Which of the following is correct for a spontaneous reaction at any temperature?

Δ H = - v e, Δ S = + v e, Δ G = - v e

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Δ H = - v e, Δ S = - v e, Δ G = - v e

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Δ H = + v e, Δ S = + v e, Δ G = - v e

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Δ H = - v e, Δ S = + v e, Δ G = + v e

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Solution

The correct option is A $Δ H = - v e, Δ S = + v e, Δ G = - v e$
For a spontaneous reaction at any temperature $Δ H = - v e, Δ S = + v e, Δ G = - v e$
For spontaneous processes, the change in Gibbs free energy is negative.
$Δ G = Δ H - T Δ S$
This is possible at all temperatures when enthalpy change is negative and entropy change is positive.

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Similar questions

For $H_{2} O (l) (1 b a r, 373 K) \to H_{2} O (g) (1 b a r, 373 K)$ .The correct set of thermodynamic parameter is

H_{2} O (l) (1 bar, 373 K) ⇌ H_{2} O (g) (1 bar, 373 K)

For the process $H_{2} O (l) (1 b a r, 373 K) \to H_{2} O (g) (1 b a r, 373 K)$ , the correct set of thermodynamic parameters is

H_{2} O (l) ⇌ H_{2} O (g) (1 b a r, 373 K),

H_{2} O_{(l)} (1 bar, 373 K) ⇌ H_{2} O_{(g)} (1 bar, 373 K)

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