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) rarr CH_(3)COCH(g)`
In acetone, six `C-H` bonds, one `C=O` bond and two `C-C` bonds are present. Enegry released in the formation of these bonds is
`=- 6 xx 413.4 - 728.0 - 2 xx 347.0 =- 3902.4 kJ mol^(-1)`
The equation of the enthalpy of formation of acetone is
`3C_(("garphite")) +3H_(2)(g) +(1)/(2)O_(2)(g) rarr CH_(3)COCH_(3)(g),DeltaH = ?`
This equation can be obtained from the following equaitons by adding:
`3C(g) +6H(g) +O(g) rarr CH_(3)COCH(g), DeltaH =- 3902.4`
`3C(s)rarr 3C(g), DeltaH = 2155.2`
`3H_(2)(g) rarr 6H(g), DeltaH = 1307.4`
and `(1)/(2) O_(2)(g) rarr O(g), DeltaH = 247.5`
`ulbar({:(3C(S)+3H_(2)(g)+(1)/(2)O_(2)(g)rarr,CH_(3)COCH_(3)(g)),(,DeltaH=-192.3 kJ mol^(-1)):})`