Use the data provided below to calculate the lattice energy of RbCl.

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 break down the formation of RbCl from its elements using Hess’s Law, which states that the total enthalpy change for a process is the sum of the enthalpy changes for the steps into which the process can be divided.

Step-by-Step Breakdown:

The formation of RbCl from its elements can be written as:

Rb (s) + 1/2 Cl2 (g) → RbCl (s)

This process can be broken into the following steps:

1. Sublimation of Rb (s) → Rb (g):
   The energy required to convert solid Rb into gaseous Rb is called the sublimation energy.
   ΔH° = +86.5 kJ/mol
2. Ionization of Rb (g) → Rb+ (g) + e-:
   The first ionization energy is the energy needed to remove an electron from a gaseous Rb atom to form Rb+.
   ΔH° = +403 kJ/mol
3. Dissociation of Cl2 (g) → 2 Cl (g):
   The bond energy of Cl2 is the energy required to break the Cl-Cl bond and form two chlorine atoms in the gas phase.
   ΔH° = +242 kJ/mol
4. Electron affinity of Cl (g) + e- → Cl- (g):
   The electron affinity of chlorine is the energy released when a chlorine atom gains an electron to form Cl-.
   ΔH° = -349 kJ/mol
5. Formation of RbCl from Rb+ and Cl- (g):
   The lattice energy is the energy released when a solid ionic lattice forms from gaseous ions. This is the quantity we want to calculate.
   ΔH° = Lattice Energy

Overall Enthalpy Change:

The enthalpy change for the entire process of forming RbCl from its elements is given as:
ΔH° = -430.5 kJ/mol

Now, applying Hess’s Law, we can sum the enthalpies of all the steps:ΔHformation=ΔHsublimation+ΔHionization+ΔHbond energy+ΔHelectron affinity+ΔHlattice\Delta H_{\text{formation}} = \Delta H_{\text{sublimation}} + \Delta H_{\text{ionization}} + \Delta H_{\text{bond energy}} + \Delta H_{\text{electron affinity}} + \Delta H_{\text{lattice}}ΔHformation​=ΔHsublimation​+ΔHionization​+ΔHbond energy​+ΔHelectron affinity​+ΔHlattice​−430.5=86.5+403+242−349+ΔHlattice-430.5 = 86.5 + 403 + 242 – 349 + \Delta H_{\text{lattice}}−430.5=86.5+403+242−349+ΔHlattice​

Simplifying the right side:−430.5=86.5+403+242−349+ΔHlattice-430.5 = 86.5 + 403 + 242 – 349 + \Delta H_{\text{lattice}}−430.5=86.5+403+242−349+ΔHlattice​−430.5=382.5+ΔHlattice-430.5 = 382.5 + \Delta H_{\text{lattice}}−430.5=382.5+ΔHlattice​ΔHlattice=−430.5−382.5=−692 kJ/mol\Delta H_{\text{lattice}} = -430.5 – 382.5 = -692 \text{ kJ/mol}ΔHlattice​=−430.5−382.5=−692 kJ/mol

Comparison with NaCl:

Lattice energy is influenced by two main factors: the charge of the ions and the size of the ions. The lattice energy increases with the charge of the ions and decreases with the size of the ions. Since Rb+ is larger than Na+ (because Rb is lower in the periodic table), the lattice energy of RbCl will be less than that of NaCl. Larger ions result in weaker electrostatic attraction between them, which means a smaller lattice energy.

Thus, the lattice energy of RbCl is -692 kJ/mol, and it is less than the lattice energy of NaCl, which is typically around -786 kJ/mol. This difference is due to the larger ionic radius of Rb+ compared to Na+, leading to a weaker ionic bond in RbCl.