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# Diamond and Graphite (tricky question)

1. Hey guys,
Here is a multiple choice question that is giving me a headache. I have tried to reason it out, but I am not too sure if my reasoning is correct. The mark scheme says that all statements are correct. I am just not too sure why they are correct.

The conversion of graphite into diamond is an endothermic reaction

C(graphite) --> C(diamond) Enthalpy change is +3kJ/mol

Which statements are correct?

1 The enthalpy change of atomisation of diamond is smaller than that of graphite.

2 The bond energy of the C–C bonds in graphite is greater than that in diamond.

3 The enthalpy change of combustion of diamond is greater than that of graphite.

Now I examine each statement:
1. Enthalpy change of atomization is the enegy change when one mole of gas atoms are formed from the element in its standard state.

If the reaction is endothermic, then it means that on a reaction pathway graph, diamond will be higher than graphite. So it means that diamond has a higher energy than graphite. So this means that, in a race, where diamond and graphite starts in their standard states and finish as gaseous atoms, then diamond will win because it has a shorter distance to reach the gaseous finish line(it is already at a higher energy).

So, following that reasoning, statement one is true?

2. If the enthalpy change of atomization is smaller for diamond, then diamond must require less energy for it to transform into gaseous state. This must mean that its bonds aren't that strong as compared to graphite. So graphite must have a greater bond energy (more energy is required to pull all the carbon atoms from eachother). This fits in with the reasoning for 1 above.

Is the reasoning for 2 true?

3. All combustion reactions are exothermic (give off heat). Now, diamond and graphite are both carbon atoms; when each fully reacts with oxygen we get the same product, carbon dioxide. So, from 1., as diamond is on a higher level on a reaction pathway graph, it will have to drop down a lot more to get to its products, i.e diamond will have to give out more heat than graphite to reach combustion products.

I am still confused and fuzzy about the type of reasoning I have applied. How would you go about figuring out a problem like this?

Thanks for the help mates!
Much love.
2. Your reasoning is fine on all counts, although in part 3 you probably could just refer to the stronger bonding in graphite and reason that in order to produce the combustion product you must first break the C-C bonds and then form the CO2 bonds. As the C-C bonds require more energy to be broken then the overall combustion energy must be greater in diamond.
3. In the Enthalpy of formation all the element in their natural condition is 0 (Kj/Mol)

So The reaction is endothermic because heat+C(graphite)----> C(diamond)

Ag (s)
Al (s)
As (s)
Au(s)
C(graphite)
S(rhombic)
4. @charco: Thank you SIR!!! Your explanation has helped to organize my cluttered mind.

@AVAINTEX: My goodness! I was confused as to how this process could be endothermic-and now I know why! Thanks mate! What is a little confusing is the list of elements and their states at the end of your posts. I am sure the answer is obvious, but just can't see it.

Thanks guys for your input. I really appreciate them. Thank God for a place like this!
5. Hi..even after reading the explanations..i still dont quite understand statement no.3..anyone mind explaning further about statement no.3 regarding the enthalpy change of combustion of diamond being higher than graphite?thanks....
6. (Original post by calvinxsaber)
Hi..even after reading the explanations..i still dont quite understand statement no.3..anyone mind explaning further about statement no.3 regarding the enthalpy change of combustion of diamond being higher than graphite?thanks....
exactly the same as what charco has mentioned.

If you think about bonding type in diamond and graphite,

diamond has sp3 hybridised carbon framework
whilst graphtie has sp2 hybridised carbon framework.

Overall, C-C bond in graphite is stronger than in diamond partly due to this sp2 hybridisation(33% s character) and also the fact that delocalisation of the free electron. As combustion involves breaking these C-C bonds in order to form CO2, then combustion of diamond would release more heat than graphite, hence more exothermic.
7. (Original post by shengoc)
exactly the same as what charco has mentioned.

If you think about bonding type in diamond and graphite,

diamond has sp3 hybridised carbon framework
whilst graphtie has sp2 hybridised carbon framework.

Overall, C-C bond in graphite is stronger than in diamond partly due to this sp2 hybridisation(33% s character) and also the fact that delocalisation of the free electron. As combustion involves breaking these C-C bonds in order to form CO2, then combustion of diamond would release more heat than graphite, hence more exothermic.
so does it mean that since the C-C bond in graphite is stronger than diamond,more energy is required to break the C-C bond..when more energy is absorbed,less energy or heat is released..hence enthalpy of combustion of graphite is lower than diamond...is this correct..thanks
8. (Original post by calvinxsaber)
so does it mean that since the C-C bond in graphite is stronger than diamond,more energy is required to break the C-C bond..when more energy is absorbed,less energy or heat is released..hence enthalpy of combustion of graphite is lower than diamond...is this correct..thanks
Yes, since you are going to form the same product, CO2 from both of them, so CO2 stays on the same energy level in your profile diagram.
9. (Original post by calvinxsaber)
so does it mean that since the C-C bond in graphite is stronger than diamond,more energy is required to break the C-C bond..when more energy is absorbed,less energy or heat is released..hence enthalpy of combustion of graphite is lower than diamond...is this correct..thanks
It's often easier to see energetics relationships if you represent them in an energy diagram...

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