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Calculate the lattice energy of NaF given the following information: the enthalpy of formation of NaF is -572.6 kJ/mol, the ionization energy of Na is 495.8 kJ/mol, the electron affinity of F is -328 kJ/mol, and the Madelung constant for NaF is 1.747.

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To calculate the lattice energy of NaF, we can use the Born-Haber cycle, which is an application of Hess's Law. The lattice energy  U  can be calculated using the following equation:U = Hf - IE Na  - EA F  - Hsub Na  - Hdiss F2 where:Hf = enthalpy of formation of NaF = -572.6 kJ/molIE Na  = ionization energy of Na = 495.8 kJ/molEA F  = electron affinity of F = -328 kJ/molHsub Na  = enthalpy of sublimation of NaHdiss F2  = enthalpy of dissociation of F2We are not given the values for Hsub Na  and Hdiss F2 , so we cannot directly calculate the lattice energy using this equation. However, we can use the Madelung constant and the Born-Lande equation to find the lattice energy:U = -A *  1 - 1/n  *  e^2 / 40 * r0 where:A = Madelung constant = 1.747n = Born exponent  typically between 5 and 12, we will use 9 as an approximation e = elementary charge = 1.602 x 10^-19 C0 = vacuum permittivity = 8.854 x 10^-12 C/ Nm r0 = distance between ions in the latticeTo find r0, we can use the ionic radii of Na+ and F-. The ionic radius of Na+ is 102 pm, and the ionic radius of F- is 133 pm. Therefore, r0 =  102 + 133  x 10^-12 m = 235 x 10^-12 m.Now we can plug these values into the Born-Lande equation:U = -1.747 *  1 - 1/9  *  1.602 x 10^-19 C  /  4 * 8.854 x 10^-12 C/ Nm  * 235 x 10^-12 m U = -1.747 *  8/9  *  2.566 x 10^-38 C  /  1.112 x 10^-10 Nm U = -1.556 * 2.566 x 10^-38 C / 1.112 x 10^-10 NmU = -4.000 x 10^-28 C / 1.112 x 10^-10 NmU = -3.598 x 10^-18 NmSince 1 Nm = 1 J, we can convert this to kJ/mol:U = -3.598 x 10^-18 J *  1 kJ / 1000 J  *  6.022 x 10^23 mol U = -2167 kJ/molSo, the lattice energy of NaF is approximately -2167 kJ/mol.

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