Calculate the standard enthalpy of formation for diamond, given?

Calculate the standard enthalpy of formation for diamond, given that


C(graphite) + O2(g) → CO2(g) ΔH0 = –393.5 kJ/mol


C(diamond) + O2(g) → CO2(g) ΔH0 = –395.4 kJ/mol





Is it 1.9 or -1.9 and why?

Calculate the standard enthalpy of formation for diamond, given?
It has to be 1.9 kJ/mol.





The reason being, you are looking for the enthalpy of FORMATION of DIAMOND.





So 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) -%26gt; 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) -%26gt; CO2 (g) , H = -393.5 kJ/mol


CO2(g) -%26gt; 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) --%26gt; C(diamond) , H = -393.5 + 395.4 = +1.9 KJ/mol





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)
Reply:If you reverse the second reaction, then add the two togther, the CO2 and O2 cancel out





C (graphite) + O2 (g) ---%26gt; CO2 (g)


CO2 (g) ----%26gt; C (diamond) + O2 (g)





and you get





C(graphite) ---%26gt; C(diamond)





By Hess's Law, you change the sign of the enthalpy for the second reaction (because you reversed it), and add it to the enthalpy of the first.





So, the enthalpy change is -393.5 kJ/mol + (+395.4 kJ/mol)


or +1.9 kJ/mol.