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Hey guys, so in this question, we're given stearic acid, and we're asked to do four different things with this.
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First, we have to find a balanced equation for its combustion reaction.
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Second, we need to find the delta h of the reaction.
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And then we had to find how much heat is produced when one gram of stearic acid is burned, and as well as comparing how a candy bar and its 11 grams of fat will, is it consistent to the.
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The amount of heat that's produced that we found in part c.
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So i will go over all of them and try my best to explain how to do each step.
00:41
So in part a, we have stearic acid, which is c -18 h -36 -o2.
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And it's combusting, so that means it's reacting with oxygen, and a combustion reaction produces carbon dioxide and water.
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And these are gases since we need to know that since this is a thermochemistry question and we need to know what state they are in.
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So to balance this out, i'm going to go ahead and put an 18 by the carbon dioxide to balance the carbons, as well as an 18 in front of the hydrogens to balance out the hydrogens.
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Now to balance out the carbon or the, excuse me, the oxygens, there are 54 oxygens on the right side of the equation.
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And to make it even on the left side, we have to put a coefficient of 26 in front of our oxygen.
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So that is our balanced equation for the combustion of stearic acid.
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Now for part b, we have to find the delta h reaction at standard conditions.
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So that is equal to our delta h products minus our delta h of our region.
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Now we just need to look for these values in the book, and we're also given the delta h formation for our stearic acid.
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So i'll go ahead and plug these into our equation.
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So for carbon dioxide, we have a coefficient of 18, so we have to include that in this equation.
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So it's 18 times negative 393 .5 plus 18.
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Times negative 241 .8 for water and that is subtracted from a negative 948 for styric acid.
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Now all of these values are in kilojoules and i just didn't write the units throughout the equation just to save some room.
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Now plugging all of this into our calculator, we find that this is equal to negative 10 ,487 kilojoules...