Question 5.6.2: Calculate enthalpy change using standard heats of formation.......
Calculate enthalpy change using standard heats of formation.
Using the standard heats of formation that follow, calculate the standard enthalpy change for the following reaction.
3 Fe_{2}O_{3}(s) + H_{2}(g) → 2 Fe_{3}O_{4}(s) + H_{2}O(g)
Compound \Delta H_{f}^{\mathrm{o}} (kJ/mol)
Fe_{2}O_{3}(s) −824.2
Fe_{3}O_{4}(s) −1118.4
H_{2}O(g) −241.8
You are asked to calculate the standard enthalpy change for a reaction.
You are given the chemical equation for the reaction and the standard heats of formation for the compounds in the equation.
The standard enthalpy change for a reaction can be calculated from the standard heats of formation of the products and reactants, each multiplied by the stoichiometric coefficient in the balanced equation. Recall that the standard heat of formation for an element in its stable state at 298 K and 1 bar is zero.
=(2\,\mathrm{mol~Fe}_{3}O_{4})(\Delta H_{f}^{\circ}[\mathrm{Fe}_{3}O_{4}({\bf s})])\,+\,(1\,\mathrm{mol~H}_{2}O)(\Delta H_{f}^{\circ}[\mathrm{H}_{2}O({\bf g})])
-\ (3\operatorname{mol}\operatorname{Fe}_{2}{\mathsf{O}}_{3})(\Delta H_{f}^{\circ}[\mathbf{Fe}_{2}{\mathsf{O}}_{3}({\mathsf{S}})])
=(2{\mathrm{~mol~}}\operatorname{Fe}_{3}O_{4})(-1118.4{\mathrm{~kJ/mol}})+(1{\mathrm{~mol~}}\mathrm{H}_{2}O)(-241.8{\mathrm{~kJ/mol}})
-\ (3\operatorname{mol}\,\operatorname{Fe}_{2}O_{3})(-824.2\ kJ/\operatorname{mol})
= -6.0 kJ