Question

1. calculate the lattice energy for nacl(s) using a born - haber cycle and the following information:

nacl(s)→na⁺(g)+cl⁻(g) δh =?
na(s)+\\(\frac{1}{2}\\)cl₂(g)→nacl(s) δh = - 411.0 kj/mol
na(s)→na(g) δh = + 107.3 kj/mol
na(g)→na⁺(g)+e⁻ δh = + 495.8 kj/mol
\\(\frac{1}{2}\\)cl₂(g)→cl(g) δh = + 121.7 kj/mol
cl(g)+e⁻→cl⁻(g) δh = - 348.6 kj/mol
a. + 34.8 kj/mol
b. + 690.3 kj/mol
c. + 787.2 kj/mol
d. + 1512 kj/mol
e. - 698.7 kj/mol

Explanation:

Step1: Write the Born - Haber cycle equation

According to the Born - Haber cycle, \(\Delta H_{f}^{\circ}(NaCl(s))=\Delta H_{sublimation}(Na)+\Delta H_{ionization}(Na)+\Delta H_{dissociation}(\frac{1}{2}Cl_{2})+\Delta H_{electron - affinity}(Cl)+\Delta H_{lattice}\)

We need to solve for \(\Delta H_{lattice}\), so \(\Delta H_{lattice}=\Delta H_{sublimation}(Na)+\Delta H_{ionization}(Na)+\Delta H_{dissociation}(\frac{1}{2}Cl_{2})+\Delta H_{electron - affinity}(Cl)-\Delta H_{f}^{\circ}(NaCl(s))\)

Step2: Substitute the given values

Given:
\(\Delta H_{f}^{\circ}(NaCl(s))=- 411.0\space kJ/mol\)
\(\Delta H_{sublimation}(Na)=+107.3\space kJ/mol\)
\(\Delta H_{ionization}(Na)=+495.8\space kJ/mol\)
\(\Delta H_{dissociation}(\frac{1}{2}Cl_{2})=+121.7\space kJ/mol\)
\(\Delta H_{electron - affinity}(Cl)=-348.6\space kJ/mol\)

Substitute these values into the equation:

$$
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$$

Answer:

C. +787.2 kJ/mol