# 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?

Nov 1, 2015

$2 \text{KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g]). " "DeltaH_text(rxn) = -"44.7 kJ}$

#### Explanation:

A thermochemical equation is simply a chemical equation that contains information about the enthalpy change of reaction, $\Delta {H}_{\text{rxn}}$.

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, ${\text{KClO}}_{3}$, undergo decomposition to form potassium chloride, $\text{KCl}$, and oxygen gas, ${\text{O}}_{2}$.

So you know that you have

$2 {\text{KClO"_text(3(s]) -> "KCl"_text((s]) + "O}}_{\textrm{2 \left(g\right]}}$

Balance this equation by multiplying the products by $2$ and $3$, respectively

$2 {\text{KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O}}_{\textrm{2 \left(g\right]}}$

Now focus on the enthalpy change of reaction. The problem tells you that the reaction given off $\text{44.7 kJ}$, which means that the reaction is actually exothermic.

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

$\Delta H = - \text{44. 7 kJ}$

As a result, the thermochemical equation for the decomposition of potassium chlorate looks like this

$2 \text{KClO"_text(3(s]) -> 2"KCl"_text((s]) + 3"O"_text(2(g]). " "DeltaH_text(rxn) = -"44.7 kJ}$