Question

Find the enthalpy of isomerisation of propene (g)
to cyclopropane (g) if B.E. (C - C) = 348 kJ/mol
and B.E. (C = C) = 615 kJ/mol. :-
(A), 162 kJ/mol
(B) - 81 kJ/mol
(C)- 267 kJ/mol
(D) - 429 kJ/mol​

Answer 1

Given :

B.E. (C - C) = 348 kJ/mol

and B.E. (C = C) = 615 kJ/mol

To find : enthalpy of isomerisation of propene (g)  to cyclopropane (g)

Solution :

enthalpy of isomerisation of propene (g)  to cyclopropane (g) = ( enthalpy formation of propene ) - ( enthalpy of formation of cyclopropane )

= B.E. ( C=C ) + B.E. ( C-C ) + 6 * B.E. ( C-H ) - 3 * B.E. ( C-C ) - 6 * B.E. ( C-H )

= ( 615 - 696 ) kJ/mol

= -81 kJ/mol

The enthalpy of isomerisation of propene (g)  to cyclopropane (g) is  -81 kJ/mol