The equilibrium constant for the reaction is 10- Calculate the value of - Go- Given R-8-314 J K-1mol-1- T-300 K-

Δ G^o= RT ln K= 2.303 RT log K. R=8.314 J K^ 1 mol^ 1; T=300 K; K=10 Δ G^o = 2.303× 8.314 J K^ 1mol^ 1× (300 K)×log 10 = 5744 J mol^ 1= 5.744 kJ mol^ 1. ...

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Standard XII

Chemistry

Enthalpy of Combustion

The equilibri...

Question

The equilibrium constant for the reaction is $10$ . Calculate the value of $Δ G^{o}$ ; (Given $R = 8.314 {J K}^{- 1} {mol}^{- 1}; T = 300 K$ ).

5.527 {kJ mol}^{- 1}

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- 5.527 {kJ mol}^{- 1}

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- 3.327 {kJ mol}^{- 1}

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3.327 {kJ mol}^{- 1}

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Solution

The correct option is B $- 5.527 {kJ mol}^{- 1}$
$Δ G^{o} = - R T l n K = - 2.303 R T log K$ .
$R = 8.314 {J K}^{- 1} {mol}^{- 1}; T = 300 K; K = 10$
$Δ G^{o} = - 2.303 \times 8.314 {J K}^{- 1} {mol}^{- 1} \times (300 K) \times log 10$
$= - 5744 {J mol}^{- 1} = - 5.744 {kJ mol}^{- 1}$ .

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The equilibrium constant for the reaction is 10. Calculate the value of ΔGo; (Given R=8.314 J K−1mol−1; T=300 K).

The correct option is B −5.527 kJ mol−1ΔGo=−RTlnK=−2.303RTlogK. R=8.314 J K−1 mol−1;T=300K; K=10 ΔGo=−2.303×8.314 J K−1mol−1×(300K)×log10 =−5744 J mol−1=−5.744 kJ mol−1.

The equilibrium constant for the reaction is $10$ . Calculate the value of $Δ G^{o}$ ; (Given $R = 8.314 {J K}^{- 1} {mol}^{- 1}; T = 300 K$ ).

The correct option is B $- 5.527 {kJ mol}^{- 1}$
$Δ G^{o} = - R T l n K = - 2.303 R T log K$ .
$R = 8.314 {J K}^{- 1} {mol}^{- 1}; T = 300 K; K = 10$
$Δ G^{o} = - 2.303 \times 8.314 {J K}^{- 1} {mol}^{- 1} \times (300 K) \times log 10$
$= - 5744 {J mol}^{- 1} = - 5.744 {kJ mol}^{- 1}$ .