when 2 moles of potassium chlorate crystals decompose to potassium chloride crystals and oxygen gas at a constant pressure and temperature, 78.0 kj of heat is given off. write a thermodynamic equation for this reaction.I don't know how to set up the equation i know that for the potassium there would have to be 2 for every one oxygen and 1 potassium in the product side but how to i balance or use the correct nomenclature

Question

when 2 moles of potassium chlorate crystals decompose to potassium chloride crystals and oxygen gas at a constant pressure and temperature, 78.0 kj of heat is given off. write a thermodynamic equation for this reaction.I don't know how to set up the equation i know that for the potassium there would have to be 2 for every one oxygen and 1 potassium in the product side but how to i balance or use the correct nomenclature

Answer 1

2KClO3 ==> 2KCl + 3O2 dH = -78 kJ

or
you could write
KClO3 ==> KCl + 3/2 O2 dH = -39 kJ/mol

Answer 2

To write a thermodynamic equation for the decomposition of potassium chlorate into potassium chloride and oxygen gas, we can follow the steps below:

Step 1: Write the unbalanced chemical equation for the reaction:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Step 2: Balance the equation:
Since there are 2 K atoms on both sides, the number of Cl atoms is already balanced. The only thing remaining is to balance the number of O atoms. Since there are 3 O atoms on the product side and only 1 O atom on the reactant side, we need to put a coefficient of 3 in front of KClO3:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Step 3: Add the physical states of the substances:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)

Step 4: Write the enthalpy change (heat given off) for the reaction:
The heat given off during the reaction is stated to be 78.0 kJ. We can include this information in the equation using the symbol ΔH:
2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -78.0 kJ

Note: In this case, since the heat is being given off, the sign of ΔH is negative.

So, the final thermodynamic equation for the decomposition of potassium chlorate is:
2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -78.0 kJ

Answer 3

To write the thermodynamic equation for the decomposition of potassium chlorate (KClO3) to form potassium chloride (KCl) and oxygen (O2) at a constant pressure and temperature, you need to balance the equation and include the appropriate coefficients.

The balanced equation for this reaction is:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

This equation indicates that two moles of potassium chlorate decompose to form two moles of potassium chloride and three moles of oxygen gas. The coefficients are used to balance the number of atoms of each element on both sides of the equation.

In terms of nomenclature, you correctly identified that there should be two potassium (K) ions on the product side since there are two KCl molecules formed. Additionally, the subscript "3" in KClO3 indicates that there are three oxygen (O) atoms in each KClO3 molecule.

Therefore, the thermodynamic equation for this reaction is:

2 KClO3(s) → 2 KCl(s) + 3 O2(g) + 78.0 kJ

The "+78.0 kJ" term indicates that 78.0 kilojoules of heat are given off during the reaction.