Question

Given $$ 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H^{\circ}=+2269.4 \mathrm{~kJ} $$ (a) What is the heat of formation of chromium(III) oxide? (b) What is \(\Delta H^{\circ}\) for the formation of \(13.65 \mathrm{~g}\) of chromium (III) oxide?

Short answer

Answer: The heat of formation for chromium(III) oxide is 1134.7 kJ/mol. The enthalpy change for the formation of 13.65 g of chromium(III) oxide is +101.87 kJ.

Step-by-step solution

Part (a): Determine the heat of formation of chromium(III) oxide

To find the heat of formation of chromium(III) oxide, we need to determine the enthalpy change per mole of the compound. Given the balanced equation: $$ 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s) \longrightarrow 4\mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H^{\circ}=+2269.4 \mathrm{~kJ} $$ The enthalpy change for the given reaction is +2269.4 kJ. As the reaction involves 2 moles of chromium(III) oxide, the heat of formation for 1 mole of chromium(III) oxide can be found by dividing the given enthalpy change by 2: $$ \text{Heat of formation of chromium(III) oxide} \ = \ \frac{2269.4 \mathrm{~kJ}}{2} \ = \ 1134.7 \mathrm{~kJ} \, \text{per mole} $$

Part (b): Calculate \(\Delta H^{\circ}\) for the formation of 13.65 g of chromium(III) oxide

Now we have to find the enthalpy change (\(\Delta H^{\circ}\)) for the formation of 13.65 g of chromium(III) oxide. First, we need to convert the mass (13.65 g) of chromium(III) oxide to moles. The molar mass of chromium(III) oxide (Cr₂O₃) is: $$ \text{Molar mass of Cr₂O₃} \ = \ 2\cdot(51.9961 \mathrm{~g/mol})\ +\ 3\cdot(16.00 \mathrm{~g/mol})\ =\ 151.99 \mathrm{~g/mol} $$ So, the number of moles of chromium(III) oxide is given by: $$ \text{moles of Cr₂O₃} \ = \ \frac{13.65 \mathrm{~g}}{151.99 \mathrm{~g/mol}}\ =\ 0.0898 \mathrm{~mol} $$ Now that we know the number of moles in 13.65 g of chromium(III) oxide, we can find the enthalpy change for its formation using the heat of formation we found in part (a): $$ \Delta H^{\circ} \ = \ 0.0898 \mathrm{~mol} \times 1134.7 \mathrm{~kJ/mol} \ = \ 101.87 \mathrm{~kJ} $$ The enthalpy change for the formation of 13.65 g of chromium(III) oxide is +101.87 kJ.