Question

(a) Write the chemical equations that are used in calculating the lattice energy of \(\mathrm{SrCl}_{2}(s)\) via a Born-Haber cycle. (b) The second ionization energy of \(\operatorname{Sr}(g)\) is \(1064 \mathrm{~kJ} / \mathrm{mol}\). Use this fact along with data in Appendix \(\mathrm{C}\), Figure 7.9 , Figure \(7.11,\) and Table 8.2 to calculate \(\Delta H_{f}^{\circ}\) for \(\operatorname{Sr} \mathrm{Cl}_{2}(s)\)

Short answer

The enthalpy of formation, ΔHf°, for SrCl2(s) can be calculated using the Born-Haber cycle and the provided data. The necessary equations for the Born-Haber cycle involve sublimation, ionization, electron affinity, and lattice energy. Using the given data values, the enthalpy of formation, ΔHf°, for SrCl2(s) is calculated to be approximately -540.85 kJ/mol.

Step-by-step solution

Write the equations for the Born-Haber cycle for SrCl2(s)

The Born-Haber cycle links the enthalpy of formation of an ionic solid to its lattice energy and the other chemical properties, such as ionization energy and electron affinity. The following equations represent the steps in the Born-Haber cycle for SrCl2(s): 1. Formation of gaseous atoms from the solid elements (sublimation/atomization): \( Sr(s) \rightarrow Sr(g) \) \( Cl_{2}(g) \rightarrow 2Cl(g) \) 2. Ionization of strontium: \( Sr(g) \rightarrow Sr^{+}(g) + e^{-} \) \( Sr^{+}(g) + e^{-} \rightarrow Sr^{2+}(g) \) (second ionization) 3. Ionization of chlorine atoms: \( Cl(g) \rightarrow Cl^{+}(g) + e^{-} \) \( Cl(g) \rightarrow Cl^{-}(g) + e^{-} \) 4. Formation of the lattice from gaseous ions (lattice energy): \( Sr^{2+}(g) + 2Cl^{-}(g) \rightarrow SrCl_{2}(s) \)

Obtain data values from given sources

Now, we need to gather the data for enthalpy changes from the given sources: 1. First ionization energy of Sr(g) (from Appendix C): 549.5 kJ/mol 2. Second ionization energy of Sr(g) (given): 1064 kJ/mol 3. Bond dissociation energy for Cl2(g) (from Appendix C): 242.3 kJ/mol 4. Electron affinity of Cl (from Appendix C): -349 kJ/mol 5. Lattice energy of SrCl2(s) (from Table 8.2): -2127 kJ/mol

Calculate enthalpy of formation, ΔHf°

Now that we have all the necessary enthalpy values from the Born-Haber cycle, we can calculate the enthalpy of formation, ΔHf°, for SrCl2(s). The overall enthalpy of formation is the sum of the enthalpy changes for all the steps in the cycle: ΔHf°(SrCl2) = Sublimation energy (Sr) + Atomization energy (Cl2) + Ionization energy (Sr) + Electron affinity (Cl) + Lattice energy (SrCl2) So: ΔHf°(SrCl2) = 1 * (Sublimation energy of Sr) + 0.5 * (Bond dissociation energy of Cl2) + (First ionization energy of Sr + Second ionization energy of Sr) + 2 * (Electron affinity of Cl) + (Lattice energy of SrCl2) Plugging in the values, we get: ΔHf°(SrCl2) = 1 * (549.5 kJ/mol) + 0.5 * (242.3 kJ/mol) + (549.5 kJ/mol + 1064 kJ/mol) + 2 * (-349 kJ/mol) + (-2127 kJ/mol) ΔHf°(SrCl2) = 549.5 kJ/mol + 121.15 kJ/mol + 1613.5 kJ/mol - 698 kJ/mol - 2127 kJ/mol ΔHf°(SrCl2) = -540.85 kJ/mol So, the enthalpy of formation, ΔHf°, of SrCl2(s) is approximately -540.85 kJ/mol.

Key concepts

Born-Haber Cycle Explained

The Born-Haber cycle is a powerful tool for understanding how the energy of an Ionic compound is influenced by its constituent parts. In essence, this cycle breaks down the formation of an ionic solid into a series of steps where each step has an associated energy change.

It begins with the atomization of the elements, transforming them from their solid state into gas atoms. Proceeding further, the atoms are ionized, meaning electrons are removed from or added to the atoms, forming ions. This involves endothermic reactions, such as ionization of metals, and exothermic processes, such as the electron affinity of non-metals. Finally, the gaseous ions combine to form a solid ionic compound, releasing the lattice energy.

Visualizing Energy Changes

To simplify, imagine each step on a ladder where you must first climb up (absorb energy) to ionize the atoms, then descend down (release energy) as ions come together to form the solid lattice. The total energy required to climb up represents the energy input, while descending the ladder represents energy released. The Born-Haber cycle is essentially the bookkeeping of these energy changes.

Enthalpy of Formation Decoded

The enthalpy of formation, or \( \Delta H_{f}^{\circ} \), is essentially the heat change that occurs when one mole of a compound is assembled from its elements in their standard states. It provides insights into the stability of the compound relative to its constituent elements.

For ionic compounds like \( SrCl_2(s) \), the enthalpy of formation can be calculated by adding up all the individual steps in the Born-Haber cycle. This involves incorporating the energy for sublimation, bond dissociation, ionization, and electron affinities. The negative value of \( \Delta H_{f}^{\circ} \) suggests that the formation of the compound from its elements is exothermic, and energy is released during the process.

Real-World Analogy

Think of building a house. The enthalpy of formation would be like the net cost (or benefit) of gathering materials and labor to construct the house from scratch. If the total cost is negative, it means you somehow saved energy (or in this analogy, money) by building the house rather than leaving the materials separate.

Ionization Energy Unpacked

Ionization energy is quantified as the minimum amount of energy required to remove an electron from an atom or ion in its gaseous state. This energy has to overcome the attraction between the negatively charged electron and the positively charged nucleus.

Higher ionization energies indicate a stronger hold of the atom on its electrons. Taking strontium as an example, its second ionization energy is always higher than the first because it's harder to remove an electron from a positively charged ion than from a neutral atom.

Ionization in Everyday Life

Ionization energy can be likened to taking a book off a shelf. The first book (electron) is easy to take, but as you remove more books, the shelf (atom) becomes more resistant to losing any more. This is because with fewer books, there's less weight pushing down, making the remaining ones more stable and difficult to remove.