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

6.20  The equilibrium constant for a reaction is 10. What will be the value of   

             \Delta G ^ \ominus ? R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K  

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M manish

Given values are,
R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K
and equilibrium constant  = 10

It is known that,
\Delta G^o = -2.303\ RT \log K_{eq}
           \\=-2.303(8.314)(300)\log 10\\ =-5744.14 Jmol^{-1}\\ =-5.744 kJmol^{-1}

Hence the value of \Delta G^o is -5.744 kJ/mol

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