Guide To Archery

Diamond Rh

enthalpy question??

Given the following enthalpies of reaction for 298 K and 1 bar,
C(Graphite) + O2(g) = CO2(g) ΔrH°1,298 = -393.5 kJ mol-1
C(Diamond) + O2(g) = CO2(g) ΔrH°1,298 = -395.4 kJ mol-1
a) Determine the enthalpy of reaction of graphite going to diamond.
b) Is this reaction endothermic or exothermic?
c) What does this imply about the chemical bonding in diamond compared to
graphite?

Question: do i just add the above 2 values to determine the enthalpy of the reaction??

if i get a negative value is exothermic and if its a positive value its endothermic??

Also, for chemical bonding would a negative enthalphy indicate thats its a weak bond??

Please comment....thanks

When you apply Hess' Law, the second equation MUST be flipped so you get this equation...remember when you reverse the reaction, you must change the sign of the enthalpy (H).

CO2(gas) -> C(diamond) + O2(gas) , H=+395.4 kJ/mol

Then you retain your graphite equation and add these two equations together.

C(graphite) + O2(g) -> CO2 (g) , H = -393.5 kJ/mol
CO2(g) -> C(diamond) + O2(g) , H = +395.4 kJ/mol

The end result when you add these two together is this equation...note that the CO2 and O2 gas drops out of the equation:

C(graphite) --> C(diamond) , H = -393.5 + 395.4 = +1.9 KJ/mol

endothermic

Remember, you flipped the second equation and not the first because you wanted to FORM , i.e. you are looking for the enthalpy of FORMATION, of C(diamond)

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enthalpy question??

Given the following enthalpies of reaction for 298 K and 1 bar,
C(Graphite) + O2(g) = CO2(g) ΔrH°1,298 = -393.5 kJ mol-1
C(Diamond) + O2(g) = CO2(g) ΔrH°1,298 = -395.4 kJ mol-1
a) Determine the enthalpy of reaction of graphite going to diamond.
b) Is this reaction endothermic or exothermic?
c) What does this imply about the chemical bonding in diamond compared to
graphite?

Question: do i just add the above 2 values to determine the enthalpy of the reaction??

if i get a negative value is exothermic and if its a positive value its endothermic??

Also, for chemical bonding would a negative enthalphy indicate thats its a weak bond??

Please comment....thanks

When you apply Hess' Law, the second equation MUST be flipped so you get this equation...remember when you reverse the reaction, you must change the sign of the enthalpy (H).

CO2(gas) -> C(diamond) + O2(gas) , H=+395.4 kJ/mol

Then you retain your graphite equation and add these two equations together.

C(graphite) + O2(g) -> CO2 (g) , H = -393.5 kJ/mol
CO2(g) -> C(diamond) + O2(g) , H = +395.4 kJ/mol

The end result when you add these two together is this equation...note that the CO2 and O2 gas drops out of the equation:

C(graphite) --> C(diamond) , H = -393.5 + 395.4 = +1.9 KJ/mol

endothermic

Remember, you flipped the second equation and not the first because you wanted to FORM , i.e. you are looking for the enthalpy of FORMATION, of C(diamond)

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