Question

At 50C50 ^ { \circ } \mathrm { C } the ion-product constant for H2OH_2O has the value Kw=5.48×1014K _ { w } = 5.48 \times 10 ^ { - 14 } Based on the change in KwK_w with temperature, predict whether ΔH\Delta H is positive, negative, or zero for the autoionization reaction of water: 2H2O(l)H3O+(aq)+OH(aq)2 \mathrm { H } _ { 2 } \mathrm { O } ( l ) \rightleftharpoons \mathrm { H } _ { 3 } \mathrm { O } ^ { + } ( a q ) + \mathrm { OH } ^ { - } ( a q )

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

dln(Kw)dT=ΔHRT2\dfrac{\mathrm{d} \ln \left(\mathrm{K}_{w}\right)}{\mathrm{dT}}=\dfrac{\Delta \mathrm{H}}{\mathrm{RT}^{2}}

\star Kw\mathrm{K}_{w} increases from 1.01011.0 \cdot 10^{-1} to 5.481015.48 \cdot 10^{-1} so the nominator term dln(Kw)d \ln \left(K_{w}\right) is positive

\star The temperature increases from 25C25^{\circ} \mathrm{C} to 50C50^{\circ} \mathrm{C} so the denominator term dT\mathrm{dT} is positive

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

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