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Introduction to Le Chatelier's Principle

The enthalpy ...

Question

The enthalpy of dissolution(or solution) of sodium chloride is $4 k J m o l^{- 1}$ and its enthalpy of hydration of ions is $- 784 k J m o l^{- 1}$ . What will be the lattice enthalpy of sodium chloride?

+ 780 k J m o l^{- 1}

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+ 394 k J m o l^{- 1}

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+ 788 k J m o l^{- 1}

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+ 398 k J m o l^{- 1}

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Solution

The correct option is B $+ 788 k J m o l^{- 1}$
$Δ H_{s o l} = Δ H_{l a t t i c e} + Δ H_{h y d}$

$4 = Δ H_{l a t t i c e} + (- 784)$

$Δ H_{l a t t i c e} = + 788 k J m o l^{- 1}$

Hence, the correct answer is option $C$ .

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Similar questions

N a C l (s)

Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}

Δ_{s o l} H = + 4 k J m o l^{- 1}

N a C l

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

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

{N a}^{+}

{C l}^{-}

3 : 2.5

K C l

K C l

2 : 1

= - 2

= + 2

N a^{+} (g) = - 390

= - 772

= - 699

N a C l (s)

- 790 k J . {m o l}^{- 1}

- 785 k J . {m o l}^{- 1}

N a C l (s)

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