GS-1 Physical Sciences Question 38

[kzteam5226]

how do we know that CO2 --> Diamond and that we must flip the equation? Why can't the answer be the other way around .. -1.8kJ/mol?

[cbaha847962]

you have to flip the equation because you want the CO2's to cancel to give you the NET reaction C(graphite) --> C(diamond). In doing so, the sign for the enthalpy changes.

[dnpgr16513]

Isn't delta H products - reactants?? By doing that I would get 788 kJ. I don't know if I totally messed up the concepts.

[jellywing_2058]

[mohsinalam5969]

For any compound you say? That might be difficult.
Depending on how common the compound is you can often look up the value for its standard enthalpy of formation in a reference table/book somewhere.
Often chemistry text books will have a decent list in the back appendixes.

Pure substances in their elemental state have enthalpies of formation equal to zero. For example, both Hydrogen gas (H2) and Oxygen gas (O2) both have an enthalpy of formation of zero, when they react,
2H2 + O2 --> 2H2O + delta H
There is a change in enthalpy (exothermic in this case). Since water is the only thing in the reaction which does not have an enthalpy of formation of zero it makes it easy to find its value. Similar methods can be employed to find the enthalpy of formation values of other substances if you know the change in enthalpy of other reactions.
For example,
C (graphite) + O2 (g) --> CO2 (g) + delta H = -393.5 kJ
C (diamond) + O2 (g) --> CO2 (g) + delta H = -395.4 kJ
Through some clever manipulation,
C (graphite) --> C (Diamond) + delta H = +1.9 kJ
In this case the graphite allotrope of Carbon has a zero enthalpy of formation (it is the most stable form of Carbon), so from this we can infer that the diamond allotrope must have an enthalpy of formation of +1.9 kJ/mol.
This value is very close to +1.8 kJ/mol.