Question

C₆₀(s) + 60 O₂(g) --> 60 CO₂(g)

The change in internal energy in the following combustion reaction of C₆₀(s) is -25968 kJ/mol (1 bar and 298.15 K)

a) What is the standard enthalpy of reaction for the above reaction? Assume that O₂(g) and CO₂(g) behave as ideal gas and use C_V,m = 5R/2 for both

b) What is the standard enthalpy of formation of C₆₀(s)

Answer 1

the combustion reaction is C60(s)+ 60O2(g) ----->60CO2(g)

change in internal energy, deltaU= -25968 Kj/mole

change in enthalpy, deltaH= deltaU+delta(PV)= deltaU+ deltan*RT

deltan= change in no moles of gaseous products= 60-60=0

hence deltaH= deltaU= -25968 KJ/mole

2. for the reaction C60+ 60O2(g)----->60CO2(g)

enthalpy data (KJ/mole): O2= 0, CO2= -393.5

enthalpy change= 60* standard enthalpy of CO2- { standard enthalpy of formation of C60+ 60* standard enthalpy of formation of O2}

=60*(-393.5) - (standard enthalpy of C60+60*0)=-25968

standard enthalpy of formation of C60 = 60*(-393.5)+25968=2358 KJ/mole