Using the bond enthalpy data given below, calculate the enthalpy of formation of acetone (g).
Bond enegry `C-H = 413.4 kJ mol^(-1)`,
Bond enegry `C-C = 347.0 kJ mol^(-1)`,
Bond enegry `C=O = 728.0 kJ mol^(-1)`,
Bond enegry `O=O = 495.0 kJ mol^(-1)`,
Bond enegry `H-H = 435.8 kJ mol^(-1)`,
`Delta_("sub")H^(Theta)C(s) = 718.4 kJ mol^(-1)`

Explanation

`3C(g)+6H(g)+O(g)toCH_(3)COCH_(3)(g)`
In acetone, six C-H bonds, one C=O bond and two C-C Bonds are present. Energy released in the formation of these bonds is
`=-6xx413.4-728.0-2xx347.0=3902.4kJ" "mol^(-1)`
The equation of the enthalpy of formation of acetone is
`3C_(("graphite"))+3H_(2)(g)+(1)/(2)O_(2)(g)toCH_(3)COCH_(3)(g),DeltaH=?` ltbr. This equation can be obtained from the following equations by adding,
`3C(g)+6H(g)+O(g)toCH_(3)COCH_(3)(g), " "DeltaH=-3902.4kJ" "mol^(-1)`
`3C(s)to3C(g)," "Delta=2155.2kJ" "mol^(-1)`
`3H_(2)(g)to6H(g)," "DeltaH=1307.4kJ" "mol^(-1)`
and `(1)/(2)O_(2)(g)toO(g)," "DeltaH=247.5kJ" "mol^(-1)`
`overline(3C(s)+3H_(2)(g)+(1)/(2)(g)toCH_(3)COCH_(3)(g),DeltaH=-192.3kJ" "mol^(-1))`