Chemistry, 16.01.2020 22:31 isabellecannuli

Consider an ionic compound, mx2 , composed of generic metal m and generic, gaseous halogen x .

the enthalpy of formation of mx2 is δ∘f=−985 kj/mol.

the enthalpy of sublimation of m is δsub=135 kj/mol.

the first and second ionization energies of m are ie1=731 kj/mol and ie2=1403 kj/mol.

the electron affinity of x is δea=−335 kj/mol. (refer to the hint).

the bond energy of x2 is be=207 kj/mol.

determine the lattice energy of mx2 .

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Answer from: angelica19carmona

$\boxed{\text{-2791 kJ/mol}}$

Explanation:

One way to calculate the lattice energy is to use Hess's Law .

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

M²⁺(g) + 2X⁻(g) ⟶ MgX₂(s); U = ?

We must generate this reaction rom the equations given.

(1) M(s) + X₂ (g) ⟶ MX₂(s); ΔHf⁰ = -985 kJ·mol⁻¹

(2) M(s) ⟶ M(g); ΔHsub = 135 kJ·mol⁻¹

(3) M(g) ⟶M⁺(g) +e⁻ IE₁ = 731 kJ·mol⁻¹

(4) M⁺(g) ⟶ M²⁺(g) + e⁻ IE₂ = 1403 kJ·mol⁻¹

(5) X(g) + e⁻ ⟶ X⁻(g) EA = -335 kJ·mol⁻¹

(6) X₂(g) ⟶ 2X(g) BE = 207 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy. Underlined species have been cancelled.

E/kJ

(7) M²⁺(g) + e⁻⟶ M⁺(g) -1403

(8) M⁺(g) + e⁻ ⟶ M(g) -731

(9) M(g) ⟶ M(s) -135

(10) M(s) + X₂(g) ⟶ MX₂(s) -985

(11) 2X(g) ⟶ X₂(g) -207

(12) 2X⁻(g) ⟶ 2X(g) + 2e⁻ +670

M²⁺(g) + 2X⁻(g) ⟶ MX₂(s) -2791

The lattice energy of MX₂ is $\boxed{\textbf{-2791 kJ/mol}}$ .

Answer from: Quest

The density would be 1.622

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.

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Consider an ionic compound, mx2 , composed of generic metal m and generic, gaseous halogen x .
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