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    1). Two enthalpy changes of reaction are shown below.

    N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \Delta H = -92kJ mol^{-1}
    N_2O_{4(g)}\rightarrow 2NO_{2(g)}\Delta H = +54kJ mol^{-1}

    What is the enthalpy change of reaction for the following:

    \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}\rightarrow NH_{3(g)}
    \frac{1}{2}N_2O_{4(g)}\rightarro  w NO_{2(g)}

    We haven't been given any equations or anything to work it out so I have no idea how to do it =\ .
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    They are equivalent...?
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    (Original post by ViralRiver)
    1). Two enthalpy changes of reaction are shown below.

    N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \Delta H = -92kJ mol^{-1}
    N_2O_{4(g)}\rightarrow 2NO_{2(g)}\Delta H = +54kJ mol^{-1}

    What is the enthalpy change of reaction for the following:

    \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}\rightarrow NH_{3(g)}
    \frac{1}{2}N_2O_{4(g)}\rightarro  w NO_{2(g)}

    We haven't been given any equations or anything to work it out so I have no idea how to do it =\ .
    As enthalpy of formation for elements in their standard states are zero, that means your enthalpy for each of the reaction is simply the enthalpy of formation of the product, ie NH3 and NO2.

    But your values in your first part are given in kJ per mol of product formed already, so the values for the second part is for 1 mol of the corresponding products and hence should be the same.

    If however, the enthalpy for the first reaction is asked, it'd be 2x(-92) because you are producing 2 mol of NH3. Hope you are getting the hang of it.
 
 
 
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