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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`.

Text Solution

Verified by Experts

The correct Answer is:
6
Given : `" "K_("eq") = 10`
`" "T= 300` K
Asked : `DeltaG^(@) =?`
Formula: `" DeltaG^(@) =-2.303` RT log `K_("eq")`
Explanation: `DeltaG^(@)` = Standard Gibb's free energy change
`" "` R= Gas constant (`R=8.314 JK^(-1) mol^(-1))`
`" "` T= Temperature in K
`" " K_("eq")` = equilibrium constant
Substitution & calculation
`" "=-2.3030 xx 8.31 xx 314` log `10` J
`" "=-6 "KJ mol"^(-1)`.
The value of `DeltaG^(@)` is -6 KJ `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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