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Electrolysis and Electrolytes

The equilibri...

Question

The equilibrium constant for a reaction is 10. what will be value of $△ G^{⊖}$ ? R =8.314 $J K^{- 1} m o l^{- 1}$ , T=300 K

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Solution

$Δ G^{0} = - 2.303 R T l o g K = - 2.303 \times 8.314 \times 300 l o g 10 = - 5.744 k J / m o l$

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Similar questions

10

Δ G^{o}

R = 8.314 {J K}^{- 1} {mol}^{- 1}; T = 300 K

The equilibrium constant for a reaction is 10. What will be the value of $Δ G^{θ}$ ? $R = 8.314 J K^{- 1} m o l^{- 1} T = 300 K$ .

{Δ G}^{o}

(R = 8.314 J K^{- 1} m o l^{- 1}

T = 300 K)

10

△ G^{o}

(R = 8 J K^{- 1} m o l^{- 1}, T = 300 K)

X (s) ⇌ Y (s) + Z (g)

l n \frac{p_{z}}{p^{0}} versus \frac{10^{4}}{T}

p_{z}

Z

T

p^{0} = 1

\frac{d (l n K)}{d (\frac{1}{T})} = - \frac{Δ H^{0}}{R}

K = \frac{p_{z}}{p^{0}}

R = 8.314 J K^{- 1} m o l^{- 1}

Δ S^{0}

J K^{- 1} m o l^{- 1}

1000 K

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