# combustion

#### SunChip

How are heats of combustion used to compare relative stabilities of isomers?

If you have EK, its on page 30 of the organic book. Thanks!

#### IntelInside

10+ Year Member
7+ Year Member
How are heats of combustion used to compare relative stabilities of isomers?

If you have EK, its on page 30 of the organic book. Thanks!
The heat of combustion (&#916;Hc) is the change in enthalpy of a combustion reaction when one mole of a compound undergoes complete combustion with oxygen.

Now since isomers have the exact same formula, but different bond to bond connectivity, we dont have to worry about different amounts of CO2 and H2O affecting the change in enthalpy. Therefore:

The larger the heat of combustion, the higher the energy level of the molecule and therefore the less stable the molecule!!!

This can be see if you look at the graph for an exothermic reaction!!

However, if we are comparing two different molecules of different sizes, for example cyclohexane vs cyclopropane. We can see that the enthalpy of the products will be greater for cyclohexane because we will generate more moles of them, and enthalpy is an extensive property. This still applies even though the sum of H for the reactants will be larger as well (the major contributor to the difference is the H for the products). From the formula:

&#916;H = sum of H for products - sum of H for reactants

We see that we will have an overall greater &#916;Hc for cyclohexane. Does this mean that we cyclohexane is less stable than cyclopentane? No it doesn't!! Why is this the case? Well, first we know that cyclohexane has the lowest ring strain out of all the cycloalkanes. So how do we account for this difference? We account for this difference by having to take the &#916;H in terms of "per CH2". Why do we do this you might ask? It's because we want to see the actual energy drop for each bond aka bond stabilitys on each CH2. We would divide our &#916;H by the number of CH2 present in each ring. After doing this we can see that the &#916;H per CH2 is actually less for cyclohexane than cyclopropane and and therefore the entire molecule has a greater stability than cyclopropane

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#### SunChip

You are awesome! I appreciate the reply! OP
S

#### SunChip

Could you explain this more? "We can see that the enthalpy of the products will be greater for cyclohexane because we will generate more moles of them." How does cyclohexane generate more moles of them? Sorry it's been awhile since I've taken organic.

Thanks in advance!

#### Morsetlis

##### I wish I were a dentist
7+ Year Member
Cyclohexane has 6 carbons and cyclopropane has 3 carbons. 6 > 3

You really need to start visualizing molecules.

#### IntelInside

10+ Year Member
7+ Year Member
Could you explain this more? "We can see that the enthalpy of the products will be greater for cyclohexane because we will generate more moles of them." How does cyclohexane generate more moles of them? Sorry it's been awhile since I've taken organic.

Thanks in advance!
First you have to write out the equation and then balance it. If you balance your reaction correctly you can see that there would be more moles of CO2 and H2O.

Or you could have just memorized the general formula for a combustion reaction: I wouldn't recommend that, however, because we have to memorize so many other formulas and this is just going to take up unnecessary space that could go to something more important. Especially, when you could just balance it out in 3 seconds

#### SKation

How are heats of combustion used to compare relative stabilities of isomers?
Hey ,
I seee #916 in random places, what does this mean? What are we dividing in this case to figure our the per CH2 energy released?

Thanks!
If you have EK, its on page 30 of the organic book. Thanks!
Hey,

The heat of combustion (&#916;Hc) is the change in enthalpy of a combustion reaction when one mole of a compound undergoes complete combustion with oxygen.

Now since isomers have the exact same formula, but different bond to bond connectivity, we dont have to worry about different amounts of CO2 and H2O affecting the change in enthalpy. Therefore:

The larger the heat of combustion, the higher the energy level of the molecule and therefore the less stable the molecule!!!

This can be see if you look at the graph for an exothermic reaction!!

However, if we are comparing two different molecules of different sizes, for example cyclohexane vs cyclopropane. We can see that the enthalpy of the products will be greater for cyclohexane because we will generate more moles of them, and enthalpy is an extensive property. This still applies even though the sum of H for the reactants will be larger as well (the major contributor to the difference is the H for the products). From the formula:

&#916;H = sum of H for products - sum of H for reactants

We see that we will have an overall greater &#916;Hc for cyclohexane. Does this mean that we cyclohexane is less stable than cyclopentane? No it doesn't!! Why is this the case? Well, first we know that cyclohexane has the lowest ring strain out of all the cycloalkanes. So how do we account for this difference? We account for this difference by having to take the &#916;H in terms of "per CH2". Why do we do this you might ask? It's because we want to see the actual energy drop for each bond aka bond stabilitys on each CH2. We would divide our &#916;H by the number of CH2 present in each ring. After doing this we can see that the &#916;H per CH2 is actually less for cyclohexane than cyclopropane and and therefore the entire molecule has a greater stability than cyclopropane