IB11 SL2A

Ib year 11 exam standard level paper 2. Question 1. Define the average or define the term average bond enthalpy. The average bond enthalpy is the energy required to break a specific type of bond measured over a range of similar molecules. So energy required to break, we should say the average energy because it's where it says average bond enthalpy here. Let's put the word average in there as well because it is the average or mean, the mean energy required to break a specific type of bond measured over a range of similar molecules. Going to highlight that in blue to show that we're answering the questions. All right, we've got the answers in blue. Part B. Deduce the balanced chemical equation for the complete combustion of Butan1o. Okay, combustion. Well, combustion means it's reacting with oxygen. So the first thing we put in is it's oxygen as O2. Let's put in. An arrow. And what are we going to get? Well, carbon from the organic molecule here is going to make carbon dioxide CO2 and the hydrogen is going to make water. H2O. This is true for all organic combustions. Let me just format that. O2, CO2, H2O format it. Okay, now, four carbon atoms in the reactant that's being burnt. So There must be four carbon atoms here, four carbon dioxides. On the left hand side we have 10 hydrogens. That must make five waters on the right hand side. Now we count up the oxygens. 4:02 is 8, 5 O's 8 and 5 is 13. So that's 13 oxygen atoms. Well, there's already one there on the left hand side. So we need six more. Sorry, we need 12 more. So we need 6:02. So our balanced equation is as written there. Okay, then we have to determine the standard enthalpy change for the complete combustion of Butan1ol using the information from table 10 of the data booklet. Well, table 10 of the data booklet. Let me just get the data booklet here. Table 10 is. Bond enthalpies and bond enthalpies. Average bond enthalpies at 298. Okay, so what we have to do is we have to say how many and what type of bonds we've got in the reactants. So reactant bonds, a bit of a strange format there. Reactant bonds. Well, we can see that there are four carbon atoms joined to nine hydrogen atoms and an OH group. So first let's have a look at reactant bonds. Well, C H bonds, there must be nine of them. CH bonds times nine, that's going to be equal. So we look at the carbon hydrogen bond enthalpy and it's 413. So it's 413 times nine. I'm going to calculate that in a moment. What else have we got? Well, if there's four carbon atoms in a line, there must be three C C bonds. C C bonds times three. And that equals just pretty pretty up a bit. C C bonds times three. But each C single bond, C is 347 kilojoules multiplied by three. And we work that out in a moment as well. What else have we got? Well, we've got one C O bond because we're joining the carbon to an oh group times 1. And the C single bond O is 358 kilojoules. Just keep it the same shape times 1. And last but not least, there's an OH bond. O H bond times one. And that equals the OH bond is given to us in the Data booklet is 464. 464 kilojoules times one. So that will allow us to work out the reactants. The energy required to break up the reactants. Apart from the oxygen. We've forgotten the oxygen. Here's the oxygen, 6:02. So we've also got to break these bonds up. So there's an O double bond O times six. That equals, well, O double bond, O is Bond Energy is 498. 498 times 6 equals. All right, so it's a question of hard slog. Get on the calculator and work those out. 413 times 9 equals 3717. 347 times 3 equals 1041. And then 358 and 464. And the last one 6 times 498. 498 times 6 equals 2988. What do we do now? Well, we just have to add them all up together to give a total. Total, total total NG equals. Okay, 2, 988. We've got plus 4, 6, 4 plus 3, 5, 8 plus plus 3717 equals. And that comes to 8, 56-88568 kilojoules. But now this was just the reactants. On the right hand side, we've also got the product bonds. So let's just scroll up a little bit. Say products bonds. The products bonds are somewhat easier because there's four CO2. Now, CO2 contains each carbon dioxide contains two CO bonds. So it's C O. There are two per carbon dioxide and there are four carbon dioxides. So we must multiply that by eight. Similarly, water contains the O H bond. Each water molecule contains two. There are five water molecules. So we multiply that by ten. All right. CO bond from the data booklet is 746 kilojoules. And the OH bond is. We already used that, haven't we? So we've got it up here somewhere. There we go. 464. All right, so it's 746 times 8 equals 4, 6. 4 times 10 equals. We can see that's going to be 4, 640. So we just multiply out 746 times 8, 7, 4, 6 times 8. That comes to 5968. And the total energy change here can be added up as well. Total energy equals 5968 plus 46404640. And that comes to 1060810608. Now, these are bonds that are being made, so they're all exothermic. So I'm going to put negative signs in here. These are all negative signs to show that it's exothermic. Okay, now to determine the standard enthalpy change, then using this information, well, what we do is we just add these two values. So we just come down here and I want to just cheat. I want to say standard enthalpy change. I'm going to copy it here. Standard change copy equals 8, 5, 68 minus 10, 6 08. That equals 8568. Negative 2040 kilo. And there is the calculation for 3 marks. Quite a long calculation for 3 marks. Let's move on to the next question. Based on the type of intermolecular forces present, explain why Butan1ol has a higher boiling point than Butan L. Well, Butan1ol Glutan1o is an alcohol. It contains O H. Whoa. What's happened here? Oh, groups. Groups which can hydrogen bond from molecule to molecule. This is intermolecular butanal. Butanal cannot do this. It's an aldehyde. Doesn't have any hydrogen attached directly to an electronegative atom. Butanal has only van der Waals forces. Forces and dipole dipole interactions. Of course, Bhutan Wanol has Van der Waals forces as well. We'll put here as well as van der Waals forces. Okay, remember, everything's got Van der waals forces. Butan 1 oil has got hydrogen bonding as well. So we get our two marks for discussing those two issues. Next question, question two. Then iron has Three main naturally occurring isotopes used using a mass spectrometer to investigate explain why very low pressure is maintained inside the mass spectrometers. Well, you've got ions flying around. You don't want them bumping into air particles so that there are no collisions with the air particles. That will do for an answer. Another reason perhaps would also avoid getting false data from the ionization of air. So isn't that there also? It avoids ionization and detection of air molecules. There's two reasons either of those would be worth your mark. Iron is the following composition by mass. Calculate the relative atomic mass of iron based on the data. Okay. Right. So there's three isotopes. So the relative mass will be the average, the weighted average of the three. So these are percentages. So it's 54 contribution from isotope 54 times 0.0595. + 56, 56 times the contribution 0.9188 +. 57 times 0.0217. We work that out, taking a note that we're going to answer it to two decimal places in the end. All right, so let's do a calculation. 54 times 0.0595 equals plus brackets. 56 times 0.9188 close brackets equals plus bracket. 57 times 0217 close brackets equals 55.90 5.90. And the two decimal places were required. So there's our calculation. Calculate the number of electrons in this iron electrons. Well, if it's look up iron in the periodic table, the atomic number of iron number Fe26. So if the two plus is showing, therefore two plus ion has two less electrons, 24 electrons. On to the next question. Graphing is an important method in the study of the rates of chemical reactions. Sketch a graph to show how the reactant concentration changes with time in a typical chemical reaction. Show the rate of reaction at a particular time. Show how the rate can be determined. Okay, so how does the concentration of a reactant change? Well, it starts off high and then it curves down to auto shape in here. So axes. Not liking. We'll start off hit somewhere up here. I'm going to put a little X. Well, I can't put a little X on here. It doesn't really matter. I'll draw a shape, shall I? That'll do me an auto shape. There we go. I'm auto shape. And let's just drag it. Start off high, it'll end up low and it will curve between the two points. Reactant concentration changes with time. This axis is going to be time along here. Okay. This axis is going to be the reactant concentration will be a curve. I don't think I can draw a curve on this change auto shape, can I? No, not particularly versatile this program. All right, so the rate of the reaction at particular time can be determined from the gradient. All right, so we put in the rate rate at time T where T is any time equals the gradient to the curve. Okay. Part B. A solution of hydrogen peroxide is added to a solution of sodium iodide acidified with hydrofluoric acid. The yellow colored iodine can be used to determine the rate of reaction. So if we look at this equation, you'll see that there is iodine appearance appearing on the right hand side. Okay. Iodine is appearing on the right hand side here. I'm going to highlight it. The experiment is repeated with some changes in the reactant conditions. For each of the changes that follow predict stating a reason its effect on the rate of reaction. Let's have a look. The concentration of H2O2 is increased at constant temperature. Well this is a reactant. You increase a reactant it's going to increase the rate of reaction. So the rate increases. We need to explain it. There's two marks. It's probably we're doing setting. The reason increases because there are more collisions per unit time. The solution is prepared from a fine powder instead of large crystals. It doesn't really matter what you make it from because it's the actual solution that's reacting. This has no effect. No effect. The solution characteristics are the same regardless of how the solution is made. Alright, so that's our 2 marks skirmish. Next question. Explain why the rate of reaction increases when the temperature of the system increases. Well, the main reason is there are more particles of the required activation energy. But also there are more collisions per unit time. More particles require activation energy. So there are more successful collisions. There are more collisions per unit time. Okay, let's just highlight that one blue and move on to the next question. Hydrogen can be made by the reaction of hydrochloric acid with magnesium. According to that equation, what mass of hydrogen is formed when 100 cm cubed of hydrochloric acid of concentration 5 reacts with excess magnesium. So this tells us that all of the hydrochloric acid is going to be used up. Alright, so first thing we need is the moles of hydrochloric acid. Moles? HCl is the volume in liters 0.1 times the molarity 5 and that comes to 0.5 moles so from our equation, you can see that two hydrochloric acids make one mole of hydrogen. So therefore, moles of hydrogen, moles of H2 created. Created, equals 0.5 divided by 2, which equals, of course 0.25 moles. And the mass of hydrogen then is the moles times the relative mass. Mass equals moles times Mr. Relative Mass. Well, the relative mass of hydrogen x2 is 2, so it's 0.25 times 2 equals 0.5. We 0.0.5 grams. Nice easy sum for 3 marks. Question 5. Compound contains 38.9% by mass of iron. So it's 38.9 divided by the relative mass of iron, which is 56 and 17. Or 16.7. Sorry, 16 point. 16.7 of carbon. Divided it by 12 to get carbon. So this first one here is iron, second one here is carbon. And what must be missing here is oxygen. So the oxygen proportion is 100 minus those other two percentages. So it's 100 minus 38.9 minus 16.7 and that comes to 44.4. 44.4. And we divide this by the relative mass of oxygen, which is 16. All right, so let's work our three sums out here, our ratios. The first one comes to 38.9 over 56. 38.9. 38.9 divided by 56. 0.695. 0.695. Carbon 16.7 over 12, 16.7 divided by 12, that comes to 1.39. 1.392. Working to three significant figures. Really? Yeah. So it's three decimal places, not significant figures. 44.4 divided by 16 and that comes to 2.775. 2.775. Now, the way to work out is now to divide through by the smallest number we can see. This is the smallest number. So let's divide everything by 0.695. Alright? So we put the 0.695 under the other two as well. And this comes to. Obviously the first one comes to one, the next one comes to, I don't know, one point. Well, you can see this by inspection. Really? 1.39 2. 1.39 2 divided by 0.69 5 and that comes to 2.00, it comes to 2. And the next one 2.77 5 divided by 0. 695 and that comes to 3.99. Well, it comes to 4. 3.99 is as close to 4 as you're going to get. So our empirical formula then. The empirical formula is equal to Fe C2,04. Those across and let's color them in so we can see it's the answer. That's for our three marks here. All right, one mole of this compound is decomposed. It's heated, it decomposes to give one mole of iron oxide. Right. So let's put our chemical in down here. This is our chemical Fe. This is going to be heated and it's going to give. Wrong way around. It's going to give feo. All right, so what else can it give? Well, you've taken one fe one out. That leaves C2. That gives one mole of carbon dioxide. We're told that as well. Plus CO2. And what else is left? Well, a carbon's left and an oxygen is left. So the other gas must be co. Just a case of counting up, that's left. Identify this other gas. Right, okay, so that's quick. Question and move on. Propanol is the structure gaseous. Propanol burns as shown in the equation. Oh, this is another one of putting together all the bonds. Well here we've got 1, 2, 3, 4, 5, 6, 76 CH bonds. So that's 6 times CH CH is 6 times 413-413 times 6. That comes to 2, 4, 7, 8 and 2 C C bonds 2 times CC equals each CC bonds 347347 times 2694. What else have we got? 1 C double bond O 1 times C double bond O equals is 736736. And don't forget there's four oxygens. So there's also 4 times O double bond O O double bond 498498 times 4 comes to 1992. So the energy change when all the bonds in the reactants are broken, total energy change equals. Well, we're breaking the bond. So it's positive. So it's plus. And we just add those together. 1992 plus 736 plus 694 plus 24782478. That comes to 590059 kilojoules. We'll highlight that to show that we've done the working. I thought so. I thought they might ask us for how many the bonds in the products. All right, well what have we got? Let's just take the products out just so I can remember them. Down here. The products are the. Didn't get that at all, did I? What's happened? Why didn't I copy that? Copy. These are our products here. Alright, so we've got six oh bonds and an oh, bond, we're told, is 464. 464 times six, that comes to 2784 kilojoules. We've got three CO2, so we've got six C double bond O. And that comes to C double Bond O is 736. So it's 736 times 6,4416 kilojoules. So the total. Total bond energy equals. In this case, we're making the bond, so it's negative, negative 4416 and 2784. 7200. 7200 kilojoules. Make this nice and blue. And the enthalpy change for the complete combustion will be the difference between those two. So it's plus minus. What do we get for that? That's going to come to the calculator. 1,300 minus 1,300 kilojoules. That's the end of section A. I'll be going section B in a moment.