For a spontaneous process.

Δ G_{s y t e m}

= +ve only

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Δ G_{s y t e m}

= zero

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Δ S_{t o t a l}

= - ve

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Δ S_{t o t a l}

= +ve

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Solution

The correct option is D $Δ S_{t o t a l}$ = +ve
$Δ S_{t o t a l} =$ + ve for a spontaneous reaction

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Similar questions

The value of ∆S for spontaneous process is

For a spontaneous reaction $△$ G, equilibrium constant (K) and $E_{c e l l}^{\circ}$ will be respectively:

E_{c e l l} = + v e

E_{c e l l} = + v e, △ G

- v e

\begin{matrix} Δ H & Δ S & Temperature & Spontaneity I. & \oplus v e & ⊖ v e & Any temperature & Non-spontaneous II. & ⊖ v e & ⊖ v e & Low temperature & Non-spontaneous III. & \oplus v e & \oplus v e & Low temperature & Spontaneous IV. & ⊖ v e & \oplus v e & Any temperature & Spontaneous \end{matrix}

\begin{matrix} Δ H & Δ S & Temperature & Spontanity I. & \oplus v e & ⊖ v e & Any temperature & Non-spontaneous II. & ⊖ v e & ⊖ v e & Low temperature & Non-spontaneous III. & \oplus v e & \oplus v e & Low temperature & Spontaneous IV. & ⊖ v e & \oplus v e & Any temperature & Spontaneous \end{matrix}

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