Enthalpy Reaction

Introduced to this concept in AP Chemistry

Equation §

$△$ H${}_{rxn}$ = $\Sigma$ n $△$ H ${}_f^0$ (products) - $\Sigma$ m $△$ H${}_f^0$ (reactants)

n represents the coefficients for the products m represents the coefficients for the reactants

1. using the table of $△$ H ${}_f^0$ values, find the value for each product and multiply it by the products coefficient
2. Add up all of the enthalpies of formation for the products
3. Using the table of $△$ H${}_f^0$ values find the value for each reactant and multiply it by the reactants coefficient
4. Add up all of the enthalpies of formation for the reactants
5. Take the difference

Sample Problem §

Problem: Calculate the enthalpy of reaction, $△$ H${}_{rxn}^0$ for the following reaction given the enthalpy of formation data provided in the table

4 FeO (s) + O${}_2$ (g) -> 2 Fe${}_2$O${}_3$ (s)

$△$ H${}_{rxn}^0$ = ?

Enthalpy Formation of FeO(s) = -272.0 $\frac{kJ}{mol}$

Enthalpy Formation of Fe${}_2$O${}_3$(s) = -824.2 $\frac{kJ}{mol}$

4 FeO(s) + O${}_2$ (g) -> 2 Fe${}_2$O${}_3$ (s)

$△$ H${}_{rxn}$ = $\Sigma$ n $△$ H ${}_f^0$ (products) - $\Sigma$ m $△$ H${}_f^0$ (reactants)

$△$ H${}_{rxn}$ = (2x824.2) - ((4x-272.0)+0) = -560.4 kJ