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Gibbs Free Energy & Spontaneity

The free ener...

Question

The free energy change for a reaction having $△ H = 31400 c a l / m o l, △ S = 32 c a l K^{- 1} m o l^{- 1} a t 727^{o} C$ is:

- 600 cal

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-420 cal

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-850 cal

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600 cal

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Solution

The correct option is A - 600 cal
Given:
$△ H = 31400 c a l / m o l △ S = 32 c a l K^{- 1} m o l^{- 1} a n d T = 727^{o} C = 727 + 273 = 1000 K$
The expression to calculate Gibbs free energy is given as
$△ G = △ H - T △ S = 31400 - 1000 \times 32 = - 600 c a l$
hence, the free energy for a reaction is $- 600 c a l$

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Similar questions

△ H = 31400 c a l / m o l, △ S = 32 c a l K^{- 1} m o l^{- 1} a t 727^{o} C

△ H = 31400 c a l / m o l, △ S = 32 c a l K^{- 1} m o l^{- 1} a t 727^{o} C

Δ H = 31400 c a l, Δ S = 32 c a l K^{- 1} {m o l}^{- 1}

1000^{o} C

Δ_{s y s} G = Δ_{s y s} H - T Δ_{s y s} S

Δ_{s y s} G < 0 (s p o n t a n e o u s)

Δ_{s y s} G = 0 (e q u i l i b r i u m)

Δ_{s y s} G > 0 (n o n - s p o n t a n e o u s)

Δ H = 31400 c a l

Δ S = 32 c a l K^{- 1} m o l^{- 1}

1000^{\circ} C

$K_{c}$ for the reaction, $F e^{2 +} + Z n \to Z n^{2 +} + F e$ is $10^{23}$ , The standard free energy change is

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