Question

The enthalpy of combustion ΔcH° of gaseous cyclopropane, C3H6, is -2091 kJ mol-1 at 25 °C.

a. What is the standard enthalpy of formation of cyclopropane?

b. What is the standard internal energy change of combustion of cyclopropane at 25 °C.

c. Using heat capacity values in the back of your book, what is the standard enthalpy of combustion of cyclopropane at 100 °C?

d. Again using heat capacity values in the back of your book, what is the standard enthalpy of formation of cyclopropane at 150 °C?

Answer 1

a. The equation for the formation of cyclopropane is

3C + 3H₂ -----> C₃H₆

ΔfH°_reaction = ΣΔcH°_products - ΣΔcH°_reactants = [ΔcH°(C₃H₆)] - [ΔcH°(3C) + ΔcH°(3H₂)]

ΔfH°_reaction = [-2091 kJ mol^-1)] - [(3*-393.5 kJ mol^-1)+ (3*-285.8 kJ mol^-1​)] = -53 kJ mol^-1

Standard enthalpy of formation of cyclopropane = -53 kJ mol^-1

b. ΔfH^∘=ΔE^∘+(Δn)RT

where ΔfH^∘ = Standard enthalpy of formation of cyclopropane = -53 kJ mol^-1

ΔE^∘ = Standard internal energy change of combustion

Δn = Difference in the number of moles = 1 - (3+3) = -5

R = Gas constant = 8.314*10⁻³ kJ/Kmol

T = Temperature = 25°C = 298 K

ΔfH^∘=ΔE^∘+(Δn)RT

-53 kJ mol^-1 = ΔE^∘+ (-5) (8.314*10⁻³ kJ/Kmol) (298 K)

-53 kJ mol^-1 = ΔE^∘+ (-12.3879 kJ mol^-1)

ΔE^∘ = -53 kJ mol^-1 - (-12.3879 kJ mol^-1) = -40.61 kJ mol^-1

Standard internal energy change of combustion of cyclopropane = -40.61 kJ mol^-1