If the equilibrium constant for a reaction is $10$ , then the value of $△ G^{o}$ will be $(R = 8 J K^{- 1} m o l^{- 1}, T = 300 K)$

+ 5.527 K J m o l^{- 1}

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

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+ 55.27 K J m o l^{- 1}

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

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Solution

The correct option is B $- 5.527 K J m o l^{- 1}$
Solution:- (B) $- 5.527 K J / m o l$
$K = 10 (Given)$
As we know that,
$Δ G^{\circ} = - 2.303 R T log K$
At $300 K$ ,
$Δ G^{\circ}$ $= - 2.303 \times 8.314 \times 300 log 10 = - 5527.2 J / m o l = - 5.527 k J / m o l$

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Similar questions

Δ G^{\circ}

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

Δ G^{\circ}

(R = 8 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)

The equilibrium constant for a reaction is 10. What will be the value of ΔG^θ? R = 8.314 JK^–1 mol^–1, T = 300 K.

2 V

300 K

Z n (s) + C u^{2 +} (a q) ⇌ Z n^{2 +} (a q) + C u (s)

300 K

(R = 8 J K^{- 1} m o l^{- 1}, F = 96000 C m o l^{- 1})

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