Mean bond enthalpies: C=C 612kJ/mol, C-C 348 kJ/mol, C-HkJ/mol, H-H 436kJ/mol, C(s) —> C(g) 715kJ/mol

Question

Mean bond enthalpies: C=C 612kJ/mol, C-C 348 kJ/mol, C-HkJ/mol, H-H 436kJ/mol, C(s) —> C(g) 715kJ/mol

1 - Calculate the enthalpy change for the dimerisation of ethene to give cyclobutane

2 C2H4(g) —> C4H8(g)

2 - Given that the mean enthalpy of the C-C and C-H bonds in cyclobutane are in fact 320kJ/mol and 412kJ/mol respectively, calculate the standard enthalpy of formation of cyclobutane.

Answer 1

Bond energy in cyclobutane:

4 C-C = 4 x 348 kJmol = 1392 kJ/mol

8 C-H = 8 x 412 kJ/mol = 3296 kJ/mol

Total = 4688 kJ/mol

Bond energy in ethene:

1 C=C = 612 kJ/mol

4 C-H = 4 x 412 kJ/mol = 1648 kJ/mol

Total = 2260 kJ/mol

Dimerization of ethene to cyclobutane:

2 ethene ==> cylcobutane

∆H = 4688 kJ - 2x2260 kJ = 4688 kJ - 4520 kJ = 168 kJ/mole

Standard enthalpy of formation of cyclobutane:

4C(s) ==> 4C(g) ∆H = 4 x 715 kJ/mol = 2860 kJ/mol

4C(g) + 4H2(g) ==> C₄H₈ (cyclobutane)

∆Hºformation = (4x320 + 8x412) - 4x715 = 1280 + 3296 - 2860

∆Hº = 1716 kJ/mol