Name: predicting ionic stability with lattice energy
Uploaded: 2020-09-03T13:38:28.000Z
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Chemistry

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PS.Chm.3.1

The Born-Haber Cycle

La formación de enlaces iónicos requiere la transferencia de electrones de un átomo metálico a otro no metálico, un proceso que a menudo es endotérmico. Sin embargo, cuando el sodio elemental y el cloro reaccionan para formar un cristal sólido de cloruro de sodio, es un proceso altamente exotérmico. Pero, de dónde viene la energía?De acuerdo con la ley de Coulomb, los cationes y los aniones se atraen entre sí mediante fuertes fuerzas electrostáticas en una matriz sólida o rejilla. La estructura reticular resultante se estabiliza disminuyendo la energía potencial, que se libera en forma de calor, una reacción exotérmica. La energía total asociada con la formación o descomposición de una red cristalina en sus constituyentes gaseosos se llama energía de red.En un compuesto iónico sólido, una gran cantidad de partículas cargadas interactúan entre sí, lo que dificulta la determinación experimental del valor exacto de la energía de la red. Sin embargo, se puede calcular utilizando la ley de Hess en una serie hipotética de pasos denominada ciclo de Born-Haber, que representa la formación de un compuesto iónico a partir de sus elementos constituyentes. Por ejemplo, el ciclo de Born-Haber para la formación de cloruro de sodio considera dos rutas alternativas, una directa y otra indirecta.La ruta directa representa la entalpía estándar de formación de NaCl a partir de sodio elemental y cloro. La ruta indirecta tiene cinco pasos. En el primer paso, el sodio en estado sólido se convierte en su forma gaseosa.A continuación, las moléculas de cloro diatómico se disocian en átomos de cloro gaseoso. Los pasos tercero y cuarto explican la transferencia de electrones para formar iones. Se extrae un electrón del sodio gaseoso para formar un catión de sodio.Luego, el cloro gaseoso absorbe el electrón para formar un anión cloruro. En el paso final, la atracción electrostática entre los iones gaseosos conduce a la formación de la estructura reticular. La ley de Hess establece que el cambio en la entalpía general de un proceso escalonado es la suma de los cambios de la entalpía de cada paso.Es decir, el valor de entalpía de la ruta directa es igual a la suma de las entalpías de los cinco pasos. Al resolver la ecuación para la energía reticular se determina un valor negativo grande, lo que implica una reacción exotérmica.

el ciclo de born-haber

<h4>Energía Reticular </h4>
 
 Un compuesto iónico es estable debido a la atracción electrostática entre sus iones positivos y negativos. La energía reticular de un compuesto es una medida de la fuerza de esta atracción. La energía de celosía (ΔHreticular) de un compuesto iónico se define como la energía requerida para separar un mol del sólido en los iones gaseosos que lo componen. 
 
 Aquí, la convención se usa donde el sólido iónico se separa en iones, significando que las energías de celosía serán endotérmicas (valores positivos). Otra manera es usar una convención equivalente, pero opuesta, en donde la energía reticular es exotérmica (valores negativos) y descrita como la energía liberada cuando los iones se combinan para formar una red. Por lo tanto, asegúrese de confirmar qué definición es usada cuando busque energías reticulares en otra referencia. 
 
 En ambos casos, una magnitud mayor para la energía de celosía indica un compuesto iónico más estable. Para el cloruro sódico, ΔHreticular = 769 kJ. Por lo tanto, se requieren 769 kJ para separar un mol de NaCl sólido en los iones gaseosos Na+ y Cl–. Cuando un mol de iones gaseoso Na+ y Cl– forman NaCl sólido, se liberan 769 kJ de calor.
 
 <h4>Determinación de la energía Reticular de un Compuesto Iónico</h4>
 
 No es posible medir la energía reticular directamente. Sin embargo, la energía reticular puede ser calculada usando un ciclo termoquímico. El ciclo Born-Haber es una aplicación de la ley de Hess que divide la formación de un sólido iónico en una serie de pasos individuales:
 
 <table style="margin-left: auto; margin-right: auto;">
 <tbody style="border: 1px solid black;">
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Entalpía de sublimación de Cs (s)</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">Cs (s) → Cs (g)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = ΔHs° = 76,5 kJ/mol</td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">La mitad de la energía de enlace de F 2</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">&frac; F2 (g) → F (g)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = &frac; D = 79,4 kJ/mol</td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Energía de ionización de Cs (g)</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">Cs (g) → Cs+ (g) + e−</td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = EI = 375,7 kJ/mol</td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Afinidad Electrónica de F</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">F (g) + e− → F− (g)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = AE = −328,2 kJ/mol</td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Negativo de la energía de celosía del CsF (s)</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">Cs+ (g) + F− (g) → CsF (s)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = −ΔHreticular = ?</td>
 </tr>
 <tr>
 <td style="color: black; border: 1px solid black; text-align: center;">Entalpía de formación de CsF (s), añadir los pasos 1–5</td>
 <td colspan="2" style="color: black; border: 1px solid black; text-align: center;">ΔH = ΔHf° = ΔHs°+&frac; D +EI + (AE) + (−ΔHreticular) 
 Cs (s) + &frac; F2 (g) → CsF (s)</td>
 <td style="color: black; border: 1px solid black; text-align: center;">ΔH = −553,5 kJ/mol</td>
 </tr>
 </tbody>
 </table>
 
 <ol>
 <li>Considere los elementos en sus estados más comunes, Cs (s) y F2 (g). </li>
 <li>El ΔHs° representa la conversión del cesio sólido en un gas (sublimación), y luego la energía de ionización convierte los átomos de cesio gaseoso en cationes. </li>
 <li>En el siguiente paso, se debe tener en cuenta la energía necesaria para romper el enlace F–F para producir átomos de flúor. </li>
 <li>Convertir un mol de átomos de flúor en iones de flúor es un proceso exotérmico, por lo que este paso emite energía (la afinidad electrónica)  </li>
 <li>Ahora, está presente un mol de cationes y un mol de aniones F. Estos iones se combinan para producir fluoruro de cesio sólido. El cambio de entalpía en este paso es el negativo de la energía de celosía, así que también es una cantidad exotérmica. </li>
 <li>La energía total implicada en esta conversión es igual a la entalpía de formación experimentalmente determinada, ΔHf°, del compuesto a partir de sus elementos. En este caso, el cambio general es exotérmico.</li>
 </ol>
 
 Las energías reticulares calculadas para compuestos iónicos son típicamente mucho más altas que las energías de disociación de enlace medidas para enlaces covalentes. Mientras que las energías de celosía suelen estar en el rango de 600 – 4000 kJ/mol (algunas incluso más altas), las energías de disociación del enlace covalente suelen estar entre 150 – 400 kJ/mol para enlaces simples. Sin embargo, tenga en cuenta que estos valores no son directamente comparables. Para los compuestos iónicos, las energías reticulares están asociadas con muchas interacciones, ya que los cationes y los aniones se empaquetan juntos en un retículo extendido. Para los enlaces covalentes, la energía de disociación de enlace está relacionada con la interacción de sólo dos átomos.
 
 <h4>Energía reticular como una función del radio y de la carga iónica </h4>
 
 La energía de celosía de un cristal iónico aumenta rápidamente a medida que aumentan las cargas de los iones y disminuyen los tamaños de los iones. Cuando todos los demás parámetros se mantienen constantes, al doblar la carga tanto del catión como del anión se cuadruplica la energía reticular. Por ejemplo, la energía de celosía de LiF (Z+ y Z– = 1) es de 1023 kJ/mol, mientras que la de MgO (Z+ y Z– = 2) es de 3900 kJ/mol (Ro = la distancia interiónica definida como la suma de los radios de los iones positivos y negativos, es casi la misma — alrededor de las 200 pm para ambos compuestos).
 
 Las diferentes distancias interatómicas producen diferentes energías reticulares. Por ejemplo, compare la energía de celosía de MgF2 (2957 kJ/mol) con la de MgI2 (2327 kJ/mol), lo que demuestra el efecto del tamaño iónico más pequeño del F– sobre la energía de celosía en comparación con el I–.
 
 <h4>Otras Aplicaciones del Ciclo de Born-Haber</h4>
 
 El ciclo de Born-Haber también puede ser usado para calcular cualquiera de las otras cantidades en la ecuación para la energía de celosía, siempre que las demás sean conocidas. Por ejemplo, si la entalpía de sublimación relevante  ΔHs°,   la energía de ionización (EI), la entalpía de disociación de enlace (D), la energía reticular ΔHreticular y la entalpía estándar de formación  ΔHf°  se conocen, el ciclo de Born-Haber se puede utilizar para determinar la afinidad electrónica de un átomo.
 
 Este texto es adaptado de<a href="https://openstax.org/books/chemistry-2e/pages/7-5-strengths-of-ionic-and-covalent-bonds">Openstax, Química 2e, Sección 7.5: Fortalezas de los Enlaces Iónicos y Covalentes.</a>

El Ciclo de Born-Haber

Energía reticular

Un compuesto iónico es estable debido a la atracción electrostática entre sus iones positivos y negativos. La energía reticular de un ...

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predicting ionic stability with lattice energy

Lattice energy helps explain why ionic compounds are stable. It measures the strength of attraction between positive and negative ions in a crystal. In one common definition, lattice energy is the energy required to separate one mole of an ionic solid into gaseous ions, so it has a positive value. Another definition uses the opposite sign and describes the energy released when gaseous ions form a lattice. When checking data, it is important to confirm which convention is being used. In either case, a larger lattice energy magnitude means a more stable ionic compound.

Sodium chloride gives a simple example. Its lattice energy is 769 kJ/mol, which means 769 kJ is needed to pull one mole of solid NaCl apart into Na+ and Cl- ions in the gas phase. The same amount of heat is released when those ions come together to form solid NaCl. This value shows how strongly the ions attract each other in the crystal.

Lattice energy cannot be measured directly. Instead, it is found with a thermochemical cycle. The Born-Haber cycle uses Hess’s law to break the formation of an ionic solid into separate steps. For cesium fluoride, these steps include sublimation of Cs(s) to Cs(g), half of the F2 bond energy to make F(g), ionization of Cs(g) to Cs+(g) plus an electron, electron affinity of F(g) to make F-(g), and formation of solid CsF from the gaseous ions. The enthalpy change for forming the lattice is the negative of the lattice energy.

For CsF, the overall reaction Cs(s) + 1/2 F2(g) -&gt; CsF(s) has a standard enthalpy of formation of -553.5 kJ/mol. The cycle combines the sublimation enthalpy of cesium, 76.5 kJ/mol, one-half of the F2 bond energy, 79.4 kJ/mol, the ionization energy of cesium, 375.7 kJ/mol, and the electron affinity of fluorine, -328.2 kJ/mol, with the lattice term. Because the total is exothermic, energy is released overall when the solid forms from the elements in their most common states, Cs(s) and F2(g).

Lattice energies for ionic compounds are usually much larger than bond dissociation energies for covalent bonds. Ionic lattice energies often fall between 600 and 4000 kJ/mol, and some are even higher. Single covalent bond dissociation energies are usually about 150 to 400 kJ/mol. These values are not directly comparable because lattice energy reflects many attractions in an extended crystal, while a covalent bond energy describes the interaction of just two atoms.

Ion charge and ion size strongly affect lattice energy. When all other factors stay the same, larger charges give much larger lattice energies, and smaller ions also increase lattice energy. Doubling the charge on both ions can quadruple the lattice energy. For example, LiF has a lattice energy of 1023 kJ/mol, while MgO is 3900 kJ/mol, even though their interionic distances are nearly the same at about 200 pm. Size also matters in compounds such as MgF2 and MgI2, where the smaller F- ion leads to a higher lattice energy than the larger I- ion.

The Born-Haber cycle can also solve for other unknowns in the lattice energy equation. If the other terms are known, the cycle can be used to find a missing quantity such as electron affinity. This makes the method useful for linking measured enthalpy values to the energy changes inside ionic compounds.

Predicting Ionic Stability with Lattice Energy

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types of chemical bonds

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Chemical bonding explains how atoms and ions join to form molecules and compounds. American chemist Gilbert N. Lewis pioneered chemical bonding theory with the Lewis model. This model helps explain why bonds form, why some are stronger than others, and why compounds have fixed formulas like H2O instead of H4O. Ionic bonding involves ions, which are atoms or molecules with an electrical charge. A cation is a positive ion formed when a neutral atom loses one or more electrons from its valence...

the born-haber cycle

Lattice energy helps explain why ionic compounds are stable. It measures the strength of attraction between positive and negative ions in a crystal. In one common definition, lattice energy is the energy required to separate one mole of an ionic solid into gaseous ions, so it has a positive value. Another definition uses the opposite sign and describes the energy released when gaseous ions form a lattice. When checking data, it is important to confirm which convention is being used. In either...

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Lattice energy shows how strongly ions hold an ionic solid together. It is the energy required to separate one mole of a solid into its gaseous ions. For sodium chloride, this process requires 769 kJ per mole. The opposite convention may also be used, where lattice energy is the energy released when gaseous ions form the solid. In either case, a larger magnitude means a more stable ionic compound. For ionic crystals, lattice energy follows a pattern based on charge and ion size. The...

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Learn how lattice energy, ion charge, ion size, and the Born-Haber cycle help explain ionic stability and calculate values for compounds like CsF.

Predicting Ionic Stability with Lattice Energy | JoVE K12

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Bond Polarity, Dipole Moment, and Percent Ionic Character

Bond Polarity

The absolute value of the difference in electronegativity (&Delta;EN) of two bonded atoms provides a rough measure of the polarity to be ...

Lewis Structures of Molecular Compounds and Polyatomic Ions

To draw Lewis structures for complicated molecules and molecular ions, it is helpful to follow a step-by-step procedure as outlined:


	Determine the ...

Resonance

The Lewis structure of a nitrite anion (NO2&minus;) may actually be drawn in two different ways, distinguished by the locations of the N-O and N=O ...

Formal Charges

In some cases, there are seemingly more than one valid Lewis structures for molecules and polyatomic ions. The concept of formal charges can be used to ...

Exceptions to the Octet Rule

Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:


	...

Analyze and interpret data to explain the mechanisms and properties of the two main types of intramolecular (ionic and covalent) bonds.

Analyze the structure of atoms and isotopes.

Predicting Ionic Stability with Lattice Energy

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