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Hess' Law

Calculate the...

Question

Calculate the electron gain enthalpy of fluorine from the data given below. $Δ H_{f}$ of KF is - 560.8 KJ/ mol, dissociation energy of $F_{2}$ is 158.9 KJ / mol. Lattice energy of KF is 807.5 KJ/mol, ionization energy of potassium is 414.2 KJ/mol and enthalpy of sublimation of $K = 87.8 K J / m o l .$

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Solution

Accordin to hess law
$Δ H_{F} = Δ H_{s u b} + \frac{1}{2} Δ H_{d i s s} + I E + E . A + Δ H_{l a t t i c e}$
$- 560.8 = 87.8 + \frac{158.9}{2} + 414.2 + E A - 807.5$
$- 560.8 = 581.45 + E A - 807.5$
$- 560.8 = - 226.05 + E A$
$E A = - 560.8 + 226.05$
$E A = - 334.75 K J$

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Δ H_{f}

K F (s)

- 563 k J / m o l

K (g)

419 k J / m o l

88 k J / m o l

F (g)

- 322 k J / m o l

F - F

158 k J / m o l

k J / m o l

K F (s)

Δ H_{l a t t i c e}

S n O_{2}

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S n O_{2}

(O_{2}) = 454 K J / m o l

(S n \to S n^{4 +})

= - 233 kJ/mol

C s = + 78 kJ/mol

C s = + 375 kJ/mol

O_{2} (g) = 494 kJ/mol

O = - 141 kJ/mol

O = + 845 kJ/mol

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

C s = + 78 k J m o l^{- 1}

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

O = + 845 k J m o l^{- 1} o f O^{1 -} i o n s

- 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

O_{2}

O = - 141 k J / m o l

O = + 845 k J / m o l

O^{-}

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