Regarding the problem above, the lattice energy of rubidium chloride is the energy change accompanying the process: RbCl(s) → Rb(g) + Cl(g) ΔHlattice Calculate the lattice energy of RbCl using the following data: Equation MH? (kJ) Eq A Rb(g) → Rb(s) 83 Eq B Rb(g) → Rb(g) 411 Eq C Cl(g) → 2Cl(g) 244 Eq D Cl(g) → Cl(g) -357 Eq E Rb(s) + Cl(g) → RbCl(s) 4433 Report the Lattice Energy as kJ/mo
The Correct Answer and Explanation is:
To calculate the lattice energy of rubidium chloride (RbCl), we can use the Born-Haber cycle, which applies Hess’s law. The lattice energy is the energy required to separate one mole of an ionic solid (RbCl) into its constituent ions in the gas phase.
The lattice energy of RbCl is given by the energy change in the following reaction: RbCl(s)→Rb(g)+Cl(g)\text{RbCl(s)} \rightarrow \text{Rb(g)} + \text{Cl(g)}RbCl(s)→Rb(g)+Cl(g)
Using the provided data, let’s break down the process step by step and apply Hess’s law:
Step 1: Sublimation of Rb(s) to Rb(g)
This is the energy required to convert solid rubidium into gas. From Eq A: Rb(s)→Rb(g)ΔH=83 kJ/mol\text{Rb(s)} \rightarrow \text{Rb(g)} \quad \Delta H = 83 \, \text{kJ/mol}Rb(s)→Rb(g)ΔH=83kJ/mol
Step 2: Ionization of Rb(g) to Rb+(g)
This is the energy required to remove an electron from rubidium in the gas phase. From Eq B: Rb(g)→Rb+(g)+e−ΔH=411 kJ/mol\text{Rb(g)} \rightarrow \text{Rb}^+(g) + e^- \quad \Delta H = 411 \, \text{kJ/mol}Rb(g)→Rb+(g)+e−ΔH=411kJ/mol
Step 3: Dissociation of Cl2(g) to 2Cl(g)
This is the energy required to break the Cl2 molecule into two chlorine atoms in the gas phase. From Eq C: 12Cl2(g)→Cl(g)ΔH=244 kJ/mol\frac{1}{2} \text{Cl}_2(g) \rightarrow \text{Cl(g)} \quad \Delta H = 244 \, \text{kJ/mol}21Cl2(g)→Cl(g)ΔH=244kJ/mol
Step 4: Electron affinity of Cl(g) to Cl-(g)
This is the energy released when chlorine gains an electron to form a chloride ion in the gas phase. From Eq D: Cl(g)+e−→Cl−(g)ΔH=−357 kJ/mol\text{Cl(g)} + e^- \rightarrow \text{Cl}^-(g) \quad \Delta H = -357 \, \text{kJ/mol}Cl(g)+e−→Cl−(g)ΔH=−357kJ/mol
Step 5: Formation of RbCl(s) from Rb+(g) and Cl-(g)
This is the energy released when the ions combine to form solid rubidium chloride. From Eq E: Rb+(g)+Cl−(g)→RbCl(s)ΔH=−4433 kJ/mol\text{Rb}^+(g) + \text{Cl}^-(g) \rightarrow \text{RbCl(s)} \quad \Delta H = -4433 \, \text{kJ/mol}Rb+(g)+Cl−(g)→RbCl(s)ΔH=−4433kJ/mol
Step 6: Applying Hess’s Law
Now, using Hess’s law, we can calculate the lattice energy. The lattice energy is the energy change for the formation of RbCl(s) from its ions in the gas phase. We sum all the enthalpy changes, but we must rearrange some of the equations to ensure that the overall reaction matches the formation of RbCl(s) from its gaseous ions. ΔHlattice=[(Eq A)+(Eq B)+12(Eq C)+(Eq D)+(Eq E)]\Delta H_{\text{lattice}} = \left[\text{(Eq A)} + \text{(Eq B)} + \frac{1}{2} \text{(Eq C)} + \text{(Eq D)} + \text{(Eq E)}\right]ΔHlattice=[(Eq A)+(Eq B)+21(Eq C)+(Eq D)+(Eq E)]
Substituting the values: ΔHlattice=83 kJ/mol+411 kJ/mol+244 kJ/mol+(−357 kJ/mol)+(−4433 kJ/mol)\Delta H_{\text{lattice}} = 83 \, \text{kJ/mol} + 411 \, \text{kJ/mol} + 244 \, \text{kJ/mol} + (-357 \, \text{kJ/mol}) + (-4433 \, \text{kJ/mol})ΔHlattice=83kJ/mol+411kJ/mol+244kJ/mol+(−357kJ/mol)+(−4433kJ/mol) ΔHlattice=83+411+244−357−4433=−4052 kJ/mol\Delta H_{\text{lattice}} = 83 + 411 + 244 – 357 – 4433 = -4052 \, \text{kJ/mol}ΔHlattice=83+411+244−357−4433=−4052kJ/mol
Final Answer:
The lattice energy of rubidium chloride (RbCl) is -4052 kJ/mol.
This value is negative because the formation of an ionic bond (the lattice energy) is an exothermic process, meaning energy is released when the ionic solid forms.
