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Gulab Bansal
Asked: 2022-11-06T17:43:26+05:30 In:

The equilibrium constant for a reaction is 10.What will be thevalue of `DeltaG^(@) ` ? ` R=8.314 JK^(-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 `.
Parabola
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  1. `DeltaG^(@) =-2.303 RT log K=-2.303 xx 8.314 JK^(-1) JK^(-1) mol^(-1) xx300 K xxlog10 = -5744.1 J`

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Charandeep Srinivasan
Asked: 2022-11-06T12:57:14+05:30 In:

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`.
A. `+5.527 KJ mol^(-1)`
B. `-5.527 KJ mol^(-1)`
C. `+55.27 KJ mol^(-1)`
D. `-55.27 KJ mol^(-1)`

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`.
A. `+5.527 KJ mol^(-1)`
B. `-5.527 KJ mol^(-1)`
C. `+55.27 KJ mol^(-1)`
D. `-55.27 KJ mol^(-1)`
Parabola
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  1. Correct Answer – B
    `Delta G^(@)=-2.303` RT log K
    `=-2.303xx8 5 10^(-3)xx300xxlog 10`
    `Delta G^(@)=-5.527 kJ mol^(-1)`

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Naina Bobal
Asked: 2022-11-04T18:06:28+05:30 In:

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`.

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`.
Parabola
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  1. From the expression,
    `DeltaG^(theta)=-2.303RT “log”K_(eq)`
    `DeltaG^(theta)` for the reaction,
    `=(2.303)(8.314 JK^(-1)”mol”^(-1)) (300K)` log10
    `-5744.14 J”mol”^(-1)`
    `-5.744 kJ “mol”^(-1)`

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Tejaswani Shroff
Asked: 2022-11-02T11:46:19+05:30 In:

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

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

Parabola
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  1. ΔG° = -2.303 RT log Kc

    = -2.303 x (8.314 mol-1 K-1) x 300 K log 10

    = -2.303 x 8.314 x 300 x 1

    = -5527 J mol-1

    = -5.527 kJ mol-1

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