Question

Calculate ΔH_r° of isomerization of propene (g) into cyclopropane (g) at 100°C

Answer 1

If you want to caluculate isomerization of propene to cyclopropane.

The balanced equation for this process is C3H6 (propene) → C3H6 (cyclopropane)

∆rH^o = ∆_fH^o (cyclopropene) − ∆_fH^o (propene)

if you find ∆_fH^o (cyclopropene) we can easily find the ∆rH^o

Standard value for cobustion of Cyclopropane is

∆rH^o = −2091 kJ mol⁻¹ cyclopropane combustion

∆rH^o = +20.42 kJ mol⁻¹propene enthalpy of formation

balanced chemical equation for combustion of cyclopropane

2C₃H₆ + 9O₂(g) → 6H₂O(l) + 6CO₂(g)    ∆rH^o = 2 × −2091 kJ mol⁻¹

For any such reaction the reaction enthalpy can related to the enthalpy of products and reactants by

∆rH^o = ∑ Products νp∆fH_p^o − ∑ Reactants νr∆fH_r^o

for this reaction is

∆rH^o = 6∆_fH^o (H2O(l)) + 6∆_fH^o CO2(g)) − 2∆_fH^o (C3H6) + −9∆_fH^o (O2(g))

Using the fact that the standard state of oxygen is O2(g) ∆_fH^o (O2(g)) = 0) the following expression for ∆_fH^o (C3H6) is obtained through rearrangement

∆_fH^o (C3H6) = 3∆_fH^o (H2O(l)) + 3∆_fH^o CO2(g)) − 1 /2 ∆_rH^o

= 3 × −285.83 kJ mol−1 + 3 × −393.51 kJ mol−1 − 1 2 × 2 × −2091 kJ mol−1

= 52.98 kJ mol−1

With this information we can now calculate the enthalpy of isomerization of propene to cyclopropane. The balanced equation for this process is C3H6 (propene) → C3H6 (cyclopropane) ∆rH^o =? The enthalpy of this reaction is simply given by

∆rH^o = ∆_fH^o (cyclopropene) − ∆_fH^o (propene)

= 52.98 kJ mol−1 − 20.42

= 32.56 kJ mol−1