Question

What will be the value of $\Delta G^o$, if the equilibrium constant of the reaction is 10 at ${27}^{0}C$?

• A. $-50.44kJmo{l}^{-1}$
• B. $-5.744kJmo{l}^{-1}$
• C. $-25.44kJmo{l}^{-1}$
• D. $-10kJmo{l}^{-1}$

Answer 1

The correct option is B $-5.744kJmo{l}^{-1}$

$\Delta G^o=-RTlnK$

$=-8.314\times 300\times 2.303\times log10$

$=-5.744kJ.mo{l}^{-1}$