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Q

Given that  and  and  . The value of log10K for a reaction,  is:

A.5

B.10

C.95

D.100

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Change in Enthalpy of the given reaction$=\Delta H^{o}=-54.07\ kJ\ mol^{-1}$

Entropy change$=\Delta S^{o}=10\ J\ mol^{-1}$

The change in Gibb's Free Energy is given as

$\\\Delta G^{o}=\Delta H^{o}-T\Delta S^{o}\\ \Delta G^{o}=-54.07\times 10^{3}-298\times 10\\ \Delta G^{o}=-57070\ J\ mol^{-1}$

The relation between equilibrium constant K and Go is given as follows

$\\2.303logK=-\frac{\Delta G^{o}}{RT}\\ logK=\frac{57070}{2.303\times 8.314\times 298}\\ logK=10$

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