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  • Lattice enthalpy and enthalpy of solution of NaCl are  <img alt="788 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { and } 4 \mathrm{~kJ} \mathrm{~mol}^{-1} \text {, }" src="/latex-image/?788%20%5Cmathrm%7B%7EkJ%7D%20%5Cmathrm%7B%7Emol%7D%5E%7B-1%7D%20%

Lattice enthalpy and enthalpy of solution of NaCl are  788 \mathrm{~kJ} \mathrm{~mol}^{-1} \text { and } 4 \mathrm{~kJ} \mathrm{~mol}^{-1} \text {, }  respectively. The hydration enthalpy of NaCl is:
Option: 1 -780 KJmol^{-1}
Option: 2 780 KJmol^{-1}
Option: 3 784 KJmol^{-1}
Option: 4 -784 KJmol^{-1}
 

Answers (1)

best_answer

Lattice enthalpy = 788kJ/mol

Enthalpy of solution = 4kJ/mol

The reaction occurs as follows: \mathrm{NaCl(s)\: \overset{788kj/mol}{\longrightarrow}\: NaCl(g)\: \overset{x}{\longrightarrow}\: Na^{+}Cl^{-}(aq)}

Enthalpy of solution = enthalpy of hydration + lattice energy

Hydration enthalpy of NaCl

= -788 + 4

= -784 kJ/mol

Therefore, Option(4) is correct.

Posted by

Deependra Verma

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