The enthalpy change for the combustion of ethanol at 298 K and 1 atm is given below CH 3- CH 2 OH l -3 O 2 g - - - - - - - 2 CO 2 g -3 H 2 O l - H -1363 Kj - Calculate the internal energy change for the above reaction-

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Standard XII

Chemistry

Introduction to Alkenes

The enthalpy ...

Question

The enthalpy change for the combustion of ethanol at 298 K and 1 atm is given below

CH₃-CH₂OH(l) + 3O₂(g)------------à 2CO₂ (g)+ 3H₂O(l) ∆H=-1363Kj.Calculate the internal energy change for the above reaction?

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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=2-3=-1(since 2 moles from CO2 and 3 from O2)
=> DelH=-1,363kJ
DelU=DelH-Del nRT
=>DelU=-1363-(-1)8.314*298
=>DelU=-1338.22428
=>DelU=-1338.22kJ

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The enthalpy change for the combustion of ethanol at 298 K and 1 atm is given below CH3-CH2OH(l) + 3O2(g)------------à 2CO2 (g)+ 3H2O(l) ∆H=-1363Kj.Calculate the internal energy change for the above reaction?

The enthalpy change for the combustion of ethanol at 298 K and 1 atm is given below

CH₃-CH₂OH(l) + 3O₂(g)------------à 2CO₂ (g)+ 3H₂O(l) ∆H=-1363Kj.Calculate the internal energy change for the above reaction?