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Hess' Law

Calculate Δ...

Question

Calculate $Δ H_{l a t t i c e}$ of $S n O_{2}$ , If $Δ H_{f}$ of $S n O_{2}$ is -588 KJ / mol, Enthalpy of Sublimation (Sn) = 292 KJ / mol, Enthalpy of Dissociation $(O_{2}) = 454 K J / m o l$ . Total Electron gain enthalpy for O = 636 KJ / mol, Ionization enthalpy $(S n \to S n^{4 +})$ = 8990 kJ / mol.

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Solution

Total gain = Total loss
$Δ H_{I . E} + Δ H_{s s} + Δ H_{s u b} = | Δ H_{F} | + | Δ H_{e g} | + L . E$
$\Rightarrow 8990 + 454 + 292 = 588 + 636 + L . E .$
$L . E = 11596 K J$

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Δ H_{f}

F_{2}

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{kJ mol}^{- 1}

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F_{2} = 75.3 kJ

= 520 {kJ mol}^{- 1}

F (g) = - 333 kJ Δ_{f} H_{overall} = - 594 {kJ mol}^{- 1}

M g F_{2}

M g (s) \to M g (g) = 146.4 k J m o l^{- 1}

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M g (g) \to M g^{2 +} (g) = 2186 k J m o l^{- 1}

2 F \to 2 F^{- 1} = 2 \times - 322.6 = - 645.2 k J m o l^{- 1}

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- 233 k J / m o l

C s = + 78 k J / m o l

C s = + 375 k J / m o l

O_{2} (g) = 494 k J / m o l

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O = - 141 k J / m o l

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O^{-}

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

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C s = + 375 k J m o l^{- 1}

O_{2} (g) = 494 k J m o l^{- 1} o f O_{2} m o l e c u l e s

O = - 141 k J m o l^{- 1} o f O a t o m s

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