Environmental Chemistry Chapter 4 Lesson 2 Energy from Coal

welcome to our second lesson in our chapter for energy from combustion we last left looking at some efficiency calculations and I wanted to pick up with one more example before we dive into the chemistry of coal so the power plant efficiency formula that’s a mouthful we know that no electric plant can be completely efficient one hundred percent efficient and therefore we’re just asked to consider what type of electrical net energy are we producing from the amount of coal that’s being placed into the power plant so remember our formula electrical energy produced / the available heat from the fuel represented as a percentage to find our net efficiency the higher that temperature of the steam the more efficient the power plant we had mentioned that some of our modern-day power plants run as well as ninety percent efficient some of the older models are anywhere from 35 to 40 percent efficient let’s take a quick look at a sapling question one of your homework questions to illustrate this concept we’re being asked to calculate the percent efficiency of a coal burning power plant that produces 79 million six hundred thousand kilojoules of energy to heat a home and the home only requires 29 million four hundred kj’s of heat so you can see what we need to produce the 29 400 and then three more zeros this is the amount of heat needed to heat a home let’s say annually but in order to provide this we actually have to burn or produce from coal 79 million six hundred thousand kilojoules of heat see this ratio is what the home needs this value but this is what we have to burn in order to provide that because we just don’t have that one hundred percent efficient let’s calculate that shall we I’ll grab my calculator please do the same make sure we get a comment sir and so 29 403 zeroes / 79 603 more zeros and multiplied by a hundred to express as a percent it looks like our power plant our coal burning power plant I’m finding about 36.9 three percent efficient which is about what we’re finding for our older models isn’t it just one more practice before we start into our next lesson the topic of coal so we understand that you know for about two centuries ago this industrial revolution began and it’s a great exploitation of fossil fuels that still continues to this day we burn fossil fuels in the in essence coal as a source of energy coal continues to provide more than fifty percent of our nation’s energy until about nineteen fourteen we start to see you know at an overtake from petroleum coal is a complex mixture of substances however in essence coal is about eighty-five percent by mass carbon so you can see a you know a chemical formula for coal C subscript 1 35 h 96 09 nitrogen sulfur eighty-five percent by mass of this compound we know as coal is just plain old carbon with some trace elements hydrogen oxygen nitrogen and sulfur were all part of the original ancient plant material as it fossilized over the course of millions of millions of years remember coal is considered to be a non-renewable resource or just simply because it takes so long to regenerate from the organic manner of plant material so in essence coal is carbon with other trace elements coming from the original plant life as it fossilized through millions of years we can see from this particular graph and top of page 163 about 1840 the majority it looks like all of our energy from burning was coming from burning wood about 1850 do you see here how we started to burn coal and coal overtook would notice here in about eighteen eighty five if you will and started to rise in early 1900s you can see right in this general area Wood has pretty much

remained constant people heat with wood for convenience but these coal-burning power plants were being produced and coal was the number one way that we heated our country and then time went by and look at the pretty average here we sell a little plummet but over an average here we see amount of coal is remained steady and a little bit of increase here in in late history here with about 2,000 but definitely you see this blue as a petroleum is starting to take over and so again you know the maximum we’ve reached cheers and between 1970 to the 2000 era petroleum is by far the most abundant resource we use to combust and release energy natural gas is also on the rise natural gas is a component of petroleum it’s the lightest gas CH 4 is its formula so just the simplest hydrocarbon but we are burning natural gas and again Cole is pretty much on the on these steady here with the natural gas consumption so we’re starting to see Cole you know decline as petroleum increases but we also have to consider you know we started with saying we don’t live on the earth we borrow it from our hands that we borrow it from our future generations that you know saying where we said we didn’t inherit this land from our ancestors we borrow it from our children thinking about what are these other non-renewable resources we need to start considering implementing things like renewable sources solar wind all of these hydroelectric power plants you can see overall they’re hardly contributing but wouldn’t it be nice to start to see them rise in their contributions so our coal coal is of course a combustion reaction where we’re looking at the components of coal and just calculating the mass of carbon I had said it’s about eighty-five percent by mass well let’s kind of show you where this number is coming from so we’re going to assume the composition of coal to be c 1 35 it has H 96 and oxygen we have nine of those we have one nitrogen and we have one sulfur in that formula we want to know the mass of carbon in tons if we started with 1.5 million tons of coal so 1.5 million is 1.5 times 10 to the 6th tons of coal out of that how many are coming just from carbon and so this kind of brings us back to our last chapter and we talked about percent composition by mass I want to know just the tons of coal of just carbon compared to the ton of coal so it’s really just a mass percentage so I know from the periodic table carbon has an atomic weight of 12 right off the periodic table do you see that but in this formula there’s 135 atoms of carbon so molar mass of carbon times the number of atoms we have I’m going to take 12 x 1 35 and when I do that I get 16 20 grams alright just a number here from the molar mass alrighty 1620 hydrogen has a molar mass of one do you see that on your periodic table it has an atomic weight of one but there’s 96 of them in our formula so we’re going to know to take 96 total grams oxygen on our periodic table has an atomic weight will say 16 but do you see how there’s nine of them in this formula nine of those at 16 of peas gives me a total weight of 1 44 grams nitrogen find it on your periodic table you’ll see it’s atomic weight is 14 and when you find sulfur on the periodic table it has an atomic weight of 32 let’s sum each one of the components I want you to add 16 20 the contribution of coal from carbon plus 96 the contribution of coal from hydrogen plus 144 that’s the contribution of oxygen and coal plus 14 from the nitrogen and 32 from the sulfur when you add all of

those together we get a grand total of 19 0 6 grams in every mole so the sum all of the atoms added together gives me 1906 now remember the carbon contributed 1620 of that total contribution see what we’ve done this little part here is what we called a conversion factor its percent by mass of carbon that’s a topic we covered together last chapter just part over whole to express a person so that’s the mass percent of carbon part over whole notice what’s happened we can calculate by cancelling our coal variable and end with the carbon so let’s hit that on our calculator you would hit one point I’ve use your scientific notation key for the x 10 value we want that to be 1.5 million so we need six zeros after that and then we want to multiply that by the ratio of 16 20 / 1906 that’s your key sequence on your calculator our original sample of coal times the percent by mass of carbon and we’ve converted into the grams or tonnes because we use tons here as our initial thing so that would be fine we end up with a ratio here of 1.3 and that’s a million so times 10 to the 6th tons of carbon 1.3 million or 1.3 times 10 to the 6th tons of carbon process with me what we just did we had sitting at the coal the power plant one and a half million tons of coal in that one and a half million the vast majority 1.3 million are coming just from the elemental form of carbon it gives us insight into saying coal by the vast majority of its composition is indeed carbon with these few other trace elements depending upon the purity of carbon so petroleum and natural gas started to overtake coal around the 1950 and we saw that in that previous graph we know that petroleum is a mixture of several thousand different compounds in the great majority of these compounds are hydrocarbons so as we begin to explore petroleum and this is now what we consider section 4.4 we start to consider the elements or the other molecules found in petroleum specifically also met natural gas let’s take a quick peek at a very important diagram this is found on page 167 we’re going to be asked to look at the names the formulas the structural formulas and the condensed structural formulas for the first ten hydrocarbons so the first hydrocarbon is the simplest of all it has just one carbon surrounded by four hydrogen’s it’s Lewis dot structure would have a tetrahedral shape if we remember from our molecular geometry where we’d have one central carbon and four hydrogen’s surrounding it we would have a non-polar molecule with polar bonds so when we begin to consider ch4 its structural formula we need to associate that with the word methane the easiest hydrocarbon of all we often hear that called natural gas we burn that in our homes most often as our source of heat in our kitchen as well as in our furnace now notice this that the boiling point and we’ll just kind of keep tab as we start looking at this the lightest of all this is definitely a gas at room temperature negative 164 degrees Celsius is its boiling point so it’s very low boiling point very easy to convince to turn

into a gas well let’s take a look at our next hydrocarbon here we had just one what if we had to hydrocarbons connected together our carbon backbone is CA connected to another see and we finished those four bonds we know that carbon has to have four bonds in order to satisfy the octet rule and so the six remaining hydrogen just create this perfectly symmetrical molecule where each one of these carbon one attached to each other and then the three hydrogen’s surround it this is a molecule we call ethane ETH a and E ethane our first one was methane here is ethane notice that its boiling point is a little bit a little bit higher negative 89 notice the increased molar mass the more atoms we have it takes a little more energy to get that molecule to boil the heavier the molecule becomes so the lightest molecule course takes the least amount of energy to boil and as the carbon chains begin to grow we’ll see the amount of energy needed to boil also increases now here’s our little trick I would one carbon two carbons three carbons to find the number of hydrogen’s I’m going to double the carbon subscript and add to doubling it makes it six plus two more gives me eight let me share with you why I have to add the two at the end so there’s my three carbons in a row that’s three carbons that’s not an E and the two that I add on for each fill that octet so that’s why I doubled and why add two well I have to have two for the terminal carbons don’t I so one two three four five six seven eight or the 8 hydrogens I’m going to be just a little lazy and leave them like this you may as well we’ve just built the molecule known as propane propane you might have on your gas grill out in your backyard you’ll often see signs for lpg liquid propane gas is what those signs are saying if you stop and fill your propane tank you have an LPG service station what’s happened to the boiling point well it’s getting a little bit closer to zero isn’t it it doesn’t you know it is approaching a higher and higher value and taking more and more energy as the molecules begin to grow c4h how many double the carbon that’s eight add two for the ends and we have ten don’t we the carbon backbone would have four carbons in a row two hydrogen’s on each of those and then to complete the octet of the end carbons there’s the plus 2 that gives us 10 total hydrogen’s in this molecule known as butane now maybe you know butane by its common name called lighter fluid typically in a disposable lighter you’ll see a liquid fuel in there it’s known as butane now notice butane is actually a liquid at room temperature so far methane ethane and propane have been gases at room temperature this has a boiling point of about zero it’s about point five degrees Celsius now we get to pentane c5 how many h’s double the ten add two we get 12 so again we’ll have our five carbons in a row we add 2 h’s to complete these so even though I’m just drawing the lines you understand they’re representing hydrogen’s at each terminal point and then of course the two at the end to complete the octet we just drew a molecule called pentane and pentane has a boiling point of 36 degrees Celsius now let’s emphasize as the molecule becomes heavier and heavier we’re adding more and more carbons and hydrogen’s making the molecules molar mass heavier it takes more energy to get that heavier molecule to boil we see a definite

increase in the boiling point as the carbon chain begins to lengthen now I’m emphasizing this because it will be very critical we look at the petroleum distillation you know and how we separate out these molecules based on their weights we’ll keep going here c 6 h 14 a molecule known as hexane you can see that you would have six carbons in a row for the prefix hex 123456 and again you’ll add on your structural hydrogen’s and we’ll call this hexane draw this with me notice its boiling point hexane is 69 degrees approximately we’re increasing boiling point as the molecules begin to increase in length heptane c 7h 16 seven carbons in a row long hydrocarbon chains one two three four five six seven we know to add on those hydrogen’s we’ve made a molecule known as heptane hear that pee in there and make an observation with me about its boiling point it’s about 98 degrees Celsius and went this far for a reason here’s one that we’re probably familiar with by name it’s octane c8 HSC 16 and 2 is 18 h carbons in a row octene is its name now where does that sound familiar octane well hopefully you see that on your gasoline tanks when you go to fill the gas station it’s the major component of gasoline and now we have other additives such as ethanol about ten percent gasoline these days is ethanol but this is the major component octane is found in gasoline it’s a mixture of many things but this is a major component octane has a boiling point of about 125 degrees Celsius nine would have a name of no name I love that no name and if we had a carbon 10 you would hear it called decane here’s the trend each time we’re adding a carbon to our hydrocarbon chain the boiling point is increasing as the molecules become longer and heavier it takes more energy to get them to boil and we use this physical property to begin to separate out the molecules through the process of petroleum distillation here’s a practice sapling question with those hydrocarbons that we just drew together and kind of just studied what they looked like this is what a straight-chain hydrocarbons would look like all of the carbons are in a straight chain all we’re being asked to do is to complete the structure and to do that we know that each hydrogen has to have an octet so I’m just going to draw in on my sapling single bonds representing the attachments to each of the hydrogen atoms so the first has to have three hydrogen’s because it’s a terminal carbon I have to have four total bonds the second carbon has to have two hydrogen’s attached left and right our carbons up and down our hydrogen’s so four bonds same is true for the third positioned carbon left and right our single bonds to carbon so up and down must be hydrogen’s and here’s the terminal end of the chain so it would require three carbons one two three four this molecule we just named a moment ago as butane you recognize that straight chain for carbon butane here we have something known as a branched chain in chemistry this branch chain means that we have carbons coming off of the main carbon backbone so we have one going up one going down so that’s okay every carbon needs to have four bonds so for

example if one carbon here at the upper position is attached to a carbon it has positions for three single bonds this carbon here is attached to three other carbons leaving one space open for the hydrogen this looks like it has three spaces open the interior carbons would have two spaces and at the terminal end we need three spaces and then I must also go back and actually add the hydrogen one at a time you’re going to click that H onto your sapling just to show how many hydrogen’s it took to actually satisfy all four bonds for the carbons this is probably faster in your sapling because you just click the element on there where I have to write it on but I want to give you that visual so you can see when you’re done drawing yours in sapling it needs to match mine here we had a straight chain adding hydrogen’s to all the carbons to make sure they have four bonds and even in a branch chain it’s the same game every carbon has to have four bonds anything missing must be the hydrogen well let’s look at an oil refinery how gasoline and other hydrocarbons are produced from petroleum and I know that I’ve given you not only a supplemental video lesson in the early start to kind of talk about this we’ll talk about it in this lesson as well this looks like a good old oil refinery the backbone of the oil industry so how are gasoline and other hydrocarbons produced from petroleum remember this petroleum as thousands of ingredient molecules in there and this process takes place at this oil refinery we know this by looking at I mean this is a very iconic look at the oil industry we take something called crude oil and separated into fractions including the gasoline fraction so if I look at these long skinny towers here in the petroleum oil refinery industry they’re known as the distillation tower and what they do is is kind of talk through what we looked at based on boiling points we can separate out the different fractions in petroleum based on the number of carbons and their boiling points and they kind of separate them out from there so we have this initial step where crude oil is delivered into a boiler as the temperature of the boiler increases notice this flame down here so the crude oil is separated into fractions but it all arrives into this boiler as the boiler increases in temperature compounds with the lowest boiling points to vaporize remember the lowest boiling points are the lightest molecule so the lightest fraction has the lowest boiling point they’re going to come out as vapors which is gases aren’t they things like our methane ch4 things like our ethane c2h6 and things like our propane oops ssbc so c3h8 these are the lightest gases and they come off first up there don’t they at the lightest gases with the lower boiling points they begin to vaporize as the temperature increases even more the boiling points begin to reach those of the heavier molecules and as the heavier molecules begin to rise you start to get things like gasoline coming out down here gasoline remember the major component is octane c 8 h 16 plus 2 is 18 even heavier in this range c 12 to 16 this is where jet fuels is distilled into 14 to 16 and the in the length of the carbon chain is our fuel called diesel down here the reformer is actually connected to our cracker and will actually talk a little bit about what a cracker does what it does is crack the molecule into smaller pieces so for instance if we have these larger molecules coming out let’s say C 15 to 18 let’s say you get a c 16 h 32 and 2 is 34 let’s say you get something coming out here what the cracker will do is actually separate it so if i crack that molecule I could get c 8 and i would get hydrogen’s you know two molecules of c

eight this is 34 so cut that in half you’d get 16 and then the other one would have 18 notice it has to still add to the 34 total we started so one of them will just be doubled the other would be doubled plus two this would be an en a just because it contains a double bond in the carbon but the idea here is is you know in just thinking about cracking taking this long molecule and cracking it in this component and letting it rise to be gasoline the cracker takes long hydrocarbon chains and cut some or cracks them in half doesn’t necessarily have to be in half but it cuts them into shorter molecules I might have had you know instead of 16 being broken even eight and eight I might have had a ten and six a nine and five and so forth 925 is at nine and seven so we’re getting an idea here at the very heavy heavy molecules like asphalt the heaviest component of Porter Oleum comes out the very lowest part of the distillation it would take an enormous amount of energy to get this to boil it doesn’t travel up this tube it comes straight out in this sludge like asphalt so for instance a sapling question might do the following rank the following based on boiling point you have pentane propane and hexane pentane we have to become familiar with five carbons propane has three carbons hexane with its six carbons the lightest molecule has the lowest boiling point the heaviest molecule has the highest boiling point so if we go from highest boiling point we would label that the hexane pentane would be right smack in the middle wouldn’t it with its five carbons and the lowest boiling point would be the lightest molecule called propane and you can see where they would come out up here the lightest molecules come out they have the lowest boiling point it takes the least amount of energy to get the lightest molecules to come up this tube it takes an incredible energy to get the heavy molecules to move through this tube so here we have an idea of a sapling question as well this is that same diagram we were just looking at isn’t it it’s representing the distillation of petroleum the diagram represents a fractional distillation tube the crude oil remember the crude oil real arrives in this furnace and the furnace begins to add heat energy and all the different molecules will separate from each other based on the amount of heat energy they absorb which component has the highest boiling point the highest boiling point and often trips up beginning students is not the highest point in the distillation to visit the highest boiling point takes the most energy what you’re looking at is saying who’s the heaviest molecule because the heaviest molecule comes out way down here the highest boiling point is the heaviest molecule to remember what we labeled that as it was asphalt wasn’t it whoops wrong slide there it is we asked fault right there which component is found in location a well up here input location a in our diagram we looked at from our text those are the lightest gases aren’t they the lightest would be methane ethane propane these are what refer to as refinery gases so a we know to put that as carbon one through four gases in the location see we had designated as that heaviest known as asphalt so just as simple matching yours might ask you for different things but we’re getting an idea of where to find things in that fractional distillation tube so how do we use each barrel of petroleum a barrel if ever hear that reported is 42 gallons of petroleum and if this represents the composition and how a barrel of petroleum is generally used you can see that nineteen point two gallons turn into gasoline about 10 gallons turn into diesel and home heating oil three point eight gallons turn into jet fuel heavy fuel oil 1.7 gas liquid refinery gas about 1.7 and other products and petroleum is

also a major component when we start creating plastics which is one of our chapters as well but eighty-seven percent by far is used for transportation and for heating so transportation and for heating the vast majority of this barrel of oil is used for that when we begin considering energy changes we’re kind of getting into our next section here how much energy is actually released through the process of combustion we know that hydrocarbon fuels like methane which we call natural gas and the presence of oxygen undergoes combustion combustion takes a hydrocarbon adds oxygen and forms carbon dioxide and energy wind specifically we know this to be heat energy and we know heat energy is measured in a unit called jewels or calories sometimes we measure in kilojoules if a large amount of energy is produced Sookie Lola courses a thousand when energy is released it’s known as an exothermic reaction energy released is XO leaving exiting exiting the system to those surroundings when you feel heat being released it’s hot to our touch because we’re in the surroundings feeling the heat an exothermic releases heat so the system where the chemistry here loses heat energy it’s actually designated with a negative sign when we write out its Delta H so the combustion of methane for instance releases 50.1 kilojoules for every gram that’s burned or 802.3 kill joules per mole of methane burn often times you’d see this as Delta H the heat of reaction and for an exothermic we’re saying it’s releasing heat so this is a lower energy the negative sign shows that so you’re going to see that written as negative 80 2.3 k JS which is kilojoules per mole let’s take this thermo chemical equation notice how I put the word thermo it’s a balanced chemical equation that includes heat content a thermo chemical equation i’m going to represent that on what’s called a heat diagram the term enthalpy is a thermodynamic quantity and chemistry it just means heat a heat diagram so let’s just take a little XY axes and I’m just going to say on the y axes let’s just put that label as heat content and heat is measured in a unit called kilojoules for an exothermic process the reactants versus the products the reactants are on the left side of the arrow and reactants point at products our reactants are of higher energy content ch4 and the two moles of the molecular oxygen when burn notice I’m still pointing at my products I’m just going to point the arrow down when they form their carbon dioxide and water all I’ve done is kind of shown on an enthalpy diagram the heat content of the reactants versus the products as this reaction proceeds it’s liberating or generating or releasing 802.3 kilojoules of heat energy for every mole that is burned an exothermic diagram releases heat energy exothermic diagram releases heat were reactants release heat down to the products giving off energy how we measure this is typically with an instrument known as a bomb calorimeter and it’s just mentioned in our text where we get the information of how much heat content / fuel we burn it in a bomb calorimeter so we place the fuel inside of this container that’s surrounded with water and when we ignite it and that’s this a little electrical switch up here when you give it a little spark of energy it will ignite and the combustion that occurs inside of this container is going to warm up the water and really what we can detect is how warm the water you know here are just a little stir how much the temperature changes of the water gives us insight into how much energy was released by this particular fuel so this is all done

in a laboratory setting we take a sample of fuel and ignite it and measure how it affects the temperature of the water so different fuels have different heat content when we look at this diagram and this is something we just drew a moment ago together we said that the energy of the reactants this is our natural gas known as methane when it combusts it releases heat the heat is gone to the system and that’s why it’s hot to our touch because we’re outside detecting the heat coming out of the chemical reaction therefore notice the sign on our reaction is negative we have a negative Delta H the Delta H is the heat of the reaction for exothermic it is negative telling us that the system which is our chemistry this reaction gave off heat to the surroundings where we are we turned our balanced chemical equation into a thermo chemical equation by just sharing that Delta energy the heat of our reaction is Delta H for heat and notice we mentioned this as well not all fuels are created equal I can see methane which is called natural gas for every mole of methane that we combust we get out about 50.1 kilojoules per gram notice it’s a very high energy content octane which is in gasoline when we burn one mole of octane it’s giving us about 44.4 jewels of grant joules per gram of energy look at where Cole is a little bit less about 32.8 kilo joules per gram ethanol which is grain alcohol c2h5oh this is our additive recently to octane it’s giving us a little bit not as efficient as octane is it it’s not as efficient and giving us the same amount of energy but here when we were just heating our homes with wood it took a lot of wood to heat our home because there’s not a lot of heat energy stored in that particular molecule so no not all molecules are created equal for fuel content different fuels release different amount of energy let’s compare that exothermic reaction to the opposite process called an endothermic endothermic is heat absorbing I think of a chemical cold pack have you ever had an injury and had to put a cold pack there’s a little clip in there you activate the reaction and it actually gets colder to our touch so when we have an endothermic reaction if this is our container we’re out here heat is going into the system from the surroundings so of course it’s colder out here because the system is absorbing the heat and if you think about just an enthalpy diagram here and this is our little heat content endothermic says as the reactants form products and I’m just going to use those generic letters is that all right whatever our reactants are they always point at the products our heat of reaction is climbing the energy axis so it will be reported as a positive value in kilojoules endothermic absorbs heat energy positive heat of reaction exothermic was the opposite EXO climbed down the energy releasing heat warm to our touch negative Delta H on our axes let’s take a peek at some sapling questions when how this would relate we’re asked to write a balanced equation for the complete combustion of methane now this is a fuel to hydrocarbon we remember the combustion pattern says take your fuel c2h6 which we know we just drew a moment ago just thinking about what ethane is we know it had two carbons attached to each other and the six hydrogen’s attached around c3 c3 bonded together when it combusts any fuel that combust requires oxygen so we add oxygen the process of combustion generates our global warming gases

carbon dioxide and water this is what we refer to as our skeleton it’s not yet balanced so our skeleton equation has the reactants forming products is just not yet balanced and remember a good strategy when we’re a beginning balancing student we’re going to have left and right of our arrow and I want to track seas then h’s then OHS right now on the left side I count two carbons I count six hydrogen’s and I count two oxygens over here I have one carbon 2 h’s and a total of two plus one more is 30 s for no particular reason I just like to start at the top let’s start with carbon two on the left one on the right to balance that I’m going to place a two on the product side to double the number of carbon but let’s repair the teach art because not only did I change the number of carbon i also changed the oxygen there’s two times two which is four plus one more in the water I’m now gotta nut up to five these are good let’s go to the h’s there’s six on the left and only two on the right I can place a 3 in front of the water and when I do so let’s prepare the tea chart 3 times 2 has repaired the h’s but it’s also changed the oxygens two times two is four plus three more 4 plus 3 is 7 so here we have a situation we have two carbons good six H’s aha and the O’s are upset I have two on the left and an odd number on the right when this happens my friends this is the best trick everything you just figured out I mean unless I want to write a fraction which I’m not allowed I can put seven halves there seven half 7 over 2 times 2 is seven that’s going to balance it but I’m not really allowed to put a fraction there I want us to just double everything we figured out so far instead of a 3 let’s put a six instead of a 4 let’s instead of a to double it to a 4 instead of the one make it a two because now what’s happened we’ve doubled all of the numbers everything is still balanced but now look at the scenario instead of 70 s on the right we have four times two is eight plus six that gave us 14 by doubling it I can put a whole number in there and balance the equation so the ratio that balances is to 274 26 to 27 426 created an equal number of atoms on both sides of the arrow I wanted to model that once it was a little tricky with that odd even number problem here’s another sapling question it says ethanol which is a bio fuel ethanol comes from corn when ethanol is burned it releases by the way releasing is another way of saying it’s exothermic it’s being given off warm to our touch 1240 KJ’s of energy right the balanced chemical equation including the heat content Delta H so here’s our strategy we have c2h5oh to combust it means to add oxygen and form carbon dioxide and water the Delta H part the heat of reaction can be added on at the end and we know that it’s negative because it’s releasing heat so I have to show that i know it’s exothermic by using this negative sign and i’m just going to put 1240 KJ’s of heat energy so this is what I refer to as my skeleton it’s not yet balanced but I have in place all of the

bones don’t I I have all the reactants making the products I’ve added on the heat energy I included that Delta H information showing I know it’s exothermic because it’s releasing heat negative value because remember our enthalpy diagram the reactants would point down at the products releasing that heat content between those chemical bonds all you have to do is balance it so again as a beginning chemistry student I like a little tea chart season HSN owes let’s just kind of dive in and see how we do on the left I count two carbons so well I guess I begin with a 2 right here balance the carbons I see a total of six combined hydrogen’s so I’m going to put a three-year see where it gets me for a total of six combined hydrogen’s Caesar good HSR good let’s see what the O’s are doing two times two is four plus three more on the water give me a total of seven oxygens three plus four seven now what’s nice is that there’s one already here in the molecule itself so if i take out the oxygen from the fuel that leaves me a total of six more than I need I can get that by placing a 3 in front of this oxygen three times two is six plus the one that was already in the fuel for a total of seven that we needed to balance the asian the ratio of 1 2 3 2 2 to 3 balanced our thermo chemical equation this is known as the combustion of ethanol which is corn green let’s see what else sapling has you doing a third problem from sapling I kind of just put them all on one slide consider the following thermo chemical equation thermo chemical equation means I know about the heat reaction I notice that it’s negative it’s telling me that it’s exothermic heat is being released from the system I have to h2s and I really should have typed that with subscripts plus a molecule of oxygen forming two water molecules so what I know a 2 to 1 to 2 is my ratio that balance the equation how much heat is transferred 11.8 mole of hydrogen reacts and will it be absorbed or released remember these are keywords absorbed means endo getting cold to our touch released means XO getting hot to our touch because it’s negative we know it’s XO we know that it’s being released so I know that one right off the bat by looking at the sign on the Delta H I know it’s negative so heat is being released but here’s what I know from my reaction see this balanced equation for every two moles of hydrogen 484 kj’s of heat is released this is going to be my ratio that I’m going to use to convert down here so let’s start by writing our number that’s given 1.8 the unit that’s given is the unit mole and of what is that of hydrogen I always just like to write my number unit and label of what given I want to set up a conversion factor and I always think about what do you want over what you’re given I want to know the kilojoules of heat right how much heat that’s a kilojoule of heat and my given quantity is from the mole this comes from the balanced equation this conversion factor is just grabbing what you know from the balanced equation the kilojoules of heat were negative for 84 from our balance thermo chemical equation when two moles of water were used I’m sorry two moles of hydrogen were used so by creating this conversion we’ve cancelled moles of hydrogen and are going to be left with kilojoules of

heat we were given 1.8 moles of hydrogen we wanted to know the kilojoules of heat when we move from left to right we set up a conversion that cancels from our balanced equation we’re just pulling what we know from the actual equation from the balanced equation we saw our kilojoules over to what is that 484 over to where’s my calculator for 84 / 2 times 1 point 8 and it will be negative because it’s a negative exothermic value so 430 five point six kilojoules of heat released exothermic due to that negative sign it’s actually warm to our touch I’ll let you process this lesson so far when ready come on back for less than 3 and we’ll wrap up our chapter together good work today gang