Which of the given relationship is incorrect?

Δ G - Δ H = - T Δ S

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Δ S_{T o t a l} = Δ S_{s y s} + Δ S_{s u r}

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Δ G = Δ G^{\circ} + R T l n Q

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Δ U = q \times W

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Solution

The correct option is D $Δ U = q \times W$
According to the first law of thermodynamics,
$Δ U = q + W$ .
So, the incorrect relationship is, $Δ U = q \times W$

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Similar questions

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

Δ S

Δ G = Δ H - T Δ S

Δ G = Δ H + T {[\frac{d (Δ G)}{d T}]}_{p}

List - I (Process)	List - II
(I) Reversible isothermal expansion of an ideal gas.	(P) $Δ S_{s y s} = 0$
(II) Reversible adiabatic compression of a real gas.	(Q) $Δ H = Δ U = Δ S_{total} = 0$
(III) Adiabatic free expansion.	(R) $Δ S_{total} > 0$
(IV) Isothermal free expansion	(S) $q = 0$
	(T) $Δ S_{surr.} = 0$
	(U) $Δ S_{s y s} > 0$

Δ G = Δ H - T Δ S

Δ G = Δ H + T {[\frac{d (Δ G)}{d T}]}_{P}

Δ G

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