Question

(a) Given the following reactions and their heats of combustion, cakculate the heat of combustion per CH2 (o) Rank the tour eycioalkanes in rcr of siabilty fron loast siable to most stabie. (scroll down for part (b) H for reacion H per CH2 group Compound Combustion Reacion klmol) (kUmol) Number Cyclopropane | C3H6+-01 3CO2 +3}1,0 -2091 Number Cyclobutane 2744 15 Number CyciopentaneCHo5CO,+5H,0 -3320 Cyclohexane 3952 Number Hint Previcus 3 Give Up & Vew Solution Check Answer Next

Answer 1

a) cyclopropane has the formula C₃H₆, as you can see this molecule have 3 CH₂ groups:

the enthalpy of reaction is - 2091 KJ / mol

divide this value by 3 to get:

-697.7 KJ / mol

b) cyclobutane is C₄H₈, this molecule has 4 CH₂ groups so:

divide -2744 by 4 to get:

-686 KJ / mol

c) cyclopentane has 5 ch2 groups so

divide -3320 by 5 to get:

-664 KJ / mol

d) cyclohexane has 6 ch2 groups so:

divide -3952 by 6 to get:

-658.66 KJ / mol

b) since cyclohexane has the smallest value among these compounds then cyclohexane is the most stable.

since cyclopropane has the largest value among these compound then cyclopropane is the least stable.

the order is:

cyclohexane.... most stable

cyclopentane.... 2

cyclobutane.....3

cyclopropane.... least stable

*hope it helps =)