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Gibbs Free Energy & Spontaneity

Δ H and Δ S ...

Question

$Δ H$ and $Δ S$ for the reaction
${B r}_{2} (l) + {C l}_{2} (l) ⟶ 2 B r C l (g)$
are $29.37 k J$ and $104.0 J K^{- 1}$ respectively. Above that temperature will this reaction become spontaneous?

A

T > 177.8 K

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B

T > 35.1 K

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C

T > 28.4 K

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D

T > 141.2 K

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Solution

The correct option is A T > 177.8 K
According to Gibbs-Helmholtz equation
$Δ G = Δ H - T Δ S$
For spontaneous process, $Δ G < 0$
i.e., $T Δ S > Δ H$
$T > \frac{Δ H}{Δ S}$
$T > \frac{29.37 \times 1000}{104}$
$T > 282.4 K$

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Similar questions

Q. For the reaction,

{B r}_{2} (l) + {C l}_{2} (g) ⟶ 2 B r C l (g)

Δ H = 29.37 k J {m o l}^{- 1}; Δ S = 104 J {m o l}^{- 1}

. Find the temperature above which the reaction would become spontaneous.

Q. The value of

Δ H

Δ S

for a reaction are respectively

30 k J

{m o l}^{- 1}

100 J K^{- 1}

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. Then temperature above which the reaction will become spontaneous is:

Q. The enthalpy and entropy change for the reaction,

B r_{2} (l) + C l_{2} (g) \to 2 B r C l (g)

7.1 k c a l

m o l

25 c a l K^{- 1} m o l^{- 1}

respectively. This reaction will be spontaneous at:

2 A + B \to C

For the above reaction at

298 K

Δ H = 400 {kJ mol}^{- 1}

Δ S = 0.2 {kJ K}^{- 1} {mol}^{- 1}

. At what temperature, will the reaction become spontaneous considering the

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Δ S

to be constant over the temperature range?

Δ H

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30.56 k J {m o l}^{- 1}

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