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Electronic Configuration

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Question

Calculate the heat of the following homogeneous gaseous reaction
$C H_{3} C O C H_{3} + 2 O_{2} \to C H_{3} C O O O H + C O_{2} + H_{2} O$
from the following data:
Bond energies (kJ):
$C - H$ = 414.49
$C - C$ = 347.92
$C = O$ = 724.32
$O = O$ = 494.04
$O - H$ = 462.64
$C - O$ = 967.13
Resonance energy:
$- C O O H$ = 117.23
$C O_{2}$ = 138.16

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Solution

$C H_{3} C O C H_{3} + 2 O_{2} \to C H_{3} C O O O H + C O_{2} + H_{2} O$

$C H_{3} - O ∥ C - C H_{3} + 2 (O = O) \to C H_{3} - O ∥ C - O - H + O = C = O + H - O - H$

$Δ H_{W . O . R} =$ Heat change without considering the resonance

$Δ H_{W . O . R} = (6. B . E_{C - H} + 2 B . E_{C - C} + B . E_{C = O} + 2 B . E_{O = O}) - (3. B . E_{C - H} + B . E_{C - C} + B . E_{C - O} + 3 B . E_{C = O} + 3 B . E_{O - H})$

$Δ H_{W . O . R} = (6 \times 414.99 + 2 \times 347.92 + 724.32 + 2 \times 494.04) - (3 \times 414.99 + 347.92 + 967.13 + 3 \times 724.32 + 3 \times 462.64)$

$Δ H_{W . O . R} = 4895.18 - 6119.4 = - 1224.22 K c a l$

$Δ H_{W . R} =$ Heat change assuming resonance is happening

$Δ H_{W . R} = Δ H_{W . O . R} + R . E p r o d - R . E_{r e a c t} = Δ H_{W . O . R} + R . E p r o d - 0$

$Δ H_{W . R} = - 1224.22 + (- 117.23 + (- 136.18)) = - 1477.63 k J$

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