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# $\Delta G^\circ$ for a reaction is 46.06 kcal/mol, $K_P$ for the reaction at 300 K is

(a) $10^{-8}$
(b) $10^{-22.22}$
(c) $10^{-33.55}$
(d) none of these

Answer: $10^{-33.55}$
$\Delta G^\circ$ = 46.06 kcal/mol = $46.06 \times 1000 \times 4.184 J/mol$
$\Delta G^\circ = -$RT ln $K_P = -2.303$ RT log $K_P$
$\Longrightarrow 46.04 \times 1000\times 4.184 = -2.303 \times 8.31 \times 300$ log $K_P$
$\Longrightarrow$ log $K_P = -33.55$
$\Longrightarrow K_P = 10^{-33.55}$