Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Given that \delta H_{r298K}= -54.07kJ \:mol^{-1} and \delta S^{\circ}_{r298K}= 10J\:mol^{-1} and R= 8.314JK^{-1}\:mol^{-1} . The value of log10K for a reaction, A\leftrightarrow B is:

A.5

B.10

C.95

D.100

Answers (1)

best_answer

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

Posted by

Sayak

View full answer

Similar Questions

Latest Question