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The standard enthalpy change for the combustion of 1 mole of ethylene is -1303.1 kJ C2H4(g)...

The standard enthalpy change for the combustion of 1 mole of ethylene is -1303.1 kJ

C2H4(g) + 3 O2(g) ----> 2 CO2(g) + 2 H2O

Calculate the change of Hf for ethylene based on the following standard molar enthalpies of formation.

molecules Change in Hf (kJ/mol)

CO2 -393.5

H2O -241.8

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Answer #1

Given,

delta H (combustion) = -1303.1 kJ for the reaction,

C2H4(g) + 3 O2(g) ----> 2 CO2(g) + 2 H2O

delta H (combustion) = delta H (products) - delta H (reactants)

delta H (products) = 2 x delta H (CO2) + 2 x delta H (H2O)

=> delta H (products) = 2 x (-393.5) + 2 x (-241.8) = -1270.6 kJ

delta H (reactants) = delta H (C2H4) + 3 x delta H (O2)

=> delta H (reactants) = delta H (C2H4) + 3 x 0 = delta H (C2H4)

(delta H (O2) = 0)

delta H (combustion) = delta H (products) - delta H (reactants)

=> -1303.1 = -1270.6 - delta H (C2H4)

=> delta H (C2H4) = 32.5 kJ

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