Using enthalpies of formation, calculate H.

Answers

Answer 1

Answer:

ΔH° = -851.5 kJ/mol given that

$\begin{array}{cc}\textbf{Species}&{\bf {\Delta H_f\textdegree{}}}\n \text{Fe}_2\text{O}_3\;(s) & -824.2\;\text{kJ}\cdot\text{mol}^(-1)\n\text{Al}_2\text{O}_3\;(s) & -1675.7\;\text{kJ}\cdot\text{mol}^(-1)\end{array}$

(Source: Chemistry Libretexts.)

Explanation

Refer to a thermodynamic data table for the standard enthalpy of formation for each species.

Don't be alerted if the data for Al (s) and Fe (s) are missing. Why?

The standard enthalpy of formation of a substance measures the ΔH required to form each mole of it from the most stable allotrope of its elements under STP.
Both Al (s) and Fe (s) are already the most stable form of their element under STP (note that the state symbol matters.) There's no need to form them again.

As a result, $\Delta H_f\textdegree{} = 0$ for both Al (s) and Fe (s).

$\displaystyle \Delta H_{\text{rxn}}\textdegree{} = \text{Sum of }\Delta H\text{ for all }\textbf{Product} - \text{Sum of }\Delta H\text{ for all }\textbf{Reactant}}\n\phantom{\Delta H_{\text{rxn}}\textdegree{}} = (1* \Delta H_f\textdegree{}(\text{Al}_2\text{O}_3\;(s)) + 1* \Delta H_f\textdegree{}(\text{Al}\;(s)) \n \phantom{\Delta H_{\text{rxn}}\textdegree{}=}-(1* \Delta H_f\textdegree{}(\text{Fe}_2\text{O}_3\;(s)) + 1*\Delta H_f\textdegree{}(\text{Fe}\;(s))$

$\Delta H_{\text{rxn}}\textdegree{}} = (1 * (-1675.7)) - (1*(-824.2)) = -851.5\;\text{kJ}\cdot\text{mol}^(-1)$ .

The number "1" here emphasizes that in case there are more than one mole of any species in one mole of the reaction, it will be necessary to multiply the $\Delta H_f\textdegree{}$ of that species with its coefficient in the equation.

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