Question

Calculate the total amount of enthalpy (heat) released from the combustion of ethane (C2H6)

Bond Enthalpies ( In units of kJ mol^-1)

C-O 358 O=O 495
C=O 799 O-H 463
C-H 412 C-C 348

Answer 1

The equation of the reaction is:

2C₂H₆ + 7O₂ --> 4CO₂ + 6H₂O

To calculate the total enthalpy, the sum of enthalpies of all formed bonds should be subtracted from the sum of enthalpies of all broken bonds. According to the equation, there are total of:

12 C-H bonds broken;

2 C-C bonds broken;

7 O=O bonds broken;

8 C=O bonds formed;

12 O-H bonds formed.

Therefore,

$\Delta{H_{reaction}}=\sum\Delta{H_{broken}}-\sum\Delta{H_{formed}}=12\times412+2\times348+7\times495-8\times799-12\times463=-2843kJ/mol$

Answer: -2843 kJ