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What is K if #DeltaG^@# = -18.0 kJ for a reaction at 25 degrees C?

1 Answer

Apr 16, 2017

#sf(K=6.97xx10^(-4))#

Explanation:

#sf(DeltaG^@=-RTlnK)#

#:.##sf(lnK=-(DeltaG^@)/(RT))#

#sf(lnK=-(18xx10^3)/(8.31xx298)=-7.268)#

#sf(K=6.97xx10^(-4))#

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