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The equilibrium constant for a reaction ...

The equilibrium constant for a reaction is `10`. What will be the value of `DeltaG^(Θ)`? `R=8.314 J K^(-1) mol^(-1), T=300 K`.

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To find the value of ΔG° (Gibbs free energy change at standard conditions) given the equilibrium constant (K) of a reaction, we can use the following equation: \[ \Delta G^\circ = -RT \ln K \] Where: - \( R \) is the universal gas constant (8.314 J K\(^{-1}\) mol\(^{-1}\)) ...
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The equilibrium constant for a reaction is 10. What will be the value of DeltaG^(@) ? R = 8.314 J K^(-1) "mol"^(-1) , T = 300 K.

The equilibrium constant for a reaction is 10.What will be thevalue of DeltaG^(@) ? R=8.314 JK^(-1)mol^(-1) , T = 300 K .

Knowledge Check

  • The equilibrium constant for a reaction is 10. what will be the value of DeltaG^(@) ? Given, R=8.314JK^(-1)mol^(-1),T=300K .

    A
    `-574.414Jmol^(-1)`
    B
    `-5744.14Jmol^(-1)`
    C
    `-57.4414Jmol^(-1)`
    D
    `57441.4Jmol^(-1)`

  • The equilibrium constant for a reaction is 100 what will be the value of DeltaG^(@) ? R=8.314JK^(-1)mol^(-1),T=300 K :-

    A
    `-11488KJ`
    B
    `-11.488KJ`
    C
    `-12KJ`
    D
    `-12000KJ`

  • If the equilibrium constant for a reaction is 10, then the value of triangleG^(@) will be: ("Given: "R= 8JK^(-1) mol^(-1) T=300K)

    A
    `+5.527" kJ mol"^(-1)`
    B
    `-5.527" kJ mol"^(-1)`
    C
    `+55.27" kJ mol"^(-1)`
    D
    `-55.27" kJ mol"^(-1)`

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