A large positive value of $∆\mathrm{G}°$ corresponds to which of these? K$\to$ equilibrium constant)

 

$\mathrm{\Delta G}={\mathrm{\Delta G}}^0+\mathrm{RTlnK}$

At equilibrium $∆\mathrm{G}=0$

$\therefore {\mathrm{\Delta G}}^0=-\mathrm{RT} \mathrm{lnK}$

$\therefore \mathrm{K}={\mathrm{e}}^{-{\mathrm{\Delta G}}^0\mathrm{RT}}=\frac{1}{{\mathrm{e}}^{{\mathrm{\Delta C}}^0\mathrm{RT}}}>0$

As a result, a big positive value of G and a small positive K are equivalent.