Question

1. Calculate the lattice energy of RbCl.

2. Calculate the lattice energy for CaF2.

Answer 1

1. the lattice energy of RbCl. the reaction is

2 Rb (s) + Cl₂ (g) ----> 2 RbCl (s) enthalpy of formation of rbcl=-430.5 kj/mole

as above reacton is for 2 mole multiply enthalpy of formation of rbcl by 2 = -861 * 10³ joules =E₁

cl2 (g)   -> 2cl bond dissociation energy of cl₂= 243 kj/mole = E₂

2 rb (s) -> 2 rb (g) sublimation energy 0f rb = 86 kj/mole for 2 mole =172 kj/mole = E₃

2 rb (g) -> 2 rb⁺ + 2e^- ionisation energy = 402 kj/mole   for 2 mole= 804 kj/mole = E₄

2cl (g) + 2e^-   -> 2cl^- electron affinity of chlorine is = -349   kj/mole    for 2 mole= -698    kj/mole  = E₅

2 rb⁺ (g) + 2cl^- (g) -> 2 rbcl (s) enthalpy of formation of lattice

= 2 * - (lattice energy for 1 mole ) = E

enthalpy of formatoin of 2 mole of rbcl = E₂ + E₃ + E₄ +  E₅ + E

-861 kj/mole = 243 kj/mole + 172 kj/mole + 804 kj/mole +   -698 kj/mole +  E

E = -1382

the lattice energy comes out to be = -1382 for 2 mole -> for 1 mole ->  -1382 /2 =-691 kj/mole

2. calculate the lattice energy for CaF₂

Ca(s) + F₂(g) → Ca(g) + F₂(g)
Enthalpy of this step is the sublimation enthalpy of calcium
E₁ = 121kJ/mol

Ca²⁺(g) + F₂(g) + 2 e⁻ → Ca²⁺(g) + F(g) + 2 e⁻
enthalpy of this step is equal to the bond energy of the fluorine molecule:
E₂ = 157kJ/mole

Ca(g) + F₂(g) → Ca²⁺(g) + F₂(g) + 2 e⁻
enthalpy of this step is sum of first and second ionization energy
E₃ = 590kJ/mol + 1145kJ/mol = 1735kJ/mol




Ca²⁺(g) + 2 F(g) + 2 e⁻ → Ca²⁺(g) + 2 F⁻(g)
enthalpy of this step is equal to the electron affinity of fluorine:
E₄ = -328 kJ/mol


Ca²⁺(g) + 2 F⁻(g) → CaF₂(s)
enthalpy of this step is equal to lattice energy:

= E

Overall reaction :
Ca(s) + F₂(g) → CaF₂(s) for 1 mole of reactant and product

enthalpy of formatoin of 1 mole of  CaF₂ (s)
= E₁ + E₂ + E₃ + E₄ + E
= -1120kJ/mole - 121kJ/mole - 1735kJ/mole - 157kJ/mole - (-328kJ/mole)
= -2805kJ/mole