## deltaG and deltaS(sys) at zero

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Clement Ng
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### deltaG and deltaS(sys) at zero

A sample of 1 mole of gas initially at 1 atm and 298 K is heated at constant pressure to 350 K, then the gas is compressed isothermally to its initial volume and is then cooled to 298 K at constant volume.

Why are deltaS(sys) and deltaG zero in this case? Which equations would you use to figure what factors are at zero?

ZoeHahn1J
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### Re: deltaG and deltaS(sys) at zero

delta S is zero because change in entropy is a state function, and the final and initial volumes are the same. delta G is zero when a reaction is at equilibrium.

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