Calculate the heat produced (in Mcal) when 3.785 litre of octane ( C 8 H 18 ) reacts with oxygen to form CO & water vapour at 25° C. The density of octane is 0.7025 gm/ml. Enthalpy of combustion o

Question

Calculate the heat produced (in Mcal) when 3.785 litre of octane $(C_{8} H_{18})$ reacts with oxygen to form CO & water vapour at 25° C. The density of octane is 0.7025 gm/ml. Enthalpy of combustion of $C_{8} H_{18}$ is $- 1302.7 k Cal / mol$ .

$Δ H_{f} ° C O_{2} (g) = - 94.05 k c a l m o l^{- 1}$

$Δ H_{f} ° CO (g) = - 26.41 k c a l m o l^{- 1}$

$Δ H_{f} ° H_{2} O (ℓ) = - 68.32 k c a l m o l^{- 1}$

$Δ H_{f} ° H_{2} O (g) = - 57.79 k c a l m o l^{- 1}$

TestHub · Accepted Answer

Enthalpy of formation of Octane ( Δ H f ∘ ​ ): Using standard combustion to CO₂(g) and H₂O(l), Δ H f ∘ ​ ( C 8 ​ H 18 ​ ) = 8 ( − 94.05 ) + 9 ( − 68.32 ) − ( − 1302.7 ) = − 64.58 k c a l / m o l . Enthalpy of incomplete combustion ( Δ H r x n ​ ): For C₈H₁₈(l) → 8CO(g) + 9H₂O(g), Δ H r x n ​ = 8 ( − 26.41 ) + 9 ( − 57.79 ) − ( − 64.58 ) = − 666.81 k c a l / m o l . Moles of Octane reacted: Mass = 3785 m L × 0.7025 g / m L = 2658.96 g , so Moles = 114 g / m o l 2658.96 g ​ = 23.324 m o l . Total heat produced (in Mcal): Heat = 23.324 m o l × 666.81 k c a l / m o l = 15552.8 k c a l = 15.55Mcal .

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