Question

# At $50 ^ { \circ } \mathrm { C }$ the ion-product constant for $H_2O$ has the value $K _ { w } = 5.48 \times 10 ^ { - 14 }$ Based on the change in $K_w$ with temperature, predict whether $\Delta H$ is positive, negative, or zero for the autoionization reaction of water: $2 \mathrm { H } _ { 2 } \mathrm { O } ( l ) \rightleftharpoons \mathrm { H } _ { 3 } \mathrm { O } ^ { + } ( a q ) + \mathrm { OH } ^ { - } ( a q )$

Solution

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The relation between equilibrium constant, temperature and change of enthalpy is:

$\dfrac{\mathrm{d} \ln \left(\mathrm{K}_{w}\right)}{\mathrm{dT}}=\dfrac{\Delta \mathrm{H}}{\mathrm{RT}^{2}}$

$\star$ $\mathrm{K}_{w}$ increases from $1.0 \cdot 10^{-1}$ to $5.48 \cdot 10^{-1}$ so the nominator term $d \ln \left(K_{w}\right)$ is positive

$\star$ The temperature increases from $25^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ so the denominator term $\mathrm{dT}$ is positive

Therefore we can conclude that $\Delta \mathrm{H}$ is positive because left had side term is positive

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