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Heats of combustion of alkanes

How to use heats of combustion to compare the stability of isomeric alkanes.

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  • leaf orange style avatar for user http://facebookid.khanacademy.org/523755526
    Why is a lower heat of combustion considered to be more stable? If less energy is needed to combust an isomer, wouldn't that mean it is less stable?
    (12 votes)
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  • piceratops ultimate style avatar for user Costel Voica
    What does it mean when a compound is stable or unstable?
    (8 votes)
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  • male robot donald style avatar for user krutanjali mudgal
    branching increases stability, is the concept according the video.
    high enthalpy means it is more difficult to break compound into co2 and h20! wont the more stable ones be more difficult to combust.?
    but branching also means lower boiling point , (which means less stability )
    (3 votes)
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    • duskpin ultimate style avatar for user rosafiarose
      Basically - stability has to do with INTRAmolecular covalent bonds, whereas melting point/boiling point has to do with INTERmolecular forces.

      Stability has to do with the energy contained in the covalent bonds between the individual atoms of carbon and hydrogen. When octane combusts, the bonds between the carbons, hydrogens and oxygens break-up and form brand new bonds and thus new molecules.

      Melting/boiling points, however, has to do with the intermolecular forces between the atoms of the octane molecule. When octane changes from solid to liquid to gas, it is still octane - i.e. the carbons and hydrogens remain bonded to each other in the same way.
      (4 votes)
  • winston baby style avatar for user hansenyangsf
    May I ask why branched alkanes are more stable than straight chain alkanes?
    (5 votes)
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    • duskpin sapling style avatar for user Aarti Thakur
      Longer chain alkanes are typically more stable (relatively, based on the number of carbons) compared with a shorter chain alkane. More branched compounds are typically more stable than straight chain alkanes with the same number of atoms. For example, 2-methylpropane is more stable than butane.
      (0 votes)
  • piceratops tree style avatar for user Tombentom
    Well, I think I must have missed some of the basic knowledge about combustion. But as I remember, combustion energy is calculated by the difference between the total energy needed to break the bonds of the reactants & the energy released when bonds in new products are formed.
    In this case we have the reactants as Octane isomers + 02 & the products as CO2 & H20. In reality, the energy released when bonds in CO2 & H20 are formed are just too much compared to when C O H atoms flying freely all over the places, and these energy is so much more than energy needed to break
    Octance or any kind of alkane!
    => the difference of energy here would be very high (-delta enthalpy).
    So if you have a quite high absolute value of (-delta enthalpy) for a combustion of alkane, u can conclude that this compound is highly unstable & vice versa . Hope this help.
    (2 votes)
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    • duskpin ultimate style avatar for user rosafiarose
      The octane isomers have a lot more energy contained within their bonds than CO2 or H20. So when octane combusts, it releases heat energy. This heat energy is lost from the system, which is why delta H is negative. The energy in the products is less than the energy in the reactants.

      The less stable the reactant, the more energy there is in its bonds. Thus when a less stable molecule combusts, it releases more energy, because there was more energy contained in its bonds to begin with.

      Think of it this way - CO2 and H20 are very stable molecules, so there is very little energy contained within its bonds compared to something like octane.
      (4 votes)
  • blobby green style avatar for user thegreatbasinrockies
    Why are more branched hydrocarbon chains more stable. Are cycloalkanes more stable than their respective straight chains, since methane has 890kJ/mol deltaH and methyl has 654 kJ/ml deltaH, I'm assuming the fewer hydrogens in a cyclic vs. straight chain, means that they have a lower heat of combustion, but then again there seems to be more instability in some cyclic rings that have less than optimal carbon angles, such as cyclobutane.
    (4 votes)
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  • blobby green style avatar for user my3shyll
    if released heat is high then the compound is said to be more stable, then octane should be most stable than its isomers right?
    (2 votes)
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  • blobby green style avatar for user 😊
    Does this mean that the compound having more number of branches has less heat of combustion?? Is it applicable for all compounds and their isomers??
    (2 votes)
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  • female robot amelia style avatar for user misbah baqir
    catalytic oxidation of alkanes produce ? alcohols or carboxylic acid? or which one is the best answer
    (2 votes)
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  • mr pants pink style avatar for user Q Rule
    Is heat of combustion the heat released by the complete combustion of one mole of a substance? (An explanation).
    (2 votes)
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Video transcript

- [Voiceover] Alkanes are very unreactive, but they do undergo combustion reactions. So if you take methane and react it with oxygen, you'll get carbon dioxide and water as the products. And for all hydrocarbons, you're gonna get CO2 and H2O as the products of a combustion reaction. Delta H zero for this reaction is negative 890 kiloJoules per one mole of methane which means one mole of methane, if you combust one mole of methane, you're gonna get 890 kiloJoules of heat. And so this is an exothermic reaction, heat is given off. So the heat of combustion is the heat that's released on the complete combustion of one mole of a substance. In this case we're talking about methane. Let's talk about hexane next. Let me draw out the structure for hexane. So we have six carbons and notice we have two CH3 groups, so here's a CH3 and here's a CH3, so CH3 and CH3. And then we have four CH2 groups. So here's one, two, three, and four. So that's why we have a four right here. So there are four CH2 groups in hexane, and here we have the heat of combustion. And now we're gonna talk about the heat of combustion as the negative change in the enthalpy. So negative delta H zero in terms of kiloJoules per mole. That gives us a positive value here, so for hexane it's 4,163 kiloJoules for every one mole of hexane that we combust. Notice what happens as we move on to heptane here. We're increasing in one CH2 group. We're going from four CH2 groups to five CH2 groups. So now we're at five CH2 groups. We get an increase in the heat of combustion, so more heat is released. And that makes sense, if you increase the number of carbons, you're gonna increase in the heat of combustion. So if you increase by one CH2 group, how much do we increase in terms of the heat of combustion? Well, if you take 4,163 and subtract that from 4,817 that's a difference of 654 kiloJoules. The pattern continues when we move on to octane. So octane, now we have 6 CH2 groups. So we've added one more. We've increased our heat of combustion to 5,471 kiloJoules per one mole. And that is also an increase of 654 kiloJoules. So for nonane, with one more CH2, we could predict, we could estimate the heat of combusion. And we could just add 654 to 5,471. So we can go ahead and set that up over here. So we have 5,471. We're adding 654 to that. So that would give us a five here, and then seven and five is 12, so we carry the one, and that gives us 11, so we get 6,125. So if I right that in here, 6,125 kiloJoules per mole is a pretty good estimation for the heat of combustion of nonane. Heats of combustion are really useful when you're trying to compare the stability of isomers. For example, here we have three isomers: octane, 2-methylheptane, and 2,2-dimethylhexane. So all these molecules have the molecular formula C8H18. So if we write out the combustion reactions, so we have plus O2, we know the products are CO2 and H2O. So we have CO2 plus H2O. Let's go ahead and balance this. We have eight carbons on the left side, so we need an eight over here on the right, we'll put that in front of the CO2. And now let's look at hydrogens. There's 18 on the left, and so we need 18 on the right. And we can do that by putting a nine here, cause nine and two give us 18 hydrogens. Now let's see how many oxygens are on the right side. Well we have eight and 2 give us 16 oxygens here, and then nine gives us nine for the water. So that's a total of 25 oxygens on the right side. So we need 25 oxygens on the left side. Now if we pretend like this two isn't here for a second, we could write a 25 right here, but there is a two, so we need to divide by two. And if we're talking about one mole of C8H18, we're gonna leave this as a fraction, so 25 over two is the amount of oxygen that's necessary to completely react with one mole of C8H18. So now we have our balanced equation for our combustion reaction, and if we compare these three isomers, octane, 2-methylheptane, and 2,2-dimethylhexane in terms of their heats of combustion, we can figure out which one is the most stable isomer. Here we have an energy diagram for our three isomers. So on the left here, we'll put increasing energy this way. All three of our isomers would produce the same products if you combust one mole of them. Each one would give us eight CO2 and nine H2O. So there's an energy level associated with our products. And we'll say this is the energy level associated with our products. It's the same for all three of our isomers. Next we look up the heat of combustion data for the isomers, and this is determined experimentally. For example, octane has a heat of combustion of about 5,471 kiloJoules per mole, so that's how much heat is released, 5,471 kiloJoules per mole. And if we know the energy level of the products, and we know how much energy was released, we can find the energy level of our reactants. So there is the energy level for octane. Next let's think about 2-methylheptane. So again, the energy level of the products is here, and if we look up the heat of combustion for 2-methylheptane we'll find a value somewhere around 5,466. So that's how much energy is released when you combust one mole of 2-methylheptane, 5,466 kiloJoules per mole. Since the energy level of the products is the same as the one before, but we have a smaller heat of combustion this time, that must mean the energy level for 2-methylheptane is lower than the energy level for octane. And finally, we move on to 2,2-dimethylexane. The energy level of the products is the same as before. The heat of combustion is 5,458, so somewhere around 5,458 kiloJoules per mole is released, right? That's how much energy, or heat, is released when you combust one mole of 2,2-dimethylhexane. And since we have a smaller value for the heat of combustion, that must mean the energy is lower for 2,2-dimethylhexane. Now we can finally compare the stabilities of our isomers. The higher the energy, the less stable the compound. So octane has the highest energy level, has the highest heat of combustion. So this is the least stable out of these three, this is the least stable. The lower the energy for our starting compound, the more stable it is. So that must meant that 2,2-dimethylhexane is the most stable isomer out of these three. So now we can think about trends. As we go from octane to 2-methylheptane to 2,2-dimethylhexane we're increasing in branching. So we're increasing in the amount of branching that we have. What happened to the heats of combustion? Alright, if we define heat of combustion as the negative of the change in the enthalpy, the heats of combustion decrease. We went from 5,471 to 5,466 to 5,458. So there was a decrease in the heat of combustion. And what about stability? We got more stable as we branched more. So increased branching in general means increased stability. So it's important to be able to analyze heat of combustion data. Just remember the lower the energy, the more stable the compound. So the compound that had the highest heat of combustion was the least stable. And the compound with the lowest heat of combustion was the most stable. So branched alkanes are lower in energy, or more stable, than straight chain alkanes.