H2g - Br2g- 2HBrg- - H- - -72-40 kJ - G- - -106-49 kJ- T - 298 KThe value - S is -

The correct option is A 114.3 J k^ 1 Δ G = Δ H TΔ S 106.49 #160;kJ = 72.40 298× Δ S 34.09 #160;kJ = 298× Δ S ...

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Gibbs Free Energy & Spontaneity

H2g + Br2g⟶ 2...

Question

$H_{2} (g) + B r_{2} (g) ⟶ 2 H B r (g); Δ H^{⊖} = - 72.40 k J$
$Δ G^{⊖} = - 106.49 k J, T = 298 K$
The value $Δ S$ is .

114.3 J k^{- 1}

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- 114.3 J k^{- 1}

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1363.6 J k^{- 1}

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Noe of these

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Solution

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Δ S

298 K

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (g)

Δ G = - 228.6 k J; Δ H = - 241.8 k J

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H - B r

k J m o l^{- 1}

Δ H

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m o l^{- 1}

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1^{\circ} C

Δ H = 35.5 k J m o l^{- 1}

Δ S = 83.6 J K^{- 1} m o l^{- 1}

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Δ S

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