{"id":46602,"date":"2025-07-02T07:44:21","date_gmt":"2025-07-02T07:44:21","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=46602"},"modified":"2025-07-02T07:44:23","modified_gmt":"2025-07-02T07:44:23","slug":"use-the-born-haber-cycles-to-determine-the-lattice-energies-of-the-two-ionic-compounds-then-use-these-answers-to-justify-the-large-energy-release-from-this-reaction","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/use-the-born-haber-cycles-to-determine-the-lattice-energies-of-the-two-ionic-compounds-then-use-these-answers-to-justify-the-large-energy-release-from-this-reaction\/","title":{"rendered":"Use the Born-Haber cycles to determine the lattice energies of the two ionic compounds, then use these answers to justify the large energy release from this reaction"},"content":{"rendered":"\n
Use the Born-Haber cycles to determine the lattice energies of the two ionic compounds, then use these answers to justify the large energy release from this reaction. Al(s) + Fe2O3(s) –> Fe(I) + Al2O3 (s)<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
To determine the lattice energies of the ionic compounds involved in the reaction Al(s)+Fe2O3(s)\u2192Fe(l)+Al2O3(s)\\text{Al}(s) + \\text{Fe}_2\\text{O}_3(s) \\rightarrow \\text{Fe}(l) + \\text{Al}_2\\text{O}_3(s)Al(s)+Fe2\u200bO3\u200b(s)\u2192Fe(l)+Al2\u200bO3\u200b(s), we will first use the Born-Haber cycle, which is a thermodynamic cycle used to calculate the lattice energy of an ionic compound.<\/p>\n\n\n\n
Step 1: Ionization of Aluminum (Al) and Iron (Fe)<\/h3>\n\n\n\n
\n
Aluminum (Al)<\/strong> is first ionized to Al3+\\text{Al}^3+Al3+, which requires three ionization energies: Al(s)\u2192Al3+(g)+3e\u2212\\text{Al}(s) \\rightarrow \\text{Al}^{3+}(g) + 3e^-Al(s)\u2192Al3+(g)+3e\u2212<\/li>\n\n\n\n
Iron (Fe)<\/strong> in Fe2O3\\text{Fe}_2\\text{O}_3Fe2\u200bO3\u200b is present as Fe3+\\text{Fe}^{3+}Fe3+, so for Fe2O3\\text{Fe}_2\\text{O}_3Fe2\u200bO3\u200b, each iron atom is ionized into Fe3+\\text{Fe}^{3+}Fe3+: Fe(s)\u2192Fe3+(g)+3e\u2212\\text{Fe}(s) \\rightarrow \\text{Fe}^{3+}(g) + 3e^-Fe(s)\u2192Fe3+(g)+3e\u2212<\/li>\n<\/ul>\n\n\n\n
Step 2: Formation of Fe2O3\\text{Fe}_2\\text{O}_3Fe2\u200bO3\u200b and Al2O3\\text{Al}_2\\text{O}_3Al2\u200bO3\u200b<\/h3>\n\n\n\n
\n
Oxygen atoms in the Fe2O3\\text{Fe}_2\\text{O}_3Fe2\u200bO3\u200b and Al2O3\\text{Al}_2\\text{O}_3Al2\u200bO3\u200b structures are ionized, forming O2\u2212\\text{O}^{2-}O2\u2212 ions: O2(g)\u21922O2\u2212(g)+2e\u2212\\text{O}_2(g) \\rightarrow 2\\text{O}^{2-}(g) + 2e^-O2\u200b(g)\u21922O2\u2212(g)+2e\u2212<\/li>\n<\/ul>\n\n\n\n
Step 3: Lattice Formation Energy<\/h3>\n\n\n\n
\n
The lattice energy is the energy required to assemble the ions into a solid ionic lattice from the individual gaseous ions.<\/li>\n\n\n\n
For Fe2O3\\text{Fe}_2\\text{O}_3Fe2\u200bO3\u200b, the ions Fe3+\\text{Fe}^{3+}Fe3+ and O2\u2212\\text{O}^{2-}O2\u2212 attract each other to form a highly stable lattice structure. The same holds for Al2O3\\text{Al}_2\\text{O}_3Al2\u200bO3\u200b, where the lattice energy can be calculated based on the electrostatic forces between Al3+\\text{Al}^{3+}Al3+ and O2\u2212\\text{O}^{2-}O2\u2212.<\/li>\n<\/ul>\n\n\n\n
Step 4: Calculation of Lattice Energy<\/h3>\n\n\n\n
The lattice energy UUU can be estimated using the formula derived from the Coulomb\u2019s Law: U=k\u22c5Q1\u22c5Q2rU = \\dfrac{k \\cdot Q_1 \\cdot Q_2}{r}U=rk\u22c5Q1\u200b\u22c5Q2\u200b\u200b<\/p>\n\n\n\n
where:<\/p>\n\n\n\n
\n
kkk is Coulomb’s constant,<\/li>\n\n\n\n
Q1Q_1Q1\u200b and Q2Q_2Q2\u200b are the charges of the ions,<\/li>\n\n\n\n
rrr is the sum of the ionic radii.<\/li>\n<\/ul>\n\n\n\n
The lattice energies for both Fe2O3\\text{Fe}_2\\text{O}_3Fe2\u200bO3\u200b and Al2O3\\text{Al}_2\\text{O}_3Al2\u200bO3\u200b are large due to the high charges on the ions and the small ionic radii of Fe3+\\text{Fe}^{3+}Fe3+ and Al3+\\text{Al}^{3+}Al3+. This leads to a very strong electrostatic attraction, resulting in high lattice energy values.<\/p>\n\n\n\n
Step 5: Justifying the Large Energy Release from the Reaction<\/h3>\n\n\n\n
The large energy release from the reaction: Al(s)+Fe2O3(s)\u2192Fe(l)+Al2O3(s)\\text{Al}(s) + \\text{Fe}_2\\text{O}_3(s) \\rightarrow \\text{Fe}(l) + \\text{Al}_2\\text{O}_3(s)Al(s)+Fe2\u200bO3\u200b(s)\u2192Fe(l)+Al2\u200bO3\u200b(s)<\/p>\n\n\n\n
can be attributed to the combination of several factors:<\/p>\n\n\n\n
\n
High Lattice Energy<\/strong>: Both Fe2O3\\text{Fe}_2\\text{O}_3Fe2\u200bO3\u200b and Al2O3\\text{Al}_2\\text{O}_3Al2\u200bO3\u200b have extremely high lattice energies due to the high charges on the metal ions and small ionic sizes.<\/li>\n\n\n\n
Reduction of Fe3+\\text{Fe}^{3+}Fe3+<\/strong>: The iron ion Fe3+\\text{Fe}^{3+}Fe3+ is reduced to elemental iron (Fe), which is a significant reduction in energy, making the reaction highly exothermic.<\/li>\n\n\n\n
Oxidation of Aluminum<\/strong>: Aluminum metal undergoes oxidation to form Al2O3\\text{Al}_2\\text{O}_3Al2\u200bO3\u200b, releasing a considerable amount of energy.<\/li>\n<\/ol>\n\n\n\n
Thus, the total energy released in the reaction is a result of the large negative lattice energies of the products, combined with the reduction and oxidation processes.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Use the Born-Haber cycles to determine the lattice energies of the two ionic compounds, then use these answers to justify the large energy release from this reaction. Al(s) + Fe2O3(s) –> Fe(I) + Al2O3 (s) The Correct Answer and Explanation is: To determine the lattice energies of the ionic compounds involved in the reaction Al(s)+Fe2O3(s)\u2192Fe(l)+Al2O3(s)\\text{Al}(s) […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-46602","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46602","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=46602"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46602\/revisions"}],"predecessor-version":[{"id":46656,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46602\/revisions\/46656"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=46602"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=46602"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=46602"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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<\/figure>\n","protected":false},"excerpt":{"rendered":"