The explosive nitroglycerin (C₃H₅N₃O₉) decomposes rapidly upon ignition or sudden impact according to the balanced equation: 4 C₃H₅N₃O₉(l) → 12 CO₂(g) + 10 H₂O(g) + 6 N₂(g) + O₂(g) ΔH°_rxn = –5678 kJ Calculate the standard enthalpy of formation (ΔH°_f ) for nitroglycerin.

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Hi, everybody. Welcome back. Our next question says the explosive nitroglycerin, which has a formula formula of C three H five N 309 decomposes rapidly upon ignition or sudden impact. According to the balanced equation, we have four C three H five M 309 in liquid form, subscript I goes to 12 CO2 gas form plus 10 C H2O in gas form plus six and two in gas forms plus 0 2 in gas form, we're given the delta H. So entropy of the reaction which is equal to negative 5678 kilojoules. And then it says, calculate the standard entropy of formation. So delta HF for Nitroglycerin, a 365.7 kilojoules per mole B negative 456.5 kilojoules per mole, C 365.5 kilojoules per mole or D 455.6 kilojoules per mole. So when we think about the entropy of formation of something, we might keep in mind that when we have a reaction with that compound in it, that the entropy of a reaction. So delta H uh we put the little degree meaning standard temperature and pressure of a reaction is equal to the sum of the op piece of formation of the products minus the sum of the enlo piece of formation of the reactants. So we have here a reaction including nitroglycerin and it is decomposing into some very simple elements that we could easily obtain by looking up their entropies of formation. So let's just write out this equation for delta hr reaction. So the sum of the entropy of formation of the products only we have to remember that we have to take into account our coefficients and our balanced equation. So we put N in terms of our coefficient there and then that's multiplied by the entropy of formation. So H formation of the products minus the sum of N times the H of formation of the reactants. And we know the delta H of the reaction. So that's known it's given to us right here. And then our products have known piece of formation as well. We have nitrogen gas and oxygen gas, which are the elements in their standard state. So they will have an P of formation of zero. Then our other two products are carbon dioxide and water. Simple basic molecules. We can look those up and then that will give us N times H of the entropy formation of our reactant. We have only one reactant, it's nitroglycerin. So this is what we're looking for. So I'm going to go ahead and highlight that in blue because that's how we can obtain, we can solve for that heated formation. So first, we need to double check is our reaction balanced. We need to use the coefficients here can't go anywhere good if we don't have a balanced equation. But we're told right in the problem that we have a balanced equation. So we don't need to figure that all out. So now let's look for our an ample piece of formation of each of our products. So as we said, we have nitrogen and oxygen gas, this will have both of these will have a zero equals their entropy of formation because there are elements in their standard state. They're sort of the baseline when we talk about Anthes of formation and then we look up the entropy of formation of carbon dioxide and of water. So we'll do that. Our entropy formation of carbon dioxide is negative 393.5 kilojoules per mole. Our entropy formation of water is negative 241.8 kilojoules per mole. So let's start plugging these into our equation. So we start with our entropy of formation of the reaction, which we know is, is right here in our problem. So that is this negative 5678 kilojoules. So we'll start our equation by saying negative 5678 kilojoules equals. Well, now we need the sum of N times the heated formation of the products. Now, quick side note here, notice that little piece of formation of molecules or elements are given in kilojoules per mole units. Our entropy of the whole reaction is given in kilojoules because it's based on a specific amount of those substances. How do we reconcile those units that comes down to those ends? The end is the coefficient, but more specifically, it's the number of moles of the product. So that's where the moles comes in and it's going to cancel out the moles in the kilojoules per mole leaving us with units of kilojoules. So don't get confused about that. We see the coefficients, they're written as numbers, but they do have a unit, it's moles. So the sum of N times the and there'll be a formation of the products. So our first is carbon dioxide, its coefficient is 12. So we have, I put in brackets, 12 moles of carbon dioxide multiplied by its heated formation, negative 393.5 kilojoules per mole. So you see how the moles are going to cancel out there. And then we add now the heated formation of our water and its coefficient is 10. So 10 moles times negative 241 or multiplied by negative 241.8 killer Joel for more. And then our other products of course have that heated formation of zero. So we end that there, close the brackets and then we have minus I'm going to continue this on the next line as I don't have any more room here. So let's continue down below minus four moles. So now we're on the sum of the heats of formation of our reactants. So four is the coefficient of nitroglycerin. So four moles times are multiplied by the heat of formation of nitroglycerin, which is what I'm looking for. So just to make this a little simpler to write and clear up, I'm going to write that in the further steps as X, we'll just highlight that in blue, X is going to mean the heated formation of nitroglycerin. So now let's simplify this equation a little bit here. So I still have negative 5678 kilojoules equals. So now let's do the math inside here and we get. So then I have negative 5678 kilojoules equals. And then note, I didn't need to have parentheses around the answers here, but because they're both negative values for my next two, Next two terms I like just as a tip to leave those negative numbers in parentheses because it can be really easy to lose a negative and then you have the wrong answer. So I put in brackets and then within parentheses, negative 4722 kilojoules plus and then in parentheses again, negative 2418 kilojoules again, that's just for bookkeeping purposes. So as not to leave negative sign and then close brackets and that's all minus four moles times X. So we'll simplify still more to get or say I'm going to simplify and rearrange. So now that I'm just to these three terms, I'm going to move my X term to the left side. So I'm going to add four most times X to both sides and subtract my enthalpy of reaction. So then on the left side, I'll have more moles times X equals and then I have my, I'll add together my two numbers in the brackets. So that is going to equal negative again, I'll put that in parenthesis just so as not to lose my negative negative 7140 kilojoules. And now to move my heat of rea entropy of reaction to the other side, I've added to both sides. So plus 5678 kilojoules, it was negative on the other side. So I've added it. So that means that four moles time multiplied by X is going to equal when I put those together negative 1462 kilojoules. And now I will solve for X by dividing both sides by four moles. And you can see I am going to come out with my correct units for heated formation kilojoules per mole. So a negative 1462 kilojoules divided by four moles gives me negative 365.5 kill a jules per mole. And there again, remember X was my heat of formation of nitroglycerin, which is what I was looking for. So now when I look at my answer choice, I see that indeed troy is negative 365.5 kilojoules per mole. So again, I use the fact that when I'm given an entropy of reaction, it's the sum of the heated formation of the products minus the sum of the heated formation of the reactants. And since I just had my desired heated formation, my desired chemical as all by itself on one side, I could use that to solve for its heated formation. Giving me that answer of choice C negative 365.5 kilojoules per mole. See in the next video.

Verified Solution

Key Concepts

Standard Enthalpy of Formation (ΔH°f)

Hess's Law

Balanced Chemical Equations

The explosive nitroglycerin (C3H5N3O9) decomposes rapidly upon ignition or sudden impact according to the balanced equation: 4 C3H5N3O9(l) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) ΔH°rxn = –5678 kJ Calculate the standard enthalpy of formation (ΔH°f ) for nitroglycerin.

Verified Solution

Key Concepts

Standard Enthalpy of Formation (ΔH°f)

Hess's Law

Balanced Chemical Equations