The lattice energy of solid NaCl is 180kcal per mol. The dissolution of the solid in water in the form of ions is endothermic to the extent of 1kcal per mol. If the salvation energies of ${ Na }^{ + } and { Cl }^{ - }$ ions are in the ratio 6:5, what is the enthalpy of hydration of sodium ions?
a.) -85.6kcal/mol
b.) -97.5kcal/mol
c.) 82.6kcal/mol
d.) +100kcal/mol
Answer
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Hint: The energy of solvation is the amount of energy associated with dissolving a solute in a solvent. If it is a positive number, the dissolving process is endothermic; if it is negative, it's exothermic.
Complete step by step solution:
Given that
${ \triangle H }_{ dissolution }=1kcal{ mol }^{ -1 }$
${ \triangle H }_{ lattice }=180kcal\quad { mol }^{ -1 }$
Let ${ \triangle H }_{ { Na }^{ + } } and { \triangle H }_{ { Cl }^{ - } }$ be the salvation energy of sodium and chloride ion respectively.
${ \triangle H }_{ hydration } = ? = { \triangle H }_{ { Na }^{ + } } + { \triangle H }_{ { Cl }^{ - } }$
Given that :
$\dfrac { \triangle { H }_{ { Na }^{ + } } }{ \triangle { H }_{ { Cl }^{ - } } } =\dfrac { 6 }{ 5 }$
${ \triangle H }_{ { Cl }^{ \_ } }=\dfrac { 5 }{ 6 } \triangle { H }_{ { Na }^{ + } }$
Therefore, ${ \triangle H }_{ hydration }={ \triangle H }_{ { Na }^{ + } }+\dfrac { 5 }{ 6 } { H }_{ { Na }^{ + } }$
$\Longrightarrow { \triangle H }_{ hydration }=\dfrac { 11 }{ 6 } { H }_{ { Na }^{ + } }$
As we know that,
${ \triangle H }_{ hydration }={ \triangle H }_{ lattice }-{ \triangle H }_{ dissolution }
\dfrac { 11 }{ 6 } { \triangle H }_{ { Na }^{ + } }=180-1$
$\Longrightarrow { \triangle H }_{ { Na }^{ + } }\dfrac { 6 }{ 11 } \times 179=97.63KJ{ mol }^{ - }\approx 97.5Kcal{ mol }^{ -1 }$
The sign will be negative as this energy is released during the process.
The enthalpy of hydration of sodium ions is $-97.5Kcal{ mol }^{ -1 }$. Therefore, option B is the required answer!
Note: The energy of solvation is the amount of energy associated with dissolving a solute in a solvent. If it is a positive number, the dissolving process is endothermic; if it is negative, it's exothermic.
Complete step by step solution:
Given that
${ \triangle H }_{ dissolution }=1kcal{ mol }^{ -1 }$
${ \triangle H }_{ lattice }=180kcal\quad { mol }^{ -1 }$
Let ${ \triangle H }_{ { Na }^{ + } } and { \triangle H }_{ { Cl }^{ - } }$ be the salvation energy of sodium and chloride ion respectively.
${ \triangle H }_{ hydration } = ? = { \triangle H }_{ { Na }^{ + } } + { \triangle H }_{ { Cl }^{ - } }$
Given that :
$\dfrac { \triangle { H }_{ { Na }^{ + } } }{ \triangle { H }_{ { Cl }^{ - } } } =\dfrac { 6 }{ 5 }$
${ \triangle H }_{ { Cl }^{ \_ } }=\dfrac { 5 }{ 6 } \triangle { H }_{ { Na }^{ + } }$
Therefore, ${ \triangle H }_{ hydration }={ \triangle H }_{ { Na }^{ + } }+\dfrac { 5 }{ 6 } { H }_{ { Na }^{ + } }$
$\Longrightarrow { \triangle H }_{ hydration }=\dfrac { 11 }{ 6 } { H }_{ { Na }^{ + } }$
As we know that,
${ \triangle H }_{ hydration }={ \triangle H }_{ lattice }-{ \triangle H }_{ dissolution }
\dfrac { 11 }{ 6 } { \triangle H }_{ { Na }^{ + } }=180-1$
$\Longrightarrow { \triangle H }_{ { Na }^{ + } }\dfrac { 6 }{ 11 } \times 179=97.63KJ{ mol }^{ - }\approx 97.5Kcal{ mol }^{ -1 }$
The sign will be negative as this energy is released during the process.
The enthalpy of hydration of sodium ions is $-97.5Kcal{ mol }^{ -1 }$. Therefore, option B is the required answer!
Note: The energy of solvation is the amount of energy associated with dissolving a solute in a solvent. If it is a positive number, the dissolving process is endothermic; if it is negative, it's exothermic.
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