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Calculate the Lattice Enthalpy of Potassium Chloride Given the Following

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Calculate the lattice enthalpy of potassium chloride given the following enthalpy data.
K(s)→K(g)+89 kJ⋅mol−1 K( g)→K+(g)+e−+418 kJ⋅mol−11/2Cl2( g)→Cl(g)+122 kJ⋅mol−1Cl(g)e−→Cl−(g)−349 kJ⋅mol−1 K( s)+1/2Cl2( g)→KCl(s)−437 kJ⋅mol−1\begin{array} { l l } \mathrm { K } ( \mathrm { s } ) \rightarrow \mathrm { K } ( \mathrm { g } ) & + 89 \mathrm {~kJ} \cdot \mathrm { mol } ^ { - 1 } \\\mathrm {~K} ( \mathrm {~g} ) \rightarrow \mathrm { K } ^ { + } ( \mathrm { g } ) + \mathrm { e } ^ { - } & + 418 \mathrm {~kJ} \cdot \mathrm { mol } ^ { - 1 } \\1 / 2 \mathrm { Cl } _ { 2 } ( \mathrm {~g} ) \rightarrow \mathrm { Cl } ( \mathrm { g } ) & + 122 \mathrm {~kJ} \cdot \mathrm { mol } ^ { - 1 } \\\mathrm { Cl } ( \mathrm { g } ) \mathrm { e } ^ { - } \rightarrow \mathrm { Cl } ^ { - } ( \mathrm { g } ) & - 349 \mathrm {~kJ} \cdot \mathrm { mol } ^ { - 1 } \\\mathrm {~K} ( \mathrm {~s} ) + 1 / 2 \mathrm { Cl } _ { 2 } ( \mathrm {~g} ) \rightarrow \mathrm { KCl } ( \mathrm { s } ) & - 437 \mathrm {~kJ} \cdot \mathrm { mol } ^ { - 1 }\end{array}

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