Consider the reaction 2 H21g2 + O21g2¡2 H2O1l2.
(a) Use the bond ...

Question

Consider the reaction 2 H21g2 + O21g2¡2 H2O1l2.
(a) Use the bond enthalpies in Table 5.4 to estimate H for
this reaction, ignoring the fact that water is in the liquid
state.

Accepted Answer

Hey everyone in this example, we have unsaturated hydrocarbons, reacting with chlorine to produce die chloral canes. We need to use a table of average bond energies to calculate reaction entropy or delta H reaction from the following chlorination reaction. So looking at our first reactant, we would recognize this as one propane. We have our chlorine gas here and then we have our product which we would recognize is one to die chlor oh, al caine. So the next step is to recall our formula for reaction entropy. So we would recall that that is going to be delta H of our reaction equal to the sum of the n therapies of the bonds of our products. Or sorry, reactant subtracted from the sum of the bond n therapies of our products and actually we can go ahead and make this a different color so that it's extra clear. So we'll go with purple for the sum of the bond entropy is of our products. So before we get into the reaction, let's draw out the molecules in this reaction. To draw out our one propane, we would recognize that this will consist of our metal group first, then leading us into our first carbon atom where we are bonded to another carbon atom which is bonded to two hydrogen atoms. And this first carbon again is a methyl group which means that we have three hydrogen is bonded to this carbon and because this is an elk in we want to go ahead and include a double bond here as well as show that this carbon is bonded to its hydrogen here for 1/4 bond to be stable. So this is our one propane. We would be reacting with chlorine gas, which we recall. The molecule for chlorine gas is just two chlorine atoms in a single bond. And then for our product. And let's go ahead and just fix this. We'll begin the molecule with a metal group where actually we have a methyl group and then we get into a second carbon bond here. So we'll do this. I'm sorry, this should be CH three. This second carbon here is bonded to one chlorine atom and then we have this carbon bonded to another carbon where we have another attachment to a chlorine atom here. So this is our one to die chloral, al caine molecule. And we drew this out so that it's easy to see the types of bonds we have for our calculation of entropy. So what we would have is that our entropy of our reaction is equal to first beginning with the types of bonds on our reactant. Looking at one propane, we would recognize that the first type of bond we have is carbons bonded to hydrogen and we have three carbons bonded to hydrogen is here. So three of those types of bonds here. Then we have another carbon hydrogen bond here for four and then we have five and six. So that would give us six carbon hydrogen single bonds. Moving on to the next type of bond in this molecule. We would recognize that we have carbon double bonded to a carbon and that would give us just a total of one of this type of bond. So we would have one carbon double bonded to another carbon bond adding up the next type of bond in this molecule we have a carbon bonded to a carbon single with a single bond. So this is just one of this type of bond here. So we would say plus one carbon single bonded to another carbon and then moving on to the next type of bond, which for our second product, chlorine gas, is a chlorine chlorine single bond where we just have one of those to make up our chlorine gas molecule. And this would complete the some of the types of bonds for our reactant. So now we're going to take the difference from the type of bonds of our products, which is the one to declare a al Kane. So, beginning this out, we would have again carbon hydrogen bonds where we have three here, we would recognize that this second carbon here to be stable, would also be bonded to A 4th Atom which is a hydrogen here. So we would count four carbon hydrogen bonds and then this carbon to be stable would be wanted to two hydrogen atoms. So we would count five and then six again. So this would be six carbon hydrogen single bonds. Then the next type of bond in this molecule we would recognize as a carbon carbon single bond where we have one, two of those carbon carbon single bonds. So we have carbon carbon single bond here, we have two of those for our product. And then the last type of bond we have in our product. One to declare al Kane is a carbon chlorine bond. So this is one carbon chlorine bond and this is the second carbon chlorine bond. So this would give us a total of two of our carbon chlorine bonds. So this completes to some of the bonds of our products. The next step is to recognize that we can cancel out what matches on both the reactant and product side. So we can cancel out the six carbon hydrogen single bonds here. And now we're just going to take the difference between the types of bonds of our reactant from products. So what we would have is that our entropy of our reaction is equal to now we have one carbon carbon double bond left over from our reactant. We also have one chlorine chlorine single bond from our reactant leftover. And when we do one carbon carbon single bond minus two carbon carbon single bonds, we would get negative two Correction. That would be -1. We would get negative one of our carbon carbon single bond. And then lastly because we have zero of the carbon chlorine bonds on our reactant, we would have just zero two of the carbon chlorine bonds on our products, which would give us zero minus two, which is negative two of our carbon chlorine single bonds. And now at this point we want to pull out our textbooks to plug in the bond energies of these types of bonds. So we would have that the entropy of our reaction is equal to When we look up the bond energy of a carbon carbon double bond, we see that it has a value of 614 kg joules per mole in our textbooks this is then added to the bond energy of a chlorine chlorine single bond which in our textbooks we could see is a value of 239 kg joules per mole. Then we have added to this. The bond energy or rather not added but subtracted because we have minus one times the bond energy of a carbon carbon single bond which in our textbooks we would see has a value of 347 kg joules per mole. And then for our carbon chlorine single bond we have minus two times the carbon or the bond energy of a carbon chlorine single bond which in our textbooks is a value of 339 kg joules per mole. So plugging these values into our calculators, we would get that the entropy of our reaction is equal to. We're going to take 6 14 plus 2 39 that's going to give us a sum of 853 kg joules per mole. And then we're going to take negative one times 3 47 to give us negative 47 minus two times 3 39. Which would give us 6 78. So taking negative 3 -678 is going to give us a value of negative 1025 kg jewels Permal. So when we take this difference we get that the entropy of a reaction is equal to a value of negative 172 kg joules per mole. And this would be our final answer to complete this example as the entropy of our given reaction. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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