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Question

Using the bond enthalpy data given below, calculate the enthalpy of formation of acetone (g)
Bond energy CH=413.4kJmol1;
Bond energy CC=347.0kJmol1;
Bond energy C=O=728.0kJmol1;
Bond energy O=O=495.0kJmol1;
Bond energy HH=435.8kJmol1;
ΔHsubC(s)=718.4kJmol1

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Solution

3C(g)+6H(g)+O(g)CH3COCH3(g)
In acetone, six CH bonds, one C=O bond and two CC bonds are present. Energy released in the formation of these bonds is
=6×413.4728.02×3470=3902.4kJmol1
The equation of the enthalpy of formation of acetone is
3Cgraphite+3H2(g)+12O2(g)CH3COCH3(g);ΔH=?
This equation can be obtained from the following equations by adding:
3C(s)+6H(g)+O(g)CH3COCH3(g);ΔH=3904kJmol1
3C(s)3C(g);ΔH=2155.2kJmol1
3H26H(g);ΔH=1307.4kJmol1
and 12O2(g)O(g);ΔH=247.5kJmol1
---------------------------------------------------------------------------------------------
3C(s)+3H2(g)+12O2(g)CH3COCH3(g);ΔH=192.3kJmol1

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