Question

The equilibrium constant for reaction is 10.What will be the value of ${\Delta G}^o$
$(R=8.314J{K}^{-1}mo{l}^{-1}$ & $T=300K)$

Answer 1

equilibrium constant = 10.

$\therefore K=10$

$\Delta G^{\circ }=-2.303\times RTlogK$

$\Delta G^{\circ }=-2.303\times 8.314\times 300log10$

$\Delta G^{\circ }=-2.303\times 8.314\times 300\times 1$

$\Delta G^{\circ }=-5.744KJ/mol$ (ans)