# 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?

##2″KClO”_text(3(s]) -> 2″KCl”_text((s]) + 3″O”_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
##2″KClO”_text(3(s]) -> “KCl”_text((s]) + “O”_text(2(g])##
Balance this equation by multiplying the products by ##2## and ##3##, respectively
##2″KClO”_text(3(s]) -> 2″KCl”_text((s]) + 3″O”_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
##2″KClO”_text(3(s]) -> 2″KCl”_text((s]) + 3″O”_text(2(g]). ” “DeltaH_text(rxn) = -“44.7 kJ”##

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