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    Given that:

    H2SO4 + KCl ---> 2HCl + K2SO4

    H2SO4 + 2KOH ---> K2SO4 + 2H20
    Enthalpy change = -342 kJ

    HCl + KOH ---> KCl + H20
    Enthalpy change = -204kJ

    Calculate the standard enthalpy change of formation of HCl gas from H2SO4 and KCl.

    Yeah I got the basics of Hess's Law, but they hadn't really been applied to such large equations. >_<

    I'm teaching myself AS, so if you could explain your reasoning, I'd be much obliged. This book doesn't offer answers.

    First thing you need to do is change the equations so you have reactants on the left hand side. This gives you:

    \mathrm{H_2SO_4 + 2KOH \longrightarrow K_2SO_4 + 2H_2O};\ \Delta H = -342

    \mathrm{KCl + H_2O \longrightarrow HCl + KOH};\ \Delta H = 204

    Then you need to add multiples of these together to give your original equation i.e for this, you need to add the first to two of the second. The enthalpy change is then the sum of the enthalpy changes of the separate equations in the correct amounts i.e. -342 + (2 x 204)
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