For the reaction \\A(l)\rightarrow 2B(g)\\\Delta U=2.1\: kcal,\Delta S=20\: cal\: K^{-1}\: at\; 300\; K\\Hence\; \Delta G\: in\: kcl\: is-----
Option: 1 -2.7 Kcal
Option: 2 3.3 Kcal
Option: 3 - 3.3 Kcal
Option: 42.7 Kcal
 

Answers (1)

We know:

\mathrm{\Delta G\: =\: \Delta H\: -\: T\Delta S}

\mathrm{\Delta H\: =\: \Delta U\: +\: 2RT}

Thus, we have:

\mathrm{\Delta G\: =\: \Delta U\: +\: 2RT\: -\: T\Delta S}

On putting the given values we get:

\mathrm{\Delta G\: =\: 2.1\: +\: 2\, x\, 2\, x\, 300\, x\, 10^{-3}\: -\: 300\, x\, 20\, x\, 10^{-3}}

\mathrm{\Delta G\: =\: 2.1\: +\:4\, x\, 300\, x\, 10^{-3}\: -\: 300\, x\, 20\, x\, 10^{-3}}

\mathrm{\Delta G\: =\: 1200\, x\, 10^{-3}\: -\: 6000\, x\, 10^{-3}}

\mathrm{\Delta G\: =\:2.1\: +\: 1.2\: -\: 6}

\mathrm{\Delta G\: =\:3.3\: -\: 6}

\mathrm{\Delta G\: =\:-2.7\ Kcal}

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