Question

Carbon dioxide is an important greenhouse gas. Calculate the enthalpy change for a 66.5g sample of CO2 heated from 0.6 C to 50.0 C at constant pressure assuming (a) that Cp,m is not a f(T) and (b) where Cp,m is a f(T). Cp,m(CO2) = 27.11 J/molK at 25 C and a = 44.22 J/mol K, b=0.00879 J/molK^2 and c=-8.62x10^5 K/mol. How does this energy change compare to the average intermolecular force?

Answer 1

When C_P is not a function of temperature, deltaH= enthalpy change= moles of CO2* specific heat of CO2*temperature difference

Moles of CO2= mass of CO2/ molar mass of CO2= 66.5/44= 1.511 moles

Enthalpy change=1.511* 27.11*(50-0.6)=2024 Joules=2024Kj/1000= 2.024 Kj

When enthalpy change is a function of temperature, T1= 0.6+273= 273.6 K

T2= 50+273= 323 K

Enthalpy change= moles of CO2*

= 1.511*

When integrated, the equation becomes

=1.511* [44.22*(323-273.6) + 0.00879*( 323²-273.6²)/2 +8.62*10⁵*(1/323-1/273.6)]=2768.4 J= 2.768 KJ

since the heating process increases the kinetic energy of molecules, while the forces of attractions are weakened. The enthalpy change suggests that intermolecular forces dominate.