraction between them. So here we're looking for the one with lowest boiling temperature. Lowest boiling temperature means if there are different number of carbons I'm looking for less carbons. If they're the same number of carbons, then I'm looking for the one with more branching that would have lower boiling temperature. So, let's take a look at what we have. We have 23 dimethile pentane. This is 23 dimethile pentane. 33 dimethile pentane.

[00:04:59]23 triile.

[00:05:05]and heptine. So first of all, if you count the number of carbons, they all have seven carbons. So there is no difference in the number of curves. So I'm looking for the one with lower boiling temperature. That means I'm looking for the one with most branching.

[00:05:22]So of course 23 trithile butane has more branching. So low west boiling temperature.

[00:05:38]Sodium chloride dissolves in water.

[00:05:40]Which diagram shows the arrangement of water molecules around a sodium ion and a chloride ion in solution? Remember, you should know that ionic compounds are soluble in water because when we dissolve them in water, the ions are surrounded by the water molecules. Water has two ends. The oxygen is slightly negative and the hydrogens are slightly positive. So around the positive ion, I should have the slightly negative end of water, which is oxygen.

[00:06:18]around the chloride ions, I should have the slightly positive end of water, which is the hydrogen. And remember that comparing the size of the ions, the chloride ion is something that has three shells or three energy levels. While the Na+ has lost its outer electron, so it has only two shells. So you should realize that the size of the chloride ion should be more than the size of the sodium ion.

[00:06:50]So which of these diagrams agrees with what we just said? You'll find that B is the correct arrangement. The smaller ion has the oxygen next to it. The larger ion has the hydrogens next to it.

[00:07:10]What is the name of the attraction between water molecules and these ions?

[00:07:16]Remember we said these are ionic compounds and they are surrounded by slightly positive end or slightly negative end. So this is actually an interaction between ions and a permanent dipole in the water.

[00:07:37]Which line on the sketch graph best shows the trend in boiling temperature for the hydrides of the elements in group six. Of course you know that group six are oxygen, sulfur, selenium, tallium and you should know that oxygen is very electro negative or compared to the others. So um a hydide with oxygen is actually H2O and the hydide of oxygen is actually water. Water is very polar and it has um the ability to do hydrogen bonding between its molecules.

[00:08:25]So water would have a much higher boiling temperature than the hydrides of the others because the electro negativity of sulfur and selenium and so on is very low. So going from water to the next one, water is much higher and then from H2S to H2S to H2 uh TE the molecule is getting larger. So stronger London dispersion forces. So boiling temperature increases. But with respect to water, we said water does hydrogen bonding between the molecules. This needs much more energy to be separated. So actually my graph is P.

[00:09:15]A mixture contains 36 g of water and 46 g of ethanol.

[00:09:23]What is the approximate percentage of water in the mixture by moles?

[00:09:30]So first of all we need to get the number of moles of each. So number of moles of course is mass over molecular mass. For water the molecular mass is 18. So this makes two moles.

[00:09:45]For ethanol the molecular mass is 46.

[00:09:49]You should be able to calculate Mr. So number of moles of ethanol is just one mole. And that means my mixture is made up of two moles of water and one mole of ethanol and we are required to get wet.

[00:10:06]The percentage of water in the mixture.

[00:10:10]So the total mixture has three moles.

[00:10:14]The percentage of water is the two out of the three times 100. So this is the percentage of water by moles.

[00:10:28]Hydrogen peroxide decomposes as shown 20 cm cubed of 2 mole per decime cube hydrogen peroxide solution completely decomposes at room temperature and pressure.

[00:10:44]What volume of oxygen is produced? So we are starting with what? 20 cm cubed of 2 mole per decime cubed hydrogen peroxide.

[00:10:54]Well I can get the number of moles. So number of moles of hydrogen peroxide.

[00:10:58]This is a solution. So that's concentration time volume. Don't forget to change the volume to decime cubed by dividing by th00and. So that gives me 0.04 mole of hydrogen peroxide.

[00:11:15]Now we are asked about what volume of oxygen. Looking at the equation 2 moles of H2O2 give 1 mole of oxygen that means the number of moles of oxygen is 0.04 / 2. Now whereas to get the volume at room temperature and pressure you should know that the volume is number of moles time 24 that gives 0.4.

[00:11:44]48 decime cubed but one in burns in oxygen as shown in a closed system. What volume of gas remains in the system when 20 cm cubed of but 1 in burns completely in 200 cm cubed of oxygen.

[00:12:14]Remember we said if we are dealing with gases we can deal with volumes directly.

[00:12:20]I don't need to change the 20 cm cube to number of moles and then compare. I can compare directly with volumes of gases.

[00:12:31]So I'm starting with 20 cm cubed of butane E.

[00:12:38]According to that balanced equation, the balanced equation says if I have one mole of butine, it will react with six moles of oxygen. That means if I have 20 moles we 20 * 6 that's we react with 120 cm cubed of oxygen and 1 mole of butine gives 4 moles of CO2 that means the 20 will give how much? The 20 will give 80 cm cubed of CO2. But how much oxygen did we add? We added 200. we used up only 120. So if he's saying what volume remains well what will remain would be the rest of the oxygen and the CO2 that was made. So the rest of the oxygen is 200 minus 120 because I used up only 120 plus the 80 of the CO2. Then the volume of the gas remaining is 160 cm cubed.

[00:13:51]Okay. Question 10 says ethanol can be oxidized by acidified dromate ions as shown. What is the change in oxidation number for the chromium atoms? And remember that when we talk about oxidation number, we talk about oxidation number of one atom. So if we have CR2, I'm looking for the oxidation number of just CR.

[00:14:19]So how do we calculate oxidation numbers? We have CR27 2 minus. Remember, we do not take into account any balancing when we're talking about oxidation number.

[00:14:34]We're just looking at CR27 2 minus.

[00:14:40]I'm trying to get the oxidation number of chromium. So, I say I have two chromiums that means 2x oxygen. Each oxygen is minus2. So, I have seven of them. That's -14. and I want the total to be overall minus2 and solve for x. So in that case x is + 6.

[00:15:03]So in the cr 207 to minus the chromium is + 6 and it changed to what? It changed to cr3 plus that means it went from + 6 to + 3.

[00:15:16]We're talking about the change. So what is the difference? It decreased by three.

[00:15:27]23 g of ethanol is oxidized by a solution of 0.1 mole per decime cubed of dromate ions. Which expression shows the volume or the minimum volume of this solution needed to oxidize all the ethanol as shown in the equation. So this is our equation and we are told we're starting with 23 g of ethyl.

[00:15:55]What is the volume um needed to oxidize all the eth? The volume of dromate which is 0.1 mole per decime cubed. What volume would be needed to oxidize the ethanol? Okay, so we have 23 g of ethanol. we get the number of moles of ethanol mass over molecular mass. This comes out to 0.5 moles. So look at the equation.

[00:16:26]The equation says 3 moles of ethanol react with 2 moles of dromate. That means the number of moles of the dromate will be 0.5 * 2 over3.

[00:16:42]That means I have 0.33 mole of dromate and I am told the concentration up there is 0.1.

[00:16:52]So the volume would be the number of moles over the concentration and that is 3.3. Now the questions or the options that I have are just showing what should I do. Okay, what did we do? We got 0.5 as number of moles of ethanol. Then we multiplied by 2 over3 not 3 over 2 over * 2 over 3 and then I divided by.1.

[00:17:23]Now if I divide by.1 that is the same as multiplying by 10.

[00:17:31]Do we understand? We're supposed to divide by the concentration.

[00:17:35]Concentration is 0.1. If I divide by 0.1 which is 10 to the minus1 I take it up it will be 10 to the power of one. So I it's as if I'm multiplying by 10.

[00:17:56]Question 11 says a student studies a graph of rate against time for both forward and backward reactions in a reversible reaction. The student makes these statements and we are asked how many of the statements made by the students are correct. So let's take a look at statement one.

[00:18:19]The reaction reaches equilibrium.

[00:18:23]Is that correct or wrong? Well, when the rate at the end becomes constant, remember that equilibrium means rate of forward reaction equals rate of backward reaction. Then this reaction is at equilibrium. So this is correct.

[00:18:44]The forward reaction is fast initially because the concentration of reactants is at its highest.

[00:18:52]That is correct for any reaction. When we start, the rate is fast because we have a lot of reactants, a lot of particles colliding at the same time.

[00:19:05]The rate of both reactions at equilibrium is zero. No, that's not right. We said at equilibrium, the rate of forward reaction equals the rate of backward reaction, but that doesn't mean it's zero. Actually it is not zero. So statements one and two are the ones that are correct. So only two of the statements are correct.

[00:19:32]The graph shows the change in concentration against time for a reaction.

[00:19:38]What is the best estimate for the value of the rate in mole per decimeter cube per minute at 4 minutes?

[00:19:48]Again, remember if we have a graph of concentration versus time and we're trying to get the rate, we do tangent at that time. So the rate at 4 minutes is the gradient of the tangent at 4 minutes. So we draw the tangent to the graph. We get the gradient that's y2 - y1 / x2 - x1 and so on. and that comes out to 0.15. That's how we get the rate from a concentration time graph.

[00:20:28]Then question 13 says, which test would not enable a chemistry student to distinguish between ethanol and ethanol?

[00:20:40]Well, ethanol is an alcohol. Ethanol, of course, is an aldahhide. So, we're looking for a test that would not be able to allow us to distinguish between them. Remember, for the test to distinguish between them, it should react with one and not the other. Or it should give the same results with both.

[00:21:02]That would allow us to distinguish. So the one that would not allow us to distinguish is if the test gives the same results with both or no reaction with both. So that's what we're looking for. So A says adding a very small piece of sodium metal to each liquid.

[00:21:25]You should know that out of these two the sodium metal would react only with the alcohol to give bubbles of hydrogen gas. So this would allow us to distinguish between them because the alcohol reacts gives bubbles of gas. The aldahhide no change or no reaction.

[00:21:46]So that's not our answer. B says adding sodium carbonate to each liquid.

[00:21:54]You should remember sodium carbonate reacts with what? Sodium carbonate actually reacts with acids. A carbonate would react with acid to give carbon dioxide gas.

[00:22:07]It would not react with either ethanol or ethanol. So when I do the test with sodium carbonate, there is no way of knowing which is which because no reaction with both. What about failing?

[00:22:22]You should remember that felling is a test for aldahhide. So it will react only with the aldahhide to give a brick red precipitate.

[00:22:34]No reaction with the alcohol. So that would allow us to distinguish between them. Adding PC5. Remember that the PCL5 will react only with the alcohol to give white fumes of HCl. it would not react with the aldahhide. So our answer is B.

[00:23:00]Okay. Question 14 says, which of the statements about CR bond O bonds found in aldahhides and ketones are correct?

[00:23:10]The first statement says the CR bond O bond is shorter than a single bond. That is actually correct. A double bond is usually or is always shorter than a single bond. The double bond O bond consists of two pi bonds.

[00:23:32]Well, you should know that any double bond is made up of one sigma and one pi.

[00:23:39]The sigma is a head-to-head overlap between S and P orbitals and the pi bond is a side to side or lateral overlap of P orbital. So actually that statement two is wrong. It is supposed to be made up of one sigma and one P. The C double bond O consists of four shared electrons.

[00:24:06]Yes, a double bond would be two electrons from the carbon shared with two electrons from the oxygen. So, statements one and three are correct.

[00:24:21]Question 15 says, which statement about these structures is correct? Okay, let's take a look at the options. A says all three structures show the same compound.

[00:24:36]In order to have to be the same compound, they have to have the same number of carbons, same number of hydrogen, same number of chlorines, same molecular formula, then they would be the same compound and all also the same name if they're the same compound. So let's see. Structure one has nine carbons.

[00:25:00]Structure two has 10 carbons. So they're definitely not the same. Structure three again has nine carbons. So to say that all of them are the same compound, that is wrong. B says structures two and three show the same compound. Well, two and three. No, because they have different uh number of carbons, different number of hydrogens.

[00:25:26]Structures one and three show the same compound. Well, one and three both have the same number of carbons, same number of hydrogens. And if we were to name them, remember we would name structure 1 3 4 dloro. This is nine carbons. Nine carbons is no name.

[00:25:48]Also structure three would have the same name. So actually both of them are the same compound.

[00:25:59]Okay. Then question 16 says which reaction has a standard enthalpy change equal to the standard enthaly change of formation of strontium flu.

[00:26:11]What is standard enthaly change of formation? should know it is the enthalpy change that takes place when one mole of a compound in its standard state. So here we're trying to make SRF2 as a solid. So these are the same for all choices.

[00:26:32]It is formed from its constituent elements in their standard state. So it's not formed from ions. So A and B are wrong. It is made from its constituent elements in their standard state. What is the standard state of strontium? You should realize that strontium is a matter. So it is a salt.

[00:26:59]A hex cycle is shown which could be used to calculate the enthalpy change of combustion of propane.

[00:27:08]Which expression is equal to the enthalpy change X shown on the diagram?

[00:27:14]So what is X there? X is actually an arrow going from water to its elements.

[00:27:21]So that is actually the minus of the delta H of formation of water. So minus the delta H of formation of water. So the arrow should go the other way towards water. But here we're forming four moles of water. So that is actually -4 times the delta H of formation of water.

[00:27:51]Which compound will produce a mass spectrum with a major peak at MZ at 31?

[00:27:58]So we need to draw each of them. propane one all propane 2 all propanal propanon and then try to find a fragment that would have uh an m / z of 31. This actually is 31 and the others would not have a fraction that is 31.

[00:28:28]Question 19 is about calcium and its compounds. A sample of calcium is burned in air forming calcium oxide. Complete the equation for this reaction including state symbols. So of course calcium is a solid. Oxygen is a gas. Calcium oxide is also a solid. And we need to balance. So we have two oxygen before the arrow. I need two in front of the calcium oxide.

[00:28:56]That means I need two calcium. State the flame color seen during the reaction. If I burn calcium, so basically we're burning calcium. What color do we get in the flame test? It is brick red. Please remember the different colors of the flame test.

[00:29:20]Explain how a flame color forms. How does a flame color form? Remember that it forms when the salt is heated. The electrons absorb energy, jump to a higher energy level. Then that is an excited state. The electron doesn't like to remain in the excited state. So it needs to go back. when the electron returns back to the lower or the ground state uh they emit the energy that they had absorbed and when this is emitted in the visible spectrum then we see a characteristic color.

[00:30:01]The calcium oxide formed is impure because during the reaction calcium reacts with another element in air. What is the other element present in large percentage in air?

[00:30:16]Remember that we have nitrogen. So when we burn calcium in air in addition to reacting with oxygen, it's going to react with nitrogen to form the nitri.

[00:30:28]And that is how we write the formula of calcium nitri.

[00:30:36]Another sample of impure calcium oxide made from heating limestone has a mass of 5.12 g. So the impure calcium oxide is 5.12 g. Impurities do not react with hydrochloric acid. The sample is added to 200 cm cubed of 1 mole per decime cubed HCl forming a solution. The HCl is in excess.

[00:31:05]The solution containing the excess acid is then transferred to a volutric flask.

[00:31:10]So basically we have a certain mass of calcium oxide. We're adding a lot of HCl. So some of it will react and the rest is left as excess. Now that excess we transfer to a volutric flask make up to a volume of 250 using the ionized water. that 25 cm cubed of this dilute solution of the HCl that did not react is titrated with.15 mole per decime cubed of sodium hydroxide and the tighter which is the volume of sodium hydroxide is 23.7 calculate the percent purity of the calcium oxide s so we reacted some HCL with calcium oxide and the rest we titrate it.

[00:32:04]So if we say we have sodium hydroxide.15 mole per decime cubed and we use 23.7 cm cubed of it I can get the number of moles of sodium hydroxide again remember to divide the volume by th000 to change it into decime cubed then we look at that equation between HCl and sodium hydroxide the number of moles of This would be the same as the number of moles that we originally had.

[00:32:39]Now 25 cm cubed is what we titrated. So the number of moles of HCl in 25 is the same. But then that 25 was actually part of the 250. So to get the number of moles in all the 250 of course we multiply by 10.

[00:33:03]So that is the number of moles of H cell that did not react. Now how much did we originally add? We added uh 200 cm cubed or 1 mole per decime cube. So that means we originally added 2 moles and what was left was the rest of the number of moles in the 250. So the number of moles that reacted is the difference between them.

[00:33:33]having number of moles of HCl. Then looking at that equation on top, the mole of calcium oxide reacts with two moles of HCl. That means if I have the number of moles of HCl, I can get number of moles of calcium oxide divided by two. And that means that is the number of moles I have. So I can get the mass of calcium oxide that actually was present in the impure.

[00:34:04]Mass is number of moles times the mR of calcium oxide. So that means that that 5.12 g of impure calcium oxide actually had only 4.61 g of calcium oxide. So I can get the percent purity. It's the actual mass over the impure time 100 that comes out to about 90%.

[00:34:35]Calcium nitrate is a white crystallin solid that decomposes when heated strongly with a bun and burn. A student suggested an explanation for this decomposition and they said the calcium ion is polarized by the nitrator.

[00:34:53]This distorts the electron cloud around the nitrate ion and so weakens the bond between the two ions. Identify two errors in this suggestion and include corrections that give an explanation of the decomposition.

[00:35:11]Now we already know that we cannot say calcium ion is the one that's polarized.

[00:35:19]It is the annion that is polarized by the metal ion. So we see that the metal ion has polarizing ability and the annion is the one that is polarized. So when we say calcium ion is polarized by the nitrate ion that is wrong. It is actually the nitrate that is polarized by the calcium.

[00:35:48]And the other error is when he said this causes the bonds between the two ions to be weakened. No, it causes the bonds between the nitrogen and the oxygen in the nitrate ion to weaken. And that is why it breaks up easily.

[00:36:14]This question is about some reactions of the halogenino alkan one brooane. Draw the mechanism for reaction one. Which one is reaction one? That's where we're replacing the Br with an O. So the Brs or the hogenino alkan reacts with the hydroxide ion and it the bromine is replaced by the O. And remember this is a nucleopilic substitution reaction in which the lone pairs of electrons on the oxygen attack the slightly positive carbon. The bond between the carbon and the slightly negative bromine breaks up and the Br is released as Br minus and the O is formed and that is the alcohol.

[00:37:08]Give the reason why chemists often use reaction two which is the formation of an nitril when synthesizing organic compounds. Why is this reaction useful?

[00:37:20]Remember that if I have a halogenino alkan and I react it with a cyanide it gives the nitril and we do this in order to do what? So that after that I can hydrayze the nitr form an acid with an extra carbon atom. So if I'm starting with three carbons I now have a compound that has four carbons. So usually we use this type of reaction to add an extra carbon atom to the chain.

[00:37:56]Give the conditions needed for reaction three. Where is reaction three? That is where we react ammonia in ethanol with the halogen alkan to replace the halogen with an amine. Remember that we have ammonia and it's already mentioned that it should be in ethanol. So the other condition that we need is it should be heated in a sealed tube.

[00:38:28]A different type of reaction occurs when the halogen alkan is heated with sodium hydroxide dissolved in ethanol. So remember that if I have a hogenino alkan I react it with sodium hydroxide in ethanol then what we have is elimination.

[00:38:50]So we remove the halogen and the neighboring hydrogen on the neighboring carbon to form an alken. That's when the sodium hydroxide is dissolved in ethanol.

[00:39:02]Explain why an hydro calcalium chloride is added to ethanol before it is used.

[00:39:09]So I need the ethanol to be dry. I don't want any water in it because if there is water then we don't have elimination. We have nucleophilic substitution. Remember that hydroxide in presence of water would give nucleophilic substitution. So it would change it into the alcohol. And remember that anhydros calcium chloride is a dehydrating agent. It removes water to prevent nucleophilic substitution reaction to form alcohol. We don't want to form an alcohol. We want the elimination reaction to happen.

[00:39:51]So draw the skeletal formula of the three isomers formed in this reaction.

[00:39:57]So basically we're trying to do elimination because we have sodium hydroxide and ethyl. Now elimination means I'm going to remove the CL and a neighboring hydrogen to form an alken.

[00:40:11]Now if the hydrogen on that side is removed then my double bond is like this. But remember that that means these two carbons attached to double bond have two different groups attached to them.

[00:40:25]So I can have the cis form or the transform in addition to the fact that I can also remove the hydrogen from the other side.

[00:40:37]So we remove the hydrogen from the cylohexane to form a double bond on that side. So these are the three possible isomers that could be formed.

[00:40:54]Question 21 says, "This question is about group seven elements and their compounds. The compound NAC3 is used in the paper industry to bleach wood pop. It is made by the reaction of hot sodium hydroxide with chlorine gas.

[00:41:11]So we should be familiar with this reaction. If we say hot sodium hydroxide then it will give sodium chloride and sodium chlorate which is Na3 and water explain in terms of oxidation numbers what type of reaction occurs what's happening in here the chlorine started as an element oxidation number zero it changed into two different compounds it formed sodium chloride in which the Cl has an oxidation number minus1 but it also formed Na3 if you calculate the oxidation number there the chlorine is + 5 now a reaction in which a certain species is both oxidized and reduced is called disproportionation reaction.

[00:42:10]So this is a disproportionation reaction because the chlorine is both oxidized.

[00:42:16]So it was oxidized from 0 to + 5 and reduced. It went from 0 to minus1.

[00:42:26]Calculate the percentage atom economy by mass for the formation of NL3 in this reaction. So we have this reaction and we want the percent atom economy by mass for formation of this compound. How do we calculate percent atom economy?

[00:42:46]It is the MR of the desired compound which in this case is NaO3 over the sum of all the MRS of all the products. And please take into account the balancing. So this reaction forms 5 NaCCl1 Na3 3 moles of H2O.

[00:43:13]So my atom economy would be involving the MRS. We can calculate the MRS first.

[00:43:19]So the MR of my desired compound is 106.5.

[00:43:24]The MR of sodium chloride is 58.5. The MR of water of course is 18. So the atom economy is the MR of the NAC3 over the total molar masses of all the products taking into account the balance. So it's 5 * 58.5 + 106.5 + 3 * 18. This gives an atom economy of 23.5%.

[00:43:59]A student tests samples of solid sodium chloride and sodium iodide with concentrated sulfuric acid. The following observations are made.

[00:44:10]Evaluate these observations to decide which halid ion is the stronger reducing agent. Support your answer with the products formed by referring to the observations. the changes in oxidation number to justify your deduction of the stronger reducing agent and ionic equation for the reaction that produces a smell of rotten eggs. State symbols not required. So we are told that the sodium chloride when we added concentrate it's sulfuric acid it gave misty fumes which form white smoke with ammon. When we added sodium iodide, we got misty fumes, forms white smoke with ammonia, purple vapor, and a smell of rotten eggs. You should remember that with chloride, the chloride reacts with sulfuric acid to form HCl.

[00:45:09]This is the white fuse that form a white smoke with ammonia.

[00:45:16]But this is the end of it. It doesn't react further because HCl is a weak reducing agent. So no further reaction occurs. The chloride formed HCl. So the oxidation number of chlorine remains the same and it was not oxidized or reduced.

[00:45:41]But with the iodide, we said the iodide reacts with the sulfuric to form HI.

[00:45:48]But then the HI goes on to react with sulfuric again because it is a very strong reducing agent. So when it reacts again with the sulfuric acid, it forms H2S which has a smell of rockmex. it forms iodine which is purple vapors and remember that the iodine changes its oxidation state. So in HIT it was min -1 it now goes to iodine which is zero and the sulfur is reduced from + 6 to -2.

[00:46:33]So this shows that HI is a strong reducing agent.

[00:46:43]Okay. Question 22 says the Sabbath process uses carbon dioxide and hydrogen to produce methane and water. This is a reversible reaction with the delta H having a negative value. So this is an exothermic reaction. The process is carried out at a temperature of 400° C and a pressure of 3080mm in the presence of a nickel catalyst. Discuss the advantages and disadvantages of using such high temperatures and pressures in the process. Remember what would be the advantage of using a high temperature?

[00:47:26]Of course, for any reaction, using high temperature increases the rate.

[00:47:33]Molecules have more kinetic energy, move faster, more frequent collision.

[00:47:39]However, this causes the point of equilibrium in this case to shift to the left because we said the forward is exo.

[00:47:46]Increasing the temperature causes the reaction to go to the side that is endo.

[00:47:52]So here it will go to the left.

[00:47:56]So low yield of methane. Also using high temperature makes the process expensive since we're using a lot of energy. Now with respect to pressure you should know in any reaction high pressure causes the reaction to go to the side that has less number of moles. So in this case using high pressure causes the point of equilibrium to shift to the right. So it goes forward to give less number of moles. So that means higher yield of methane. So that's the advantage. But using high pressure in any reaction is expensive.

[00:48:43]State how a catalyst increases the rate of a reaction. Remember what does a catalyst do? The catalyst provides an alternative pathway with lower activation energy. So the activation energy with catalyst shifts to the left and that means more molecules have energy higher than the activation energy. So more collisions are successful.

[00:49:15]In section C, the question is about alcohols and related compounds. Propane 2 all is a secondary alcohol. One of its uses is as a solvent to clean the surfaces of vinyl records. Propane 2 all can be oxidized to form ketone which is propanon in the laboratory. However, most propanone produced in industry is made using the cumine process where hydrocarbons from oil are used to synthesize cumine. So for example, benzene plus propene is used to make cumine.

[00:49:56]Cumine is then oxidized directly to form phenol and propanol. Propane 1ol is also a solvent and can be used as a fuel. The fuel biopropanol which is mainly propane one allol is formed from glycerol a waste product of biodiesel production from plant oils. The reaction used is called hydrogenolysis and requires catalyst containing nickel tungsten and zirconium.

[00:50:30]First state what is meant by the term secondary alcohol. Remember we said alcohols can be uh regarded as primary, secondary or tertiary depending on the O bonded to a carbon and that carbon is bonded to what? So we said if that carbon to which the O is bonded is bonded to only one carbon that's primary. two carbons, that's secondary.

[00:50:59]Three carbons, that's tertiary. So for a secondary alcohol, the O group is attached to a carbon that is bonded to two other carbon atoms.

[00:51:13]The oxidizing agent used to oxidize propane tool in a laboratory is acidified solution containing dromate ions. The reaction mixture is placed in a pear-shaped flask and heated for 30 minutes. Give the name of the additional piece of equipment needed when heating the reaction mixture for this length of time of time without any loss of the mixture. Remember, we're heating propane to all alcohols are volatile. So if I want to oxidize a secondary alcohol to a ketone, I need to heat it with acidified potassium dromate under reflux so that the alcohol does not evaporate. So the other additional piece of equipment is condense. Describe the color change that occurs during this heating. Remember we are heating potassium dromate. It's acting as an oxidizing agent. That means it changes from orange to green.

[00:52:26]Describe how the IR spectrum of the reaction mixture would change during this reaction. Your answer should consider only the organic reactant and product. Use information from the data booklet to support your answer. So, what are we doing? We're oxidizing the alcohol to a keto. So basically the O of the alcohol should disappear and the C double bond O of a ketone should appear.

[00:52:54]So that's what we're trying to uh explain and we get the exact um positions of the peaks from the data booklet. So we're starting with an alcohol and alcohol should have a peak for O at 3750 to 3200.

[00:53:12]Now once the alcohol has been oxidized that peak should disappear and it should become a ketone. So we'll have a peak for the couble bond or the ketone at 1720 to700 that should appear. Of course also remember the fingerprint region is the part to the right of this uh spectrum.

[00:53:36]The part to the right is called the fingerprint region and it would change from that of an alcohol to that of a ketone.

[00:53:48]The cumin process is shown using molecular formula. Calculate the mass of benzene needed to produce 2,000 decime cubed of propanon using the cuming process. You should assume that the overall yield of the process is 85%.

[00:54:10]So we have what we have 2,00 decime cubed of propanon and we have the density.

[00:54:20]Density is uh 0.784 g per cm cubed. And you should know that the mass of the propanon would be the density times the volume. But take note the density is in g per cm cubed while the volume given is 2,00 decime cubed. So I need to multiply that by a,000 and that will give the mass of propanon. Number of moles would be the mass over the molecular mass of propan.

[00:54:54]But take into account that these number of moles are actually uh 85% yield. That means if I had 100 if I started with 100 moles of benzene I would end up with only 85 moles of benzene. So if I have 85 moles of propanon. So if I have this number of moles of propan known what would be the original number of moles of benzene. So that would be over 85 * 100 that would be the number of moles original number of moles of benzene.

[00:55:34]Then the mass would be the number of moles times it's mr. This gives the mass in g. We are required to give it in kilog. So we divide that by a th00and.

[00:55:51]Most chemists would say that step two of the cumine process has an overall percentage atom economy by mass of 100%.

[00:56:01]Even though there are two products, we have two products phenol and propanon.

[00:56:08]But chemists regard the overall um atom economy 100%.

[00:56:16]state what can be reduced. Well, that just means that all the products are desired. So, phenol is a desired product in addition to the propano.

[00:56:29]Data about some physical properties of propane to all and propen shown. Compare and contrast these physical properties.

[00:56:39]justify your answer in terms of intermolecular forces. Descriptions of how intermolecular forces form are not required.

[00:56:50]So propane to all is an alcohol.

[00:56:54]Alcohols can do hydrogen bonding with water and that is why they are fully missable in water. Propanon also can do hydrogen bonding with water. So that is why both of them are fully missable because each of them can do hydrogen bonding with water. But then with respect to boiling temperatures, propane 2 all has higher boiling temperature and that okay so both are missible in water since they can both form hydrogen bonds with water. But then why is it that propin tool has higher boiling temperature? This is because in propin to all we can have hydrogen bonding between the molecules themselves between they can do hydrogen bonding with each other. So this needs a lot of energy to be broken in addition to the weak London forces and the permanent dipole dipole interactions. But with respect to propanon propanon molecules cannot form hydrogen bonding with each other. So there is only the permanent dipole dipole and the weak London forces and both these are weaker than the hydrogen bonding in propane 2.

[00:58:16]A copper calorimeter containing 200 cm cubed of water at 20° C is heated using a spirit burner containing biopropanol.

[00:58:28]A mass of 1.02 02 g of biopropanol is burned during the experiment. Calculate the maximum temperature in degrees C of the water at the end of the experiment.

[00:58:42]You should assume that there is complete combustion and that all energy produced is transferred to the water. So we have 1.02 02 g of bipropanol heating 200 cm cubed of water. So you should realize that we know that um delta h is qn. We are given the delta h of combustion minus 2021.

[00:59:13]So this is equal to q n and we want to get q. So we need to get molecular mass of propanol that's 60. So the number of moles is the 1.02 over the MR. Then you can put it into that equation to get Q. Q would be delta H that is given. Remember that this is given in kilo jewel. So we need to put it in jewels. Multiply by,000 times the number of moles. This gives you Q in jewles. Now what do we want? We want the maximum temperature of the water. That means we basically need delta T. So we know Q and we know the mass of the water that's 200 times the C 4.18. So delta T will be Q / MC. The change in temperature is 41.1° C. But we are required to get the maximum temperature. So if we're starting with 20 then the maximum temperature is 20 + 41.1.

[01:00:26]That means the maximum temperature reached is 61.1.

[01:00:34]Biopropanol has a higher octane number than bioethanol.

[01:00:39]Name the process carried out on alkanes obtained from crude oil to increase the octane number of a fuel. Remember increasing the octane number of a fuel means we want something that is more flammable and what we do is we use reforming. Reforming is the process in which we convert straight chain hydrocarbons into branched or cyclic.

[01:01:11]This makes it more useful as a fuel.

[01:01:15]This is called reforming.

[01:01:20]And that's the end of this paper. I hope this was useful to you. Thank you for listening.