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    Could someone please explain how I would answer this question:

    Calculate the enthalpy of formation of ethanol (C2H5OH) given the following enthalpies of combustion.

    C(s) -393
    H2(g) -286
    C2H5OH(l) -1371


    I've got the answer as -273 but I'm not too sure
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    (Original post by lucindaaa)
    Could someone please explain how I would answer this question:

    Calculate the enthalpy of formation of ethanol (C2H5OH) given the following enthalpies of combustion.

    C(s) -393
    H2(g) -286
    C2H5OH(l) -1371


    I've got the answer as -273 but I'm not too sure
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    Reverse the enthalpy of combustion equation for ethanol, change this equation to make it so that CO2 changes to C and H2O changes to H2. Add together all the enthalpies needed to do this.
    e.g.
    C_2H_5OH+3O2 \longrightarrow 2CO_2+3H_2O \Delta H^{\circ}=-1371 kJ mol^{-1}

    becomes

    2CO_2+3H_2O \longrightarrow C_2H_5OH+3O2 \Delta H^{\circ}=+1371 kJ mol^{-1}

    Then

    Make it so

     2C+3H_2+1/2O_2 \longrightarrow C_2H_5OH
 
 
 
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