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Hess' Law

The Δ H f of ...

Question

The $Δ H_{f}$ of $K F (s)$ is $- 563 k J / m o l$ . The ionisation energy of potassium $K (g)$ is $419 k J / m o l$ and enthalpy of sublimation of potassium is $88 k J / m o l$ . The electron affinity of $F (g)$ is $- 322 k J / m o l$ and $F - F$ bond enthalpy is $158 k J / m o l$ . Calculate the absolute value of lattice energy (in $k J / m o l$ ) of $K F (s)$

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Solution

Born-Haber cycle for the formation of $K F$ is:

Now, using Hess's law :
$Δ H_{f} = Δ H_{1} + Δ H_{2} + Δ H_{3} + Δ H_{4} + Δ H_{5}$
$- 563 = 88 + 419 + 79 - 322 + Δ H_{5}$
$Δ H_{5} = - 827 k J / m o l$

$∴$ Lattice energy of $K F (s)$ is $- 827 k J / m o l$

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O^{-}

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