Answer:
- 775.5 kJ/mol
Explanation:
Given that:
The enthalpy of formation of MX is [tex]\Delta_f^0[/tex] = - 421 kJ/mol
The enthalpy of sublimation of MM is [tex]\Delta H_{sub}[/tex] = 155 kJ/mol
The ionization energy of MM is = 401 kJ/mol
The electron affinity of XX is [tex]\Delta H_{EA}[/tex] = -321 kJ/mol
The bond energy of X2 is BE = 239 J/mol
The process for each above component is shown below:
[tex]M_{s} \to M_{(g)} ---> ( \Delta H_{sub} = 155 \ kJ/mol})[/tex]
[tex]\dfrac{1}{2}X_{2(g)} \to X_{(g)} ---> ( BE = \dfrac{239}{2} kJ/mol ) = (119.5 kJ/mol )[/tex]
[tex]M_{(g)} \to M^+_{(g)} + e^- ---> ( IE_1 = 401 \ kJ/mol )[/tex]
[tex]X_{(g)} + e^- --> X^-_{(g)} ---> ( \Delta H_{EA }= -321 \ kJ/mol )[/tex]
[tex]M^+_{(g)} +\dfrac{1}{2}X^-_{(g)} \to MX ---> ( \Delta H^0_f = -421 \ kJ/mol )[/tex]
Thus, the overall reaction is:
[tex]M_{(s)}+ \dfrac{1}{2}X_{2(g)} \to MX_{(s)}[/tex]
The overall energy is:
[tex]\mathtt{\Delta H^0_f = U + \Delta H_{sub} + IE + BE + \Delta H_{EA }}[/tex]
- 421 kJ/mol = U + 155 kJ/mol + 401 kJ/mol + 119.5 kJ/mol + (-321 kJ/mol)
- 421 kJ/mol = U + 354.5 kJ/mol
- U = 421 kJ/mol + 354.5 kJ/mol
- U = 421 kJ/mol + 354.5 kJ/mol
- U = 775.5 kJ/mol
U = - 775.5 kJ/mol
Thus, the lattice energy of MX (U) = - 775.5 kJ/mol
