ThermochemistryHard

Question

(i) Cis - 2 - butene → trans - 2 - butene,             ᐃH1
(ii) Cis - 2 - butene → 1 - butene,                       ᐃH2
(iii) Trans - 2 - butene is more stable than cis - 2 - butene.
(iv) Enthalpy of combustion of 1-butene,                        ᐃH = - 649.8 kcal/mol
(v) 9ᐃH1 + 5 ᐃH2 = 0
(vi) Enthalpy of combustion of trans 2 - butene,              ᐃH = - 647.0 kcal/mol.
The value of ᐃH1 & ᐃH2 in Kcal/mole are

Options

A.- 1.0, 1.8
B.1.8, -1.0
C.-5, 9
D.-2, 3.6

Solution

1-butene → trans -2-butene
ᐃH = ᐃH1 - ᐃH2 = - 649.8 + 647.0 = -2.8 kcal/mole
also    9ᐃH1 + 5 ᐃH2 = 0.
solving    ᐃH1 = - 1.0 kcal/mole ; ᐃH2 = 1.8 kcal/mole

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