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If the E°cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of \Delta G^ \circ and Keq?

  • Option 1)

    \begin{array}{*{20}c} {\Delta G^ \circ > 0;} & {K_{\text{eq}} < 1} \\ \end{array}

  • Option 2)

    \begin{array}{*{20}c} {\Delta G^ \circ > 0;} & {K_{\text{eq}} > 1} \\ \end{array}

  • Option 3)

    \begin{array}{*{20}c} {\Delta G^ \circ < 0;} & {K_{\text{eq}} > 1} \\ \end{array}

  • Option 4)

    \begin{array}{*{20}c} {\Delta G^ \circ < 0;} & {K_{\text{eq}} < 1} \\ \end{array}

 

Answers (1)

 

Gibbs energy of the reaction -

\Delta_{r}G=-nFE_{cell}

- wherein

\Delta _{r}G = gibbs energy of the reaction

E_{cell}= emf of the cell

nF = amount of charge passed

 

 \Delta G^{\circ}=-nFE^{\circ}_{cell}

If E^{\circ}_{cell}=-ve then \Delta G^{\circ}=+ve

i.e. \Delta G^{\circ}>0

\Delta G^{\circ}=-nRT log Keq

For \Delta G^{\circ}=+ve , log Keq = -Ve

i.e. Keg < 1


Option 1)

\begin{array}{*{20}c} {\Delta G^ \circ > 0;} & {K_{\text{eq}} < 1} \\ \end{array}

this is the correct option

Option 2)

\begin{array}{*{20}c} {\Delta G^ \circ > 0;} & {K_{\text{eq}} > 1} \\ \end{array}

this is the incorrect option

Option 3)

\begin{array}{*{20}c} {\Delta G^ \circ < 0;} & {K_{\text{eq}} > 1} \\ \end{array}

this is the incorrect option

Option 4)

\begin{array}{*{20}c} {\Delta G^ \circ < 0;} & {K_{\text{eq}} < 1} \\ \end{array}

this is the incorrect option

Posted by

Vakul

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