Palmitic acid (C₁₆H₃₂O₂) is a dietary fat found in beef and butter. The caloric content of palmitic acid is typical of fats in general. Write a balanced equation for the complete combustion of palmitic acid and calculate the standard enthalpy of combustion. What is the caloric content of palmitic acid in Cal/g? The standard enthalpy of formation of palmitic acid is -208 kJ/mol and that of sucrose is -2226.1 kJ/mol. [Use H₂O(l) in the balanced chemical equations because the metabolism of these compounds produces liquid water.]

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Hi everyone for this problem, we need to write a balanced equation for the complete combustion of linda lake acid, which is a poli unsaturated omega six fatty acid found in olive oil. We're going to assume that the water produced is in liquid phase. We need to calculate the standard entropy of combustion and killing joules per mole and the calorie content of linoleic acid and calories program were given the standard entropy change for the reaction, Which is negative 634.7 killer jewels per mole. So let's go ahead and get started. There are three things we need to do for this problem. The first is right the balanced equation and then we're going to calculate the entropy, the standard entropy of combustion and lastly the calorie content and calories per gram. So for the balanced chemical equation, we have our linoleic acid, C 18 H oh to and because it's a combustion, it's going to react with oxygen and this is a solid, so we're going to react with oxygen gas. But we don't know how many moles. We need to balance this and then this is going to produce carbon dioxide gas and water and we're told that it's going to be in the liquid phase. So let's go ahead and balance this out, which is step one. So if we take a look, we see that on the left side, We have 32 hydrogen. So that means on the right side, we're going to need 32 hydrogen. So we can go ahead and put 16 here On the left side, we have 18 carbons. And so that means on the right side, we need 18 carbons. And lastly, for our oxygen, we calculate on the right side that we have 26 02s. And so that means we're going to need 26 here. Okay, so that is our balanced equation. Part two is to calculate the standard entropy of combustion. And so our standard entropy of our reaction is going to be the sum of our products, minus the sum of our reactant. And I'll explain what that means. We're going to multiply the moles of our products and react ints by their standard heats of formation. And these are values that we can look up. So when we look up our standard heats of formation for our liquid water, our standard heat of formation is negative. Let me move this a little bit to the left, it's going to be negative. 200 and 85. killer jewels per more that cut off a little bit so let me move it down here. So for liquid water it's negative 285.8 kg joules per mole for our carbon dioxide Gas. Our standard heat of formation is negative 0.5 killer jewels per mole. And for anything in its elemental state. So here we have oxygen gas in its elemental state. Our standard heat of formation is going to be zero. Okay, and believe that is it. Okay, so let's go ahead and calculate. So for standard entropy of combustion, we have Let's take a look at our products, we have 16 waters, 16 moles. So we're going to multiply the moles of water by its standard heat of formation, which we said is negative 285. Killer joules per mole. And I'm going to and the units here, um and then we're going to add. So we're doing some of our products, we have 18 moles of carbon dioxide And for a balanced equation, this should be ACO two there. So we have 18 moles of carbon dioxide and the standard heat of formation is negative 393 five killer jewels per mole. So that's the sum of our products minus the sum of our reactant. We have 26 moles of oxygen gas, which we set is a standard heat of formation of zero. So that's time zero joules per mole Plus we have one mole of little lake acid. And we were told in the problem that the standard heat of formation here was negative 634 Killer joules per mole. Okay, so now we just do this calculation, we can go ahead and cross this out because we know that is going to be zero. And so we'll get a standard he of combustion, or standard entropy of combustion to be negative 11655. killer jewels, Nope, Sorry, negative. 11, 21 . Killer Jewels her more. So that is the answer to part two. And the last part of the question asked us for the caloric content and calories per gram. So if our standard heat of our standard entropy change of our combustion Is negative 11,021 0.1 kg joules per mole, let's rewrite this so that it is written this way. Okay, so this is per one more. Okay, and so in order for us to do the calorie content, we're going to say that in one calorie there is 4.184 killed Jules. So are kill jules cancel. And now we're in calories per mole. But it asked us for calories per gram. And so that means we need the molar mass of linoleic acid and We can stay in one mole of linoleic acid. We can calculate the molar mass and the molar mass is going to be 280.5 g. And from there our moles cancel. So now you see we're left with calories per gram. And so that leaves us with 9. calories per gram for our caloric content. And then we said that our standard entropy of combustion was negative. 11,021 kg jewels Permal. So those are the three parts of this problem. We wrote the balanced equation. We completed the standard entropy of combustion and the caloric content. That is the end of this problem. I hope this was helpful.

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Key Concepts

Combustion Reactions

Enthalpy of Combustion

Caloric Content