Calculate - H0 for the reaction between ethene and water to form ethyl alcohol from the following data- -cH0C2H5OHl - -1368 kJ -cH0C2H4 - -1410 kJDoes the calculated - H0 represent the enthalpy of formation of liquid ethanol-

The heat of combustion of ethanol and ethane is given: C_2H_5OH + 3O_2 → 2CO_2 + 3H_2O Δ _cH^0 = 1368kJ ...(1) C_2H_4 + 3O_2 → 2CO_2 + 2H_2O # ...

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Heat of Formation

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Question

Calculate $Δ H^{0}$ for the reaction between ethene and water to form ethyl alcohol from the following data:
$Δ_{c} H^{0} C_{2} H_{5} O H (l) = - 1368 k J$
$Δ_{c} H^{0} C_{2} H_{4} = - 1410 k J$
Does the calculated $Δ H^{0}$ represent the enthalpy of formation of liquid ethanol?

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Δ H^{o}

Δ_{c} H^{o} C_{2} H_{5} O H_{(l)} = - 1368 k J

Δ_{c} H^{o} C_{2} H_{4 (g)} = - 1410 k J

Δ H^{o}

C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)

Δ_{r} H^{\circ} = - 1368.0 k J

C (s) + O_{2} (g) \to C O_{2} (g)

Δ_{r} H^{\circ} = - 393.5 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)

Δ_{r} H^{\circ} = - 286.0 k J

Δ H^{0} = 77.2 k J m o l^{- 1}

Δ S^{0} = 122 J K^{- 1} m o l^{- 1}

P C l_{5 (g)} \to P C l_{3 (g)} + C l_{2 (g)}

C_{2} H_{4}

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