2 H₃C−CH₃(g) + 7O₂(g) → 4 CO

Question

2 H₃C−CH₃(g) + 7O₂(g) → 4 CO₂(g) + 6 H₂O(g)

Accepted Answer

welcome back everyone. We need to estimate the heat of combustion and kill jules for propane C three H eight. Using the falling bond association energies. So we're going to need to begin by writing out the reaction for the combustion of propane. And we want to recall that combustion reactions involve our reactant which in this case is propane gas reacting with oxygen gas and our products are always going to be carbon dioxide and water. And sorry, water is just H 20 here. So our next step is to balance this reaction out. And based on the atoms on both sides of our equation, we're going to have to add a coefficient of five in front of our oxygen, a coefficient of four in front of water and a coefficient of three in front of carbon dioxide in order to have a balanced reaction. Now that we have our bounced reaction our next episode of our structures of our reactant and products. So beginning with propane, we want to recall that we have three atoms of carbon where we have a total of eight hydrogen atoms surrounding these carbon atoms so filling that in, we would just make our connections to each of our atoms here. And this is our structure of propane where we can count that we have a total of eight of our carbon hydrogen single bonds for propane and and sorry, we're missing a bond here between our carbon atoms. So we have a total of two of our carbon carbon single bonds. Now drawing out our structure of oxygen, we have two oxygen atoms bonded to one another, where these oxygen atoms are connected by a double bond and our oxygen atoms have two sets of lone pairs on themselves. Because we recognize that oxygen based on its position in the periodic table, is located in group six A and should have six valence electrons where we have to in these Covalin bonds and then the other two as lone pairs as we drew in. And now for our products, drawing out our structure of carbon dioxide, we have Carbon in the center surrounded by two oxygen atoms, where we have two bonds to carbon to give carbon four bonds and then our oxygen atoms still have those two lone pairs on each of these oxygen atoms. As we stated, we have three moles of carbon dioxide and five moles of oxygen, meaning that we have a total of five of our oxygen double bonded to oxygen bonds and three of our, sorry correction, that would be six because we have two sets of our oxygen carbon double bonds here, so six of our oxygen carbon double bonds. And then lastly, we have our second product, water, we know our molecule of water is just oxygen bonded to two hydrogen with single bonds where oxygen has two sets of lone pairs on itself. And because we have four moles of water as a product, we would have a total of eight of our oxygen hydrogen single bonds. Now that we've listed out our structures and determined are number of different types of bonds for our re agents in our reaction. We're going to now determine our heat of combustion for propane by finding the entropy change of our reaction, which we recall is delta H. And we would recognize that it can be related to the bond association energy of our reactant bonds subtracted from the bond association energy of our product bonds. So simplifying this, we can say that we have our entropy of our reaction equal to where we begin our brackets. We have two moles multiplied by our bond association energies of our carbon carbon single bonds, which is added to eight moles of our or sorry, eight moles multiplied by our bond dissociation energies of our carbon hydrogen single bonds then added to five moles, multiplied by the bond dissociation energies of our oxygen oxygen double bond. We would close off our brackets for the bond association energy of our reacting bonds. And now begin by subtracting by the bond dissociation energies of our product bonds, which we'll use in purple. So we begin our brackets and we have six moles multiplied by the bond dissociation energies of our carbon oxygen double bonds added to eight moles, multiplied by the bond dissociation energy of our oxygen hydrogen single bonds and that we can close off our brackets for the bond association energy of our product bonds. And let's make that clear, this is the end of this bracket. So now let's just plug in our values given from the chart in our prompt. So we can say that our entropy of our reaction is equal to beginning our brackets. Two moles times the bond association energy of a carbon carbon single bond which from the prompt is given us 350 kg joules per mole. This is then added to eight moles, multiplied by the bond dissociation energies of our carbon hydrogen single bonds given in the prompt as 410 kg joules per mole. And then added to this, we have five moles multiplied by the bond association energies of our oxygen, oxygen double bond given in the prompt as 498 kg joules per mole. So closing off our brackets and beginning by subtracting from our brackets for the bond dissociation energies of our products. We have six moles multiplied by the bond association energy of our carbon, oxygen double bond given in the prompt as 804 kg per mole. Which is then added to eight moles multiplied by the bond dissociation energies of our oxygen, hydrogen single bonds given in the prompt as sorry we should just right killer joules per mole. So we can close off our brackets. And now we want to simplify both brackets so that now we can have in our next line that our entropy of our reaction is equal to for our bond association energies of our reactant. We should simplify to cancel all of our units of moles. And we're gonna be left with kg jewels. And now subtracting this from the bond dissociation energies of our products, which will simplify to a value of 8504 kg jewels. And so taking the difference between these two values, we would find that our entropy of our reaction is equal to a value of negative 2034 killing tools. And this is going to be our final answer for the heat of combustion and kill a jewels for propane. Using our bond association energy is given. I hope everything that I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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2 H₃C−CH₃(g) + 7O₂(g) → 4 CO₂(g) + 6 H₂O(g)

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2 H3C−CH3(g) + 7O2(g) → 4 CO2(g) + 6 H2O(g)

2 H₃C−CH₃(g) + 7O₂(g) → 4 CO₂(g) + 6 H₂O(g)