Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than the lattice energy of NaCl? Explain. Electron affinity of Cl = -349 kJ/mol 1st ionization energy of Rb = 403 kJ/mol Bond energy of Cl2 = 242 kJ/mol Sublimation energy of Rb = 86.5 kJ/mol ΔH°[RbCl (s)] = -430.5 kJ/mol Rb (s) + 1/2 Cl2 (g) → RbCl (s) This equation can be written as the sum of the following: Rb (s) → Rb (g) sublimation ΔH° = +86.5 kJ/mol Rb (g) → Rb+ + e- 1st ionization energy ΔH° = +403 kJ/mol Cl2 (g) → 2 Cl bond energy of Cl2 ΔH° = +242 kJ/mol Cl + e- → Cl- electron affinity of Cl ΔH° = -349 kJ/mol Rb+ + Cl- → RbCl lattice energy ΔH° = -222 kJ/mol ΔH° = ΔH° + ΔH° + ΔH° + ΔH° + ΔH° Lattice energy ΔH° = ΔH° – (ΔH° + ΔH° + ΔH° + ΔH°) = -430.5 – (86.5 + 403 + 242 – 349) = -692 kJ
The Correct Answer and Explanation is:
To calculate the lattice energy of RbCl, we can use the provided data and apply the Born-Haber cycle, which is a series of steps that break down the formation of an ionic compound from its elements in their standard states.
Step-by-Step Calculation:
- Sublimation of Rb(s): This step involves converting solid rubidium (Rb) to gaseous rubidium atoms.
- Sublimation energy of Rb: +86.5 kJ/mol
- Ionization of Rb(g): The next step is the ionization of gaseous rubidium, which means it loses an electron to form Rb⁺.
- First ionization energy of Rb: +403 kJ/mol
- Dissociation of Cl₂(g): Chlorine (Cl₂) is broken into two chlorine atoms (Cl) in the gas phase.
- Bond energy of Cl₂: +242 kJ/mol (for one mole of Cl₂ molecules, we get two moles of chlorine atoms)
- Electron affinity of Cl: Chlorine atoms gain an electron to form chloride ions (Cl⁻).
- Electron affinity of Cl: -349 kJ/mol
- Formation of RbCl(s): Finally, the Rb⁺ and Cl⁻ ions come together to form solid RbCl. This is the lattice energy, which we need to calculate.
The enthalpy of formation (ΔH°) of RbCl (s) is given as:
- ΔH°[RbCl (s)] = -430.5 kJ/mol
Applying the Born-Haber Cycle:
The enthalpy change for the formation of RbCl(s) is the sum of the steps: ΔH=Sublimation energy of Rb+Ionization energy of Rb+12Bond energy of Cl₂+Electron affinity of Cl+Lattice energy\Delta H = \text{Sublimation energy of Rb} + \text{Ionization energy of Rb} + \frac{1}{2} \text{Bond energy of Cl₂} + \text{Electron affinity of Cl} + \text{Lattice energy}ΔH=Sublimation energy of Rb+Ionization energy of Rb+21Bond energy of Cl₂+Electron affinity of Cl+Lattice energy
Substitute the known values: −430.5=86.5+403+12(242)+(−349)+Lattice energy-430.5 = 86.5 + 403 + \frac{1}{2}(242) + (-349) + \text{Lattice energy}−430.5=86.5+403+21(242)+(−349)+Lattice energy −430.5=86.5+403+121−349+Lattice energy-430.5 = 86.5 + 403 + 121 – 349 + \text{Lattice energy}−430.5=86.5+403+121−349+Lattice energy −430.5=261.5+Lattice energy-430.5 = 261.5 + \text{Lattice energy}−430.5=261.5+Lattice energy
Solving for lattice energy: Lattice energy=−430.5−261.5=−692kJ/mol\text{Lattice energy} = -430.5 – 261.5 = -692 kJ/molLattice energy=−430.5−261.5=−692kJ/mol
Conclusion:
The lattice energy of RbCl is -692 kJ/mol.
Comparison with NaCl:
The lattice energy of NaCl is greater than that of RbCl. This is because NaCl forms from smaller Na⁺ and Cl⁻ ions compared to RbCl, leading to stronger electrostatic attraction between the ions in NaCl. Smaller ions allow the ions to be packed more closely, increasing the lattice energy. In contrast, Rb⁺ is larger than Na⁺, which results in a lower lattice energy for RbCl. Therefore, the lattice energy of RbCl is less than the lattice energy of NaCl.
