Calculate the standard internal energy change for the reaction at 25- C - C 2 H 4 g -3 O 2 g - - - - - - 2 CO 2 g -2 H 2 O l DH for C 2 H 4 g -52-30 kJ - moleCO 2-393-5 kJ - mole H 2 O -286 kJ - mole R -8-314 J - K - mole

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Hess' Law

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Question

Calculate the standard internal energy change for the reaction at 25⁰C:-

C₂H₄(g) + 3O₂(g)----------à 2CO₂ (g) + 2H₂O (l) DH for C₂H₄ (g) = 52.30 kJ/mole,
CO₂ = - 393.5 kJ/mole H₂O = - 286 kJ/mole R = 8.314 J/K/mole.

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Solution

From first law of Thermodynamics,
DelH=DelU+Deln RT
Del H =Heat of reaction
DelU=Internal energy change
where Deln is the change in the no.of moles of gaseous substance=2-4=-2(since 2 moles from CO2 and 4 from C2H4 and O2)
From thermodynamics heat of reaction = DelH of products-DelH of reactants
DelH of the reaction = 2delH of CO2+2DelH of H2O -(delH of C2H4+3DelH of O2)
Del H=2-393.5+2-286-(52.30+0) ( since O2 is a pure element and has heat of formation as zero)
=> DelH=-1,411.3kJ
DelU=DelH-Del nRT
=>DelU=-1411.3-(-2)8.314298
=>DelU=-1411.3+4.955=-1406.344kJ
=>DelU=-1406.344kJ

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Calculate the standard internal energy change for the reaction at 25⁰C:- C2H4(g) + 3O2(g)----------à 2CO2 (g) + 2H2O (l) DH for C2H4 (g) = 52.30 kJ/mole, CO2 = - 393.5 kJ/mole H2O = - 286 kJ/mole R = 8.314 J/K/mole.

Calculate the standard internal energy change for the reaction at 25⁰C:-

C₂H₄(g) + 3O₂(g)----------à 2CO₂ (g) + 2H₂O (l) DH for C₂H₄ (g) = 52.30 kJ/mole,
CO₂ = - 393.5 kJ/mole H₂O = - 286 kJ/mole R = 8.314 J/K/mole.