AQA Chemistry - Energetics Revision

[Music] hello and welcome to this video this is a revision video for AQA so specifically for all those people studying AQA chemistry um my name&#39;s Chris Harris I&#39;m from alerts.com and we&#39;re basically going to look at energetics um so about this video is basically just to give you just a quick overview of energetics uh just a quick revision Aid um but I&#39;ve got to point out that these the Powerpoints that I&#39;m going to use on here um you can actually purchase them uh and um if you like a copy it&#39;s great for revision print up notes Etc then um you can um get them if you just click on the link in the description box for this video um and you can get them there okay so like I say these are very specific to AQA um they&#39;re matched to the specification so um if studying AQA this is what you need to know all right so let&#39;s start with what is enthalpy so this is when a chemical reaction occurs and we normally get a change in energy um it&#39;s very very common um basically we can we can measure this and we call this the enthropy change so the enthropy change is basically the reaction um is when the um you get a heat change in reaction at a constant pressure it&#39;s given the symbol Delta H it has the units of kog per mole and we normally write it like this so the we have the Greek letter Delta means change in this is the change in n P we have the little underground symbol here I like to call it uh basically this is the symbol it means standard state standard conditions Etc um obviously we need to know what they are so it&#39;s 100 kilopascals of pressure 298 Kelvin which is 25° C okay right so endothermic and exothermic reactions okay so these are key terms that we need to know reactions can get out heat or they can take it in so let&#39;s look at endothermic reactions first so these reactions that absorb energy from the surroundings so these feel colder when you touch them so looking at the energy profile of these the reactants are lower in energy than the products you can see there the Delta H the enthropy change is always positive for these um for that reason um here&#39;s an example um so you might have calcium carbonate um this is the decomposition of calcium carbonate this is an endothermic reaction takes energy takes more heat energy um and the products are actually less stable than um the uh reactant that we started with this is the value that we give it plus 178 KJ per mole this is the enthropy change of combustion uh exothermic reaction these are the opposite these release heat energy to the surroundings so these will feel warmer when you touch them reactants are higher in energy than the products so the products are lower in energy the enthropy change is negative you can see because it&#39;s dropping there&#39;s the enthropy value uh products are low energy example is most fuels all fuels when you burn them obviously they give out heat you&#39;ll know that by now um this is ethane burning ethane and oxygen we produce carbon dioxide and water very exothermic obviously because it gives out a lot of heat minus 1560 K per mole it&#39;s an example of an exothermic reaction okay bond breaking and making this is what it&#39;s all about bonds are broken and made during a reaction okay so when we break bonds H we need energy to do that um so this is energy being absorbed so breaking bonds BS are an endothermic process so Delta H is always positive when we&#39;re breaking bonds um and basically if more energies needed to break the bonds than energy given out when the bonds are being formed when we reform them to make the products the reaction will be endothermic and that&#39;s how we can link it in with um the bond enthalpies when we make bonds though um energy is released and we release a bit of energy and this is always exothermic so Delta H is always negative for these reactions and so again if we use the reaction profile we&#39;ve seen before if we get more energies released when we form in the bonds to make the products then what was needed to break the initial bonds the reaction will always be exothermic so it&#39;s basically just this subtraction away from them and we can actually calculate this using data as well this is called the mean Bond enthropy so the bonds of um this is really important actually before before I move on the the bond enthropy is obviously the amount of energy require requ ired to break that Bond or the amount of energy given out when the bond is formed so it&#39;s the energy that&#39;s stored in that Bond but it&#39;s really difficult because bonds all have different depending on where the bond is in the molecule depend on how much energy is needed to break it so in science we use something called in chemistry we use called a mean Bond eny so we take an average so and this is what the statement is saying here so bonds of the same type don&#39;t all have the same amount of energy which may seem a little bit odd so for example so if we&#39;ve got four CH Bonds in me thing okay and they all have slightly different Bond enthalpies so the energy needed to break the first CH Bond will be different to the energy needed to break the second CH Bond um and obviously this is going to pose a little bit of a problem so what we can do is we can measure the enthropy of breaking all four bonds and then just divide by four to get a mean Bond enthropy of the CH Bond so for example the uh total in this case is 1,662 K for four CH bonds to be brok in methane in particular so the mean bond eny is basically just 1662 divid by four that is the mean Bond eny of a CH it&#39;s an average it&#39;s not this not all the bonds in the methane are that value so enthropy changes using mean Bond enthropy right the equation we use is the eny change is basically the total energy needed to break the bonds minus total energy released from forming them so let&#39;s have a look at an example so calculate the energy change for the combustion of methane so here here it is here methane oxygen carbon dioxide and water here&#39;s our bond enthropy data this is the amount of energy needed to either break the bond or energy given out when we form it for C double bond o um we&#39;ve also got CH h o double bond o and O so these bonds are all involved in these in this reaction here so let&#39;s break all the bonds we assume we break all of them we&#39;ve got four CH bonds um so the CH is 435 four lots of 435 is 1740 two lots of oxygen cuz we got two oxygen there so two lots of 498 is 996 add them up total amount of energy to break the bonds is 2736 K per mole now we need to see how much energy is given out when we form them so here they are four lots of O because we&#39;ve got two water molecules that&#39;s four lots of 464 is 1856 and two lots of C double bond o uh is two lots of 805 that&#39;s 1 1610 add them up we get 3466 now it&#39;s not rocket science really to work this one out that you&#39;re getting more energy is given out when you&#39;re forming the bonds and what&#39;s needed when you take them in so overall this reaction is exothermic so we do this total energy breaking bonds minus that number and basically we get minus 730 K per mole okay calorimetry is a method in which we can actually work some of this stuff out so we can work out the enthropy change of combustion of a fuel so that&#39;s pretty useful because we can use this to determine which fuels are best so we&#39;ve got a fuel that&#39;s burnt to raise the the temperature of a water by a specific amount so generally weigh the fuel beforehand put not on a on our balance uh and we basically measure how much of the fuel has been burned and we&#39;ll come back to that in a minute so the energy for the fuel this is transferred into the water there it is there&#39;s the fuel into the water um and obviously this heats the uh the water in there but you&#39;ve got to be wary that this this thing&#39;s flicking around all over the place so some of the Heat&#39;s going over here some of it&#39;s going over here some of it&#39;s been absorbed by the tripod so not all of the energy in this fuel has been absorbed by this water so there&#39;s a lot of kind of um bad practices here obviously we&#39;ve got a lid on the top there so that&#39;s good but you just got to be wary that the amount of energy needed might isn&#39;t always going into that water okay so yeah we put the lid on there um you might want to put windshields around here as well to stop this from flickering um you basically want to try and get all this heat energy transferred in there it&#39;s really difficult to do but and the energy transferred we can calculate it using qal MC delta T and we&#39;re going to look at that in a minute um and uh basically from that we can then work out the end thpy change you&#39;ve got to be careful there&#39;s a difference between energy and enthalpy enthalpy is kilog per mole energy is just kilog okay so you got to know the difference between them okay so let&#39;s have a look see how we can work this thing out so energy from a calorimetry experiment and we can calculate this so there&#39;s our cual MC delta T let&#39;s look at what the different parts mean so Q is just energy gained uh or energy lost sorry or gained um this is in Jewels measure this in Jews this is the mass of water or the solution that we&#39;re heating C is the specific heat capacity you be given this 4.18 and temperature change this is measured in kelvin uh right here&#39;s an example 100 gr of water was heated from 23° C to 57° C by 1.8 G of ethanol calculate the energy transferred and hence the enthropy change of the fuel so we&#39;re going to use this equation first we need to work out the energy transferred so Q is the mass we&#39;re using 100 G of water so that&#39;s 100 G Times by the heat capacity 34 that&#39;s the temperature change here temperature change in degrees Celsius is the same in kelvin so that&#39;s fine 34 uh stick the numbers in and we should get Q is 14212 jewles or converted into kles okay so we need to calculate the energy first for this then we need to work up the number of moles of fuel used because we need to work out the enthropy and remember enthropy is KJ per mole so we need to work out the moles of fuel so the moles of ethanol is just Mass over Mr we used 1.8 gr of ethanol remember this is why we&#39;re weighing it before and after to work out how much fuel is used divided by the Mr of ethanol which is 46 and we work out the number of moles to be 0.039 once you we&#39;ve done that then we just take the energy in kog and divide it by moles kog per mole to work out the eny change so enthropy is energy divided by moles minus 14212 this is minus because this is exothermic it&#39;s a fuel it&#39;s producing heat energy at 0.039 and that gets us an eny of minus 364.5 KJ per mole and that&#39;s the enthropy of this fuel okay let&#39;s look at a different one this time so instead of using a fuel we&#39;re going to do a calorimetry experiment of a solution so here we have a poyene cup okay so here&#39;s our poyene cup and we have um our lid on it and we have our solution inside and a thermometer again the poly cup is there to prevent heat loss we don&#39;t want heat being lost the surroundings we want to try and contain that heat within the beaker we add our acid first and we measure the temperature so we don&#39;t mix it we just add our acid then we add an alkalite or solid or whatever it is and we stir and measure the temperature change normally we use acid alkal reactions because they they um show a reasonable temperature change so we just measure the temperature first and then we measure it afterwards so here&#39;s an example so let&#39;s say we had 25 C cubed of one mum cubed HCL had a temperature of 20° C okay so that was in our container then 25 cm cubed of 1 M um 1 mole per DM cubed of sodium hydroxide was then added and this raised the temperature to a maximum of 26° c um so it&#39;s raised it by 6° C calculate the enthropy change of neutralization for hydrochloric acid all right so same thing Q equal MC delta T but look the mass this time we&#39;re going to have the mass of both liquids because we&#39;ve mixed them mass of acid plus the mass of your alkaline now we&#39;re assuming that the density of this is um is of these liquids is just 1 G per cm cubed so that&#39;s the Assumption we&#39;re taking here so that&#39;s why we using these as masses so 50 g Times by 4.18 * by 6 is the temperature change this gives an energy change of 1,254 Jew or 1.254 kilog very important both liquids are used for the mass then we need to work out the uh number of moles of HCL because we&#39;re looking at the enthropy neutralization of hydrochloric acid so moles of HCL is concentration times volume so the concentration is one because doing one mole per DM cubed times by the volume 25 now remember the volume is Times by 10us 3 because we have to convert that into decimeters cubed we&#39;ve been given it in centimet cubed so that&#39;s why we do that um that&#39;s just the same as divide by th000 it&#39;s up to you what how you work that out as long as that&#39;s in decimet cubed that gives us the moles of 0.0250 and then once you&#39;ve got that you basically just work out the enthropy change enthropy remember from the last one is energy divided by the number of moles so the again this is an exothermic reaction because we can see the temperatur increased so we do minus 1.254 divided by 0.0250 this is the number of moles just worked out there and that&#39;s give enthropy change of minus 5016 K per mole so pretty straightforward just don&#39;t forget to take the mass of both liquids all right hess&#39;s law right Hess he basically come up with this brilliant law um and basically it was Jermaine hes and he basically came up with this law to work out eny changes that you can&#39;t find out by doing an experiment and he basically said that the total eny change of a reaction is independent of the roote taken you must remember that and they could ask you to State his law and he came with This brilliant thing it&#39;s a cycle and we call it Hess cycle uh there&#39;s two cycles um we have um a formation cycle and a combustion cycle what you&#39;re looking for is the data that they&#39;ve given you if the data they&#39;ve given you is formation then you use a formation cycle so you can see this data that we&#39;ve got here is a formation cycle so we draw it as a formation so the cycle looks a little bit like this we have reactants and products and we can either work out the combustion or the enthropy of reaction and we always have elements in their standard States at the bottom so that&#39;s the elements that make up reactants and your products we call this an enthalpy of formation Delta FH this is this bit here okay so this is what we&#39;ve written in blue um and we always set our cycle out like this make sure everything&#39;s balanced as well um it&#39;s very very important everything&#39;s balanced but your cycle for formation should have arrows up from elements in their standard States okay so remember as well what you&#39;re going to do is just substituting these numbers here um for the formation symbols and we multiply the number of moles in the equation so we&#39;ll look at an example so let&#39;s um look look at this one for formation but I suppose the first thing we need to look at is if we&#39;re going with an arrow okay and I&#39;ll show you this in the cycle and we keep the signs here if we go against the arrow we change the sign so let&#39;s look at this example calculate the enthropy of combustion of methanol burning completely in oxygen to make carbon dioxide and water now look at the data they&#39;ve given us the data is formation data so we draw a formation cycle like I showed you before so here&#39;s the reaction ethanol uh methanol sorry uh burning an oxygen to form carbon dioxide of water make sure it&#39;s balanced look then look on this side here&#39;s the elements look these are the elements that make up these here okay so we need one carbon 2 o2s and two H2S all in their standard States gas gas solid okay so we&#39;re going to look at this one first eny of formation is minus 234 because this is of uh methanol we&#39;ve only got one mole so that&#39;s fine enthropy of formation of elements is zero um that&#39;s why we don&#39;t have have a number here for oxygen going from O2 here to O2 there&#39;s no change so it&#39;s just zero add all that up and the total length of your formation of methanol is - 234 look on the other side of the cycle um again we&#39;re looking at what we&#39;ve got enthropy of formation of carbon dioxide minus 394 we&#39;ve got the data there that&#39;s the enthropy of formation the enthropy of formation of water is because we got two of them it has to be 2 * - 286 that&#39;s going to be- 5 72 so we add all that up and we get a total length to be a formation of - 966 going up there okay right so this is the enthropy of combustion and that&#39;s basically what we want to work out which is that bit at the top so dead easy what we have to do is we need to work out this bit here so um imagine this a bit like a roadblock we want to get from here to here but we can&#39;t so we have to take a diversion so we&#39;re going to go down here and then up here so when we go against the arrow we flip the sign keep the number the same and when we go with the arrow we keep the sign the same so in this case we&#39;re going to do plus 234 and then we&#39;re going to do minus 966 cuz we&#39;re going with the arrow here put that in our calculator and the enthropy of combustion is minus 732 that&#39;s this value here and a good way of checking is if you put all the numbers in they should add up to zero so put in that that number and that number um if as you go along your cycle it should add up to Z zero okay so that&#39;s pretty important if it does you&#39;ve probably got it right okay&#39;s look at combustion so this is a different cycle again you&#39;re looking for combustion data here we&#39;ve got combustion data look out for that um combustion Cycles are a little bit different we have reactants um reactants and products or we could have elements on this side as well um this could be a formational enthropy of reaction but combustion this time the arrows Point downwards always have combustion products in the bottom CO2 and water again make sure everything&#39;s balanced as well so remember when we&#39;re using these numbers we&#39;re just substituting these numbers for the combustion side as well uh multiply by the number of moles as well in the equation again we&#39;ll look at the example in the next slide okay so let&#39;s look at an example of how we can use the combustion cycle again go with the arrow keep the sign the same against the arrow you change the sign don&#39;t forget that rule so let&#39;s have a look calculate the eny of formation of pentane from their elements and their standard States here&#39;s the data that we&#39;ve being given okay here&#39;s the reaction this is um the enthropy of formation so here&#39;s the elements look here and the standard States there&#39;s our fuel and there&#39;s oxygen there as well now remember even though it says formation our cycle has got to be combustion because we&#39;re using combustion data okay so we need to draw owls down to combustion products carbon dioxide water again make sure it&#39;s balanced enthropy of combustion of five lots of carbon is 5 * - 394 this is -1970 we then need to do the enthropy of combustion of six lots of hydrogen this is 6 * - 286 that&#39;s - 1716 oxygen remember we don&#39;t need to do okay because we&#39;ve got oxygen left and right so actually these can cancel out and we can effectively just ignore them total length be of combustion on this side here is - 3686 on to the other side and of combustion of the fuel C5 h10 is - 3509 we&#39;ve only got one mole of it so that&#39;s fine again oxygen we can ignore because we can cancel them out left and right total is obviously the same then we want to work out the enthropy of formation which is this bit here just like before imagine we want to go from here to here because we need to work this value out but we can&#39;t imagine it&#39;s like a road block so we have to take a diversion so we go from down here and we go up here with the AR there against the arrow there so let&#39;s have a look so this one we&#39;re going with the arrow it&#39;s going to be minus 3686 keep the sign the same because we&#39;re going against the arrow going to our products we change the sign so it&#39;s going to be plus 3509 there it is and put that on your calculator we should get minus 177 K per mole this thing is exothermic okay again check it put all your numbers in um go around the cycle going with the arrow keep the sign the same against the arrow change the sign put all that in it should add up to zero so that&#39;s it uh that&#39;s energetics so um make sure that you can draw your H cycle and work them out plenty of marks up for grabs there again um you can get a hold of these PowerPoints um if you just click on the link in the description box it&#39;ll take you to the um um to the place where you can uh purchase this uh PowerPoint really great for a vision all right see you later byebye

[Music] hello and welcome to this video this is a revision video for AQA so specifically for all those people studying AQA chemistry um my name&amp;#39;s Chris Harris I&amp;#39;m from alerts.com and we&amp;#39;re basically going to look at energetics um so about this video is basically just to give you just a quick overview of energetics uh just a quick revision Aid um but I&amp;#39;ve got to point out that these the Powerpoints that I&amp;#39;m going to use on here um you can actually purchase them uh and um if you like a copy it&amp;#39;s great for revision print up notes Etc then um you can um get them if you just click on the link in the description box for this video um and you can get them there okay so like I say these are very specific to AQA um they&amp;#39;re matched to the specification so um if studying AQA this is what you need to know all right so let&amp;#39;s start with what is enthalpy so this is when a chemical reaction occurs and we normally get a change in energy um it&amp;#39;s very very common um basically we can we can measure this and we call this the enthropy change so the enthropy change is basically the reaction um is when the um you get a heat change in reaction at a constant pressure it&amp;#39;s given the symbol Delta H it has the units of kog per mole and we normally write it like this so the we have the Greek letter Delta means change in this is the change in n P we have the little underground symbol here I like to call it uh basically this is the symbol it means standard state standard conditions Etc um obviously we need to know what they are so it&amp;#39;s 100 kilopascals of pressure 298 Kelvin which is 25° C okay right so endothermic and exothermic reactions okay so these are key terms that we need to know reactions can get out heat or they can take it in so let&amp;#39;s look at endothermic reactions first so these reactions that absorb energy from the surroundings so these feel colder when you touch them so looking at the energy profile of these the reactants are lower in energy than the products you can see there the Delta H the enthropy change is always positive for these um for that reason um here&amp;#39;s an example um so you might have calcium carbonate um this is the decomposition of calcium carbonate this is an endothermic reaction takes energy takes more heat energy um and the products are actually less stable than um the uh reactant that we started with this is the value that we give it plus 178 KJ per mole this is the enthropy change of combustion uh exothermic reaction these are the opposite these release heat energy to the surroundings so these will feel warmer when you touch them reactants are higher in energy than the products so the products are lower in energy the enthropy change is negative you can see because it&amp;#39;s dropping there&amp;#39;s the enthropy value uh products are low energy example is most fuels all fuels when you burn them obviously they give out heat you&amp;#39;ll know that by now um this is ethane burning ethane and oxygen we produce carbon dioxide and water very exothermic obviously because it gives out a lot of heat minus 1560 K per mole it&amp;#39;s an example of an exothermic reaction okay bond breaking and making this is what it&amp;#39;s all about bonds are broken and made during a reaction okay so when we break bonds H we need energy to do that um so this is energy being absorbed so breaking bonds BS are an endothermic process so Delta H is always positive when we&amp;#39;re breaking bonds um and basically if more energies needed to break the bonds than energy given out when the bonds are being formed when we reform them to make the products the reaction will be endothermic and that&amp;#39;s how we can link it in with um the bond enthalpies when we make bonds though um energy is released and we release a bit of energy and this is always exothermic so Delta H is always negative for these reactions and so again if we use the reaction profile we&amp;#39;ve seen before if we get more energies released when we form in the bonds to make the products then what was needed to break the initial bonds the reaction will always be exothermic so it&amp;#39;s basically just this subtraction away from them and we can actually calculate this using data as well this is called the mean Bond enthropy so the bonds of um this is really important actually before before I move on the the bond enthropy is obviously the amount of energy require requ ired to break that Bond or the amount of energy given out when the bond is formed so it&amp;#39;s the energy that&amp;#39;s stored in that Bond but it&amp;#39;s really difficult because bonds all have different depending on where the bond is in the molecule depend on how much energy is needed to break it so in science we use something called in chemistry we use called a mean Bond eny so we take an average so and this is what the statement is saying here so bonds of the same type don&amp;#39;t all have the same amount of energy which may seem a little bit odd so for example so if we&amp;#39;ve got four CH Bonds in me thing okay and they all have slightly different Bond enthalpies so the energy needed to break the first CH Bond will be different to the energy needed to break the second CH Bond um and obviously this is going to pose a little bit of a problem so what we can do is we can measure the enthropy of breaking all four bonds and then just divide by four to get a mean Bond enthropy of the CH Bond so for example the uh total in this case is 1,662 K for four CH bonds to be brok in methane in particular so the mean bond eny is basically just 1662 divid by four that is the mean Bond eny of a CH it&amp;#39;s an average it&amp;#39;s not this not all the bonds in the methane are that value so enthropy changes using mean Bond enthropy right the equation we use is the eny change is basically the total energy needed to break the bonds minus total energy released from forming them so let&amp;#39;s have a look at an example so calculate the energy change for the combustion of methane so here here it is here methane oxygen carbon dioxide and water here&amp;#39;s our bond enthropy data this is the amount of energy needed to either break the bond or energy given out when we form it for C double bond o um we&amp;#39;ve also got CH h o double bond o and O so these bonds are all involved in these in this reaction here so let&amp;#39;s break all the bonds we assume we break all of them we&amp;#39;ve got four CH bonds um so the CH is 435 four lots of 435 is 1740 two lots of oxygen cuz we got two oxygen there so two lots of 498 is 996 add them up total amount of energy to break the bonds is 2736 K per mole now we need to see how much energy is given out when we form them so here they are four lots of O because we&amp;#39;ve got two water molecules that&amp;#39;s four lots of 464 is 1856 and two lots of C double bond o uh is two lots of 805 that&amp;#39;s 1 1610 add them up we get 3466 now it&amp;#39;s not rocket science really to work this one out that you&amp;#39;re getting more energy is given out when you&amp;#39;re forming the bonds and what&amp;#39;s needed when you take them in so overall this reaction is exothermic so we do this total energy breaking bonds minus that number and basically we get minus 730 K per mole okay calorimetry is a method in which we can actually work some of this stuff out so we can work out the enthropy change of combustion of a fuel so that&amp;#39;s pretty useful because we can use this to determine which fuels are best so we&amp;#39;ve got a fuel that&amp;#39;s burnt to raise the the temperature of a water by a specific amount so generally weigh the fuel beforehand put not on a on our balance uh and we basically measure how much of the fuel has been burned and we&amp;#39;ll come back to that in a minute so the energy for the fuel this is transferred into the water there it is there&amp;#39;s the fuel into the water um and obviously this heats the uh the water in there but you&amp;#39;ve got to be wary that this this thing&amp;#39;s flicking around all over the place so some of the Heat&amp;#39;s going over here some of it&amp;#39;s going over here some of it&amp;#39;s been absorbed by the tripod so not all of the energy in this fuel has been absorbed by this water so there&amp;#39;s a lot of kind of um bad practices here obviously we&amp;#39;ve got a lid on the top there so that&amp;#39;s good but you just got to be wary that the amount of energy needed might isn&amp;#39;t always going into that water okay so yeah we put the lid on there um you might want to put windshields around here as well to stop this from flickering um you basically want to try and get all this heat energy transferred in there it&amp;#39;s really difficult to do but and the energy transferred we can calculate it using qal MC delta T and we&amp;#39;re going to look at that in a minute um and uh basically from that we can then work out the end thpy change you&amp;#39;ve got to be careful there&amp;#39;s a difference between energy and enthalpy enthalpy is kilog per mole energy is just kilog okay so you got to know the difference between them okay so let&amp;#39;s have a look see how we can work this thing out so energy from a calorimetry experiment and we can calculate this so there&amp;#39;s our cual MC delta T let&amp;#39;s look at what the different parts mean so Q is just energy gained uh or energy lost sorry or gained um this is in Jewels measure this in Jews this is the mass of water or the solution that we&amp;#39;re heating C is the specific heat capacity you be given this 4.18 and temperature change this is measured in kelvin uh right here&amp;#39;s an example 100 gr of water was heated from 23° C to 57° C by 1.8 G of ethanol calculate the energy transferred and hence the enthropy change of the fuel so we&amp;#39;re going to use this equation first we need to work out the energy transferred so Q is the mass we&amp;#39;re using 100 G of water so that&amp;#39;s 100 G Times by the heat capacity 34 that&amp;#39;s the temperature change here temperature change in degrees Celsius is the same in kelvin so that&amp;#39;s fine 34 uh stick the numbers in and we should get Q is 14212 jewles or converted into kles okay so we need to calculate the energy first for this then we need to work up the number of moles of fuel used because we need to work out the enthropy and remember enthropy is KJ per mole so we need to work out the moles of fuel so the moles of ethanol is just Mass over Mr we used 1.8 gr of ethanol remember this is why we&amp;#39;re weighing it before and after to work out how much fuel is used divided by the Mr of ethanol which is 46 and we work out the number of moles to be 0.039 once you we&amp;#39;ve done that then we just take the energy in kog and divide it by moles kog per mole to work out the eny change so enthropy is energy divided by moles minus 14212 this is minus because this is exothermic it&amp;#39;s a fuel it&amp;#39;s producing heat energy at 0.039 and that gets us an eny of minus 364.5 KJ per mole and that&amp;#39;s the enthropy of this fuel okay let&amp;#39;s look at a different one this time so instead of using a fuel we&amp;#39;re going to do a calorimetry experiment of a solution so here we have a poyene cup okay so here&amp;#39;s our poyene cup and we have um our lid on it and we have our solution inside and a thermometer again the poly cup is there to prevent heat loss we don&amp;#39;t want heat being lost the surroundings we want to try and contain that heat within the beaker we add our acid first and we measure the temperature so we don&amp;#39;t mix it we just add our acid then we add an alkalite or solid or whatever it is and we stir and measure the temperature change normally we use acid alkal reactions because they they um show a reasonable temperature change so we just measure the temperature first and then we measure it afterwards so here&amp;#39;s an example so let&amp;#39;s say we had 25 C cubed of one mum cubed HCL had a temperature of 20° C okay so that was in our container then 25 cm cubed of 1 M um 1 mole per DM cubed of sodium hydroxide was then added and this raised the temperature to a maximum of 26° c um so it&amp;#39;s raised it by 6° C calculate the enthropy change of neutralization for hydrochloric acid all right so same thing Q equal MC delta T but look the mass this time we&amp;#39;re going to have the mass of both liquids because we&amp;#39;ve mixed them mass of acid plus the mass of your alkaline now we&amp;#39;re assuming that the density of this is um is of these liquids is just 1 G per cm cubed so that&amp;#39;s the Assumption we&amp;#39;re taking here so that&amp;#39;s why we using these as masses so 50 g Times by 4.18 * by 6 is the temperature change this gives an energy change of 1,254 Jew or 1.254 kilog very important both liquids are used for the mass then we need to work out the uh number of moles of HCL because we&amp;#39;re looking at the enthropy neutralization of hydrochloric acid so moles of HCL is concentration times volume so the concentration is one because doing one mole per DM cubed times by the volume 25 now remember the volume is Times by 10us 3 because we have to convert that into decimeters cubed we&amp;#39;ve been given it in centimet cubed so that&amp;#39;s why we do that um that&amp;#39;s just the same as divide by th000 it&amp;#39;s up to you what how you work that out as long as that&amp;#39;s in decimet cubed that gives us the moles of 0.0250 and then once you&amp;#39;ve got that you basically just work out the enthropy change enthropy remember from the last one is energy divided by the number of moles so the again this is an exothermic reaction because we can see the temperatur increased so we do minus 1.254 divided by 0.0250 this is the number of moles just worked out there and that&amp;#39;s give enthropy change of minus 5016 K per mole so pretty straightforward just don&amp;#39;t forget to take the mass of both liquids all right hess&amp;#39;s law right Hess he basically come up with this brilliant law um and basically it was Jermaine hes and he basically came up with this law to work out eny changes that you can&amp;#39;t find out by doing an experiment and he basically said that the total eny change of a reaction is independent of the roote taken you must remember that and they could ask you to State his law and he came with This brilliant thing it&amp;#39;s a cycle and we call it Hess cycle uh there&amp;#39;s two cycles um we have um a formation cycle and a combustion cycle what you&amp;#39;re looking for is the data that they&amp;#39;ve given you if the data they&amp;#39;ve given you is formation then you use a formation cycle so you can see this data that we&amp;#39;ve got here is a formation cycle so we draw it as a formation so the cycle looks a little bit like this we have reactants and products and we can either work out the combustion or the enthropy of reaction and we always have elements in their standard States at the bottom so that&amp;#39;s the elements that make up reactants and your products we call this an enthalpy of formation Delta FH this is this bit here okay so this is what we&amp;#39;ve written in blue um and we always set our cycle out like this make sure everything&amp;#39;s balanced as well um it&amp;#39;s very very important everything&amp;#39;s balanced but your cycle for formation should have arrows up from elements in their standard States okay so remember as well what you&amp;#39;re going to do is just substituting these numbers here um for the formation symbols and we multiply the number of moles in the equation so we&amp;#39;ll look at an example so let&amp;#39;s um look look at this one for formation but I suppose the first thing we need to look at is if we&amp;#39;re going with an arrow okay and I&amp;#39;ll show you this in the cycle and we keep the signs here if we go against the arrow we change the sign so let&amp;#39;s look at this example calculate the enthropy of combustion of methanol burning completely in oxygen to make carbon dioxide and water now look at the data they&amp;#39;ve given us the data is formation data so we draw a formation cycle like I showed you before so here&amp;#39;s the reaction ethanol uh methanol sorry uh burning an oxygen to form carbon dioxide of water make sure it&amp;#39;s balanced look then look on this side here&amp;#39;s the elements look these are the elements that make up these here okay so we need one carbon 2 o2s and two H2S all in their standard States gas gas solid okay so we&amp;#39;re going to look at this one first eny of formation is minus 234 because this is of uh methanol we&amp;#39;ve only got one mole so that&amp;#39;s fine enthropy of formation of elements is zero um that&amp;#39;s why we don&amp;#39;t have have a number here for oxygen going from O2 here to O2 there&amp;#39;s no change so it&amp;#39;s just zero add all that up and the total length of your formation of methanol is - 234 look on the other side of the cycle um again we&amp;#39;re looking at what we&amp;#39;ve got enthropy of formation of carbon dioxide minus 394 we&amp;#39;ve got the data there that&amp;#39;s the enthropy of formation the enthropy of formation of water is because we got two of them it has to be 2 * - 286 that&amp;#39;s going to be- 5 72 so we add all that up and we get a total length to be a formation of - 966 going up there okay right so this is the enthropy of combustion and that&amp;#39;s basically what we want to work out which is that bit at the top so dead easy what we have to do is we need to work out this bit here so um imagine this a bit like a roadblock we want to get from here to here but we can&amp;#39;t so we have to take a diversion so we&amp;#39;re going to go down here and then up here so when we go against the arrow we flip the sign keep the number the same and when we go with the arrow we keep the sign the same so in this case we&amp;#39;re going to do plus 234 and then we&amp;#39;re going to do minus 966 cuz we&amp;#39;re going with the arrow here put that in our calculator and the enthropy of combustion is minus 732 that&amp;#39;s this value here and a good way of checking is if you put all the numbers in they should add up to zero so put in that that number and that number um if as you go along your cycle it should add up to Z zero okay so that&amp;#39;s pretty important if it does you&amp;#39;ve probably got it right okay&amp;#39;s look at combustion so this is a different cycle again you&amp;#39;re looking for combustion data here we&amp;#39;ve got combustion data look out for that um combustion Cycles are a little bit different we have reactants um reactants and products or we could have elements on this side as well um this could be a formational enthropy of reaction but combustion this time the arrows Point downwards always have combustion products in the bottom CO2 and water again make sure everything&amp;#39;s balanced as well so remember when we&amp;#39;re using these numbers we&amp;#39;re just substituting these numbers for the combustion side as well uh multiply by the number of moles as well in the equation again we&amp;#39;ll look at the example in the next slide okay so let&amp;#39;s look at an example of how we can use the combustion cycle again go with the arrow keep the sign the same against the arrow you change the sign don&amp;#39;t forget that rule so let&amp;#39;s have a look calculate the eny of formation of pentane from their elements and their standard States here&amp;#39;s the data that we&amp;#39;ve being given okay here&amp;#39;s the reaction this is um the enthropy of formation so here&amp;#39;s the elements look here and the standard States there&amp;#39;s our fuel and there&amp;#39;s oxygen there as well now remember even though it says formation our cycle has got to be combustion because we&amp;#39;re using combustion data okay so we need to draw owls down to combustion products carbon dioxide water again make sure it&amp;#39;s balanced enthropy of combustion of five lots of carbon is 5 * - 394 this is -1970 we then need to do the enthropy of combustion of six lots of hydrogen this is 6 * - 286 that&amp;#39;s - 1716 oxygen remember we don&amp;#39;t need to do okay because we&amp;#39;ve got oxygen left and right so actually these can cancel out and we can effectively just ignore them total length be of combustion on this side here is - 3686 on to the other side and of combustion of the fuel C5 h10 is - 3509 we&amp;#39;ve only got one mole of it so that&amp;#39;s fine again oxygen we can ignore because we can cancel them out left and right total is obviously the same then we want to work out the enthropy of formation which is this bit here just like before imagine we want to go from here to here because we need to work this value out but we can&amp;#39;t imagine it&amp;#39;s like a road block so we have to take a diversion so we go from down here and we go up here with the AR there against the arrow there so let&amp;#39;s have a look so this one we&amp;#39;re going with the arrow it&amp;#39;s going to be minus 3686 keep the sign the same because we&amp;#39;re going against the arrow going to our products we change the sign so it&amp;#39;s going to be plus 3509 there it is and put that on your calculator we should get minus 177 K per mole this thing is exothermic okay again check it put all your numbers in um go around the cycle going with the arrow keep the sign the same against the arrow change the sign put all that in it should add up to zero so that&amp;#39;s it uh that&amp;#39;s energetics so um make sure that you can draw your H cycle and work them out plenty of marks up for grabs there again um you can get a hold of these PowerPoints um if you just click on the link in the description box it&amp;#39;ll take you to the um um to the place where you can uh purchase this uh PowerPoint really great for a vision all right see you later byebye

Transcript for:AQA Chemistry - Energetics Revision

Transcript for:
AQA Chemistry - Energetics Revision