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Electrochemical Cells, Galvanic Cells

The lattice e...

Question

The lattice enthalpy of solid $N a C l$ is $772 k J {m o l}^{- 1}$ and enthalpy of solution is $2 k J {m o l}^{- 1}$ . If the hydration enthalpy of ${N a}^{+}$ and ${C l}^{-}$ ions are in the ratio of $3 : 2.5$ , what is the enthalpy of hydration of chloride ion?

- 140 k J {m o l}^{- 1}

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- 350 k J {m o l}^{- 1}

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- 351.81 k J {m o l}^{- 1}

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None

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Solution

The correct option is A None
$Δ H_{l a t t i c e} = 772 k J m o l^{- 1} Δ H_{s o l u t i o n} = 2 k J m o l^{- 1} Δ H_{h y d r a t i o n} = - Δ H_{s o l u t i o n} + (- Δ H_{l a t t i c e}) = - 2 - 772 = - 774 k J m o l^{- 1}$
The hydration enthalpy of soldium and chloride ions in the ratio of 3:2.5,so
$3 x + 2.5 x = - 774 x = - 141$
The enthalpy of hydration of chloride ion is=-141*2.5=-352 kJmol-1

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