What's the solution?

Question

A chemist measures the enthalpy change ΔH during the following reaction:

2KCl+3O2→2KClO ΔH=78KJ

Use this information to complete the table below. Round each of your answers to the nearest kJ/mol.

2KClO3→2KCl+3O2

4KClO3→4KCl+6O2

KCl+3/2O2→KClO3

Answer 1

ΔH value indicates the amount of heat associated with the reaction involving the number of

moles of reactants and products

So, this equation 2KCl+3O2→2KClO3 ΔH=78KJ indicates that when 2 moles of KCL and 3

moles of oxygen gas at some temperature and pressure change to 2 moles of KCIO, 78 kJ of heat is released to the surroundings.

If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor.

Therefore,

1) 2KClO3→2KCl+3O2

This equation is reversed so multiply ΔH by -1. Thus, ΔH for this eq is -78 KJ.

2) 4KClO3→4KCl+6O2

This equation is reversed and multiplied by 2. So multiply ΔH by -2. Thus ΔH = 78*(-2) = -156 KJ

KCl+3/2O2→KClO3

3) This equation is divided by 2. So Divide ΔH by 2. Thus ΔH = 78/2 = 39 KJ