The change of internal energy cU° in the combustion of C60 (s) is -25,968 kJ/mol

Question

The change of internal energy cU° in the combustion of C60 (s) is -25,968 kJ/mol @ 25°C. To do these problems each requires a balanced chemical equation. It is assumed that you can figure out the equation from the specific description of enthalpy. a. What is the enthalpy of combustion, cH° of C60 (s)? ans. -25,968 kJ/mol. b. What is the enthalpy of formation, fH° of C60 (s)? ans. 2357 kJ/mol. c. What is the enthalpy of vaporization of C60 (s) to C (g) per mole of C (g)? ans. 677.4 kJ/mol. d. How does this compare with the enthalpy of vaporization of graphite and diamond to C (g)? C(graphite) vaporization = 716.7 kJ/mol C(diamond) vaporization = 714. 8 kJ/mol. For data to compare, look up fH° of carbon (diamond) and carbon (graphite) from the Table.

Explanation / Answer

C60(s)+ 60O2(g) ---> 60 CO2(g)

a)Delta H of the reaction = internal energy = -25,968 KJ/mol

b) Enthalpy of formation = delta H = 60* delta H of CO2 - delta Hf C60

= -25,968 KJ/mol = 60*(-393.5) - delta Hf C60

= +2358 KJ/mol

c) Enthalpy of vaporization of C60

C60(s)------> 60 C(g)

delta H = 60*716.7 - 2358 = 40644 KJ/mol

Now we want per mole so = 40644 KJ/mol / 60 = 677.4 KJ

d) Graphite ----> C(g) delta H = 716.7 Kj/mol

delta H = delta H of C - delta H of graphite

= 716.7 = 716.7 - delta H of graphite

= delta H of graphite = 0

Diamond ----> C(g)

delta H = delta H of C - delta h of diamond

= 714.8 = 716.7 - delta H of diamond

= delta H of diamond = 1.9 KJ/mol

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