Heats of Reaction

Ajouté : 2026-05-15

[00:00:01]all right guys in this video we're going to look at new ways of calculating our heat of reaction um and here we're calculating Delta H KN so anytime you see that degree symbol it is representing Delta H KN and what that degree symbol means is that we are under standard conditions for our reaction and standard conditions if we're deing with a gas it means we're kind of under ideal conditions so we're going to be at that standard pressure of one atmosphere if we're dealing with solids and liquids they are going to be pure they're not going to be mixtures if we're dealing with Solutions we're going to say that they are at one molar if we are under standard conditions and temperature- wise we tend to be about room temperature which is that 25° C so we will kind of have to account for those and those temperatures and pressures may show up later on in other calculations but understand that degree symbol means that we're under standard conditions there are going to be three ways for calculating Delta H knot the first one is going to be using hey to formation values these are values that are in a table in your textbook or they will be provided for the specific questions and the specific substances that you need we're also going to be looking at learning hess's law um this deals with some of those reactions that have mechanisms where it's multiple reactions that combined to make the overall reaction and then we can also find Delta H knot by using Bond Association energies this is going to look at the energy from bonds broken and bonds formed throughout the course of a reaction so let's look at heat of formation first so this looks at the change in enthalpy that accompanies formation of one mole of a compound from its elements with all substances in standard States now note that this produces one mole of a substance so if it is a reaction that produces is 2 moles it is not a heat of formation reaction if it has compounds that produce one mole of a substance it is not a heat of formation reaction it has to have elements forming that compound and only one mole of that compound so you might have a heat of formation reaction um typ typically like when we produce water so hydrogen oxygen combined to make water and balanced reaction wise there are two waters and two hydrogens this would not be a heat offormation reaction this would have to be simplified to only produce one mole of water so we would have to cut all of our coefficients in half so this is an instance where you may see coefficients that are fractions if they are heat of formation reactions so you may just be asked to recognize which of the following is a heat offormation reaction we can actually calculate the values of heat of formation reactions um something to note is the value for a free element in its standard state is zero so if you have elemental oxygen or Elemental iron the value will be zero for those um but we will have certain values for compounds and like I said that will be in a table in your textbook or it will be given in context of a specific question so heat of formation values are in appendix 4 of your textbook which is near the back but not all the way at the back um I believe it's page A5 they start at those are going to be different values depending on which compound we are dealing with so let's look at how this is to find heat of formation or heat of reaction from heat of formation values we have to do the sum of the heat of formation of the products minus the sum of the heat of formation values for the reactants and you are going to have to account for coefficients in these but to find the overall heat you're looking at the products minus the reactants for these so let's look at this example if we want to calculate the enthalpy change remember enthalpy is just heat so calculate the enthalpy change for the oxidation of iron 2 oxide into iron 3 oxide and here it gives us the equation of we have four iron 2 oxides and oxygen producing two iron 2 oxides so if we're looking at the heat of the reaction remember we said heat of reaction is the sum of heat of the products minus the sum of heat of the reactants so whenever we actually lay this out we have our product is the iron 3 ox oxide and note that there are two of them that is our only reactant or excuse me our only product so that whole thing is going to be our sum of our products that subtracts the sum of our Heats of reaction for the reactants notice that oxygen is an element its heat of formation value will be zero so we can ignore oxygen here but our other reactant is the iron 2 oxide and there are four of them so we need that total sum and that will be the sum of our reactance like I said these values are in that appendix in the back of your textbooks but these will typically be given in a table or in context with the specific questions if we look in the back of our textbook the value for iron 3 oxide is8 24.2 and that's K per mole so we would need that total to be the sum of our products iron 2 oxide when we look in the back of the book is - 272.50 ultimately so simplifying this you can put this in the calculator as is but the sum for our products would be 1, 16484 K the sum for our reactants would be 1,088 K and whenever we add these together it's going to be 56.4 K and since all of these are constants from our textbook we're not going to round with these we're going to leave it as is because we're dealing with constants so let's look at another one following the same kind of pattern we're going to calculate the heat of reaction for the following process and we have sulfur dioxide with water making sulfur trioxide so yet again we have to do the sum of heat for the products minus the sum of heat for the reactants so our product is the sulfur trioxide and there are two of them that's our only product that is going to subtract the sum of Heats for our reactants well we have sulfur dioxide and there are two of them and then we also have oxygen remember that will be zero so we don't have to include that whenever we're looking at it when we use the values from our textbook sulfur trioxide is 395.8 our value for sulfur dioxide is- 296.33 k for the products the reactants will be 5936 K so when we subtract that we get 97.8 Kil so this would be an exothermic process overall because it is releasing heat all right so let's look at hess's law like said hess's La is very similar to looking at our mechanisms like we did in kinetics where there are multiple steps to make up the overall reaction so with hessa's law the heat of reaction is going to end up being the same whether it takes place in one step or a series of steps so if we want to know the heat of the overall reaction we can combine the Heats of reaction for the individual steps that make up that overall reaction um this is true because enthalpy is a state function that means it does not change based off of outside conditions so let's look at how this works here we're going to determine the heat of reaction for the following reaction we have copper 2 oxide and hydrogen producing copper and water and it gives us two different thermochemical equations the first one has copper 2 oxide making copper and half of an oxygen molecule and it gives us the heat of reaction for that specific reaction and then we also have water producing hydrogen and half of an oxygen molecule and we have the heat of reaction for that one as well we're going to have to look at how can we use these thermochemical equations to get our overall equation and you may have to flip a reaction and reverse it you may have to multiply a reaction by a factor and anytime you change the reaction it will change our Delta H value in the same way so looking at this looking at the first reaction that was given with the thermochemical equations the copper 2 oxide making copper in half of an oxygen if you notice we have a copper 2 oxide on the reactant side in both reactions we have copper on the products in both reactions so and the coefficients are the same so we do not need to change that reaction at all but if we look at the second reaction that's water producing hydrogen noxion notice that water is a reactant in our thermochemical equation and a product in our overall equation so that means that this reaction is reversed so we need to flip this reaction so I'll rewrite it flipped so we have h2+ half of an oxygen produces water and what happens is if we flip the reaction it changes the sign of our Delta H equation so is now going to be 242 K so I'm just going to Mark out that original one so I don't get confused now we have water on the product side we have H2 on the reactant side and if we were to add these together the half o2s would cancel meaning they are not in our overall reaction so we have rearranged the equations in the order that we need them them so what we do at this point is we now add together the Delta H values to get the overall heat for the reaction so these added together would be netive 87 K so this whole process would still be exothermic and would produce the same amount of heat whether it was done as one step or in multiple steps let's look at another one this one's a little bit trickier here we have magnesium oxide and hydrochloric or hydrogen chloride gas producing magnesium chloride and water so here we have three thermodynamic equations and we need to see what we need to do to these equations to make sure they can equal our overall equation so from the first thermochemical equation that's given magnesium is not in there at all so we're hoping it will cancel at some point in time but we have two hydrogen chloride gas as a reactant two hydrogen chloride gases reactant so and we have one magnesium chloride as a solid one magnesium chloride as a solid so we are not going to have to change that reaction at all so let's look at another one and change to red for the second equation looking at the second equation we have magnesium which is not in our overall reaction half an oxygen is not in our overall reaction but we have this magnesium chloride or magnesium oxide excuse me that is a product but in our reaction it is a reactant so this reaction will need to flip so I'm going to rewrite this down below so if we have the magnesium oxide plus magnesium plus half oxygen and remember that means that we are going to have to change the sign of our Delta H value so I'm going to mark that one out since we changed it so now the magnesium oxide is in the correct spot and it has the correct coefficient so we're not going to have to worry about that I'm going to swap to Green for the third equation looking at the third equation we have water producing hydrogen and half of an oxygen water is a reactant in the equation given but our overall reaction water is a product so this reaction also needs to flip so we're going to have hydrogen plus the half oxygen producing water that also will change the sign of our Delta H value for that reaction so that will cancel out and notice that now water is on up the product side and is the correct coefficient we know these are going to add together correctly because now we have magnesium on the product and reactant side that will can cancel and we have the half oxygen on the reactant and product side that will cancel and hydrogen on the reactant and product side that will cancel so everything else that we needed is Left Behind so now that we've rearranged it properly we can add our Delta H values together and I'm using the ones that I wrote that way I changed the sign of them when we add these together it should be 141.ir but we are going to do an example where the coefficients do not match whereas in this one all of our coefficients matched once we flipped the reactions so let's look at this one um we have two carbon in the graphite form reacting with hydrogen to produce ethine the C2 H2 and we have three thermochemical equations and we need to see what we're going to do to them to get our overall equation so notice that we do have graphite on our reactant side in both reactions but in our overall reaction there are two graphites whereas our thermochemical equation only has one so we are going to have to multiply this reaction by two meaning we are going to multiply our H value by two as well so if you multiply the reaction you also have to multiply that Delta H value when we multiply that this is going to change to - 787 K and I'm just going to put a two in front of all of our substances that will help us out a little bit looking at the second reaction we have hydrogen as a reactant and hydrogen is a reactant in our original one and they are the same coefficient so we don't have to change that second reaction anymore but if we look at the third reaction we have two C2 H2S as a reactant and we have one C2 H2 as a product in our overall reaction so we have to flip this reaction and we have to divide by two because we need the coefficients to match up so I'm going to rewrite this flit and divide it by two so it's going to be two co2's of water producing a c2h2 and 2.5 oxygens now since we flipped the reaction that does change our sign of Delta H and since we divided by two we also have to divide this by two so our new Delta H value for this reaction is going to be 1, 12994 K so I'm going Mark out that equation so that I see my redone one we can always double check that it is going to cancel out properly so we have two co2's in the first reaction and in the third reaction on both sides water in the green one we have 2.5 Waters we have two from the first one and half from the second one so those will cancel or excuse me oxygen not water we have water will cancel from the second and third reaction and then then everything else that is left is correct so we can now add together our Heats of reaction and this is going to be 226.50 K so this is an endothermic reaction it requires energy for it to occur so that was looking at hessa's law and like I said be careful that the substances are on the correct side are they on the react side or product o side and make sure your coefficients match with those the third way that we can solve for heat of reaction values is by looking at bond dissociation energy so there's energy stored within bonds so energy is required to break a chemical bond so energy is used to break a bond and is released whenever a bond forms so so we're going to be looking at the energy used to break bonds compared to the energy used to form those new bonds so four heat of reaction with bond dissociation energy and this is different than the other heat of formation one we did we're going to look at the bond dissociation energy for the reactants minus the bond dissociation energy for the products so this when the reactants and products are switched because we're looking at the energy used whenever the bonds are broken versus release whenever the bonds are formed and that's the difference we're looking for um and they updated our textbook so there is a table of bond dissociation energies on page 422 of your textbook and you do have to pay attention to the actual structure is it a single Bond it is a double bond or a triple bond because remember triple bonds are stronger it requires more energy to break apart a triple bond than it does a single or double bond so these you may have to draw out the structures that way you know what bonds are being broken and formed so let's look at a problem with this and yet again like I said the these are in the textbook on page 422 in terms of an actual problem um like on a quizzer test you would be given values to use for these so looking at this reaction um we have ethine producing ethane whenever mixed with hydrogen and notice the way that it is drawn out this is a very basic Dot Structure you may need to expand this um but it does tell us we have a double bond in the first structure so this is a carbon with two hydrogens double bonded to another carbon with two hydrogens that is with hydrogen which remember has a single Bond and that is making the ethane so the carbon with three hydrogens single Bond if it doesn't show a double or triple bond you assume it's single bonded going to a carbon with three hydrogens and like we said we're going to be doing the energy of the reactants minus the energy of the products for this one this one is backwards this is the only one that we do reactants minus products so we're going to have to take into account what specific bonds are being broken for the reactants so in the reactants we have these CH bonds that are single bonded and if you count there are four of them so we have four CH bonds we have this double bonded carbon to carbon and then we also have this single bonded hydrogen to hydrogen so those are all of the ones that are on our reactant side remember that is going to subtract our product side stud part this Interruption students that will announ please take your things and go to the cafeteria at this time you will remain in the cafeteria stud we have those single bonded carbon hydrogen bonds so that you can purchase lunch and remain there and there are six of them and we also have this single bonded carbon carbon sorry about that so these are the values that you would be given or look up in your textbook so I went ahead and looked them up in a textbook our carbon to hydrogen single bond is 414 kog and like I said remember there are four of them a carbon double bonded to another carbon is six 11 hydrogen single Bond de hydrogen is 436 so that would give us the total for our reactants um on the product side that carbon single bonded hydrogen was 414 carbon single bonded to carbon is 347 so we would get that total for our products and we could have simplified this um where we have the same number of Bonds on both sides um but people tend to make mistakes sometimes doing that so our total for our reactants is 273 K our total for our products is 2831 K so whenever we subtract these it is netive 128 K so this overall reaction would release energy more energy is going to be released when the bonds are formed than we're used to actually break those bonds so those are the three different ways that we can calculate heat of reaction from those and we'll do some practice problems with this later on