Question 5.6.1: Use and interpret standard heats of formation. a. Write the ......
Use and interpret standard heats of formation.
a. Write the balanced chemical equation that represents the standard heat of formation of KClO_{3}(s) at 298 K.
b. The standard enthalpy change for the following reaction is 2261 kJ at 298 K.
2 Na_{2}CO_{3}(s) → 4 Na(s) + 2 C(graphite) + 3 O_{2}(g) ∆Hº_{rxn} = 2261 kJ
What is the standard heat of formation of Na_{2}CO_{3}(s)?
a. You are asked to write an equation for the standard heat of formation of a compound.
You are given the compound formula.
The standard heat of formation (or standard enthalpy of formation) of a substance in a specified state at 298 K is the enthalpy change for the reaction in which one mole of the substance is formed from the elements in their stable forms at 1 bar and 298 K. Potassium is a solid and chlorine and oxygen are diatomic gases at 1 bar and 298 K.
K(s) + ½ Cl_{2}(g) + 3/2 O_{2}(g) → KClO_{3}(s)
Is your answer reasonable? The equation shows the formation of only one mole of product and the reactants are all elements in their most stable forms at 1 bar and 298 K.
b. You are asked to determine the standard heat of formation of a compound.
You are given the enthalpy change under standard conditions for a reaction involving that
compound.
The equation that represents the standard heat of formation of Na_{2}CO_{3}(s) is
2 Na(s) + C(graphite) + 3/2 O_{2}(g) → Na_{2}CO_{3}(s)
Reversing the equation in the problem and multiplying by ½ results in the equation for the standard heat of formation of Na_{2}CO_{3}, so
∆H_{f}º = −½(∆Hº_{rxn}) = −½(2261 kJ) = −1131 kJ/mol
Is your answer reasonable? The given equation shows sodium carbonate as a reactant, and the standard heat of formation is the energy change for the formation of one mole of a compound. The given reaction is endothermic as written, so the standard heat of formation for this compound is exothermic.