When 2 moles of potassium chlorate crystals decompose to potassium chloride crystals and oxygen gas at constant temperature and pressure 44.7 kJ are given off. What is the thermochemical equation?

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    ##2KClO_text(3(s]) -> 2KCl_text((s]) + 3O_text(2(g]). DeltaH_text(rxn) = -44.7 kJ##
    A thermochemical equation is simply a chemical equation that contains information about the change of reaction ##DeltaH_rxn##.
    The first thing to do when writing a thermochemical equation is to make sure that you get the balnced chemical equation right.
    In your case you know that 2 moles of potassium chlorate ##KClO_3## undergo to form potassium chloride ##KCl## and oxygen gas ##O_2##.
    So you know that you have
    ##2KClO_text(3(s]) -> KCl_text((s]) + O_text(2(g])##
    Balance this equation by multiplying the products by ##2## and ##3## respectively
    ##2KClO_text(3(s]) -> 2KCl_text((s]) + 3O_text(2(g])##
    Now focus on the enthalpy change of reaction. The problem tells you that the reaction given off ##44.7 kJ## which means that the reaction is actually .
    As you know energy changes are being represent from the point of view of the system. This implies that for an exothermic reaction the enthalpy change will carry a negative sign since the system is losing heat to the surroundings.
    Therefore the enthalpy change of reaction will be
    ##DeltaH = -44. 7 kJ##
    As a result the thermochemical equation for the decomposition of potassium chlorate looks like this
    ##2KClO_text(3(s]) -> 2KCl_text((s]) + 3O_text(2(g]). DeltaH_text(rxn) = -44.7 kJ##

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