$2$

$3$

$4$

$=5$

$6$

$8$

$-=9$

$=10$

$11$

$12$

$=13$

$=14$

The equilibrium constant $(K_c)$ for the reaction

$2HCl(g)H_2(g)+C{l}_2(g)$

is $4\mathrm{.}17{10}^{-34}$ at $25C.$ What is the equilibrium constant for the reverse reaction at the same temperature? Round your answer to $3$ significant digits.

$H_2(g)+C{l}_2(g)2HCl(g)$

$K_c^{\text{'}}=$