Answer
H$_2$ + I$_2$--> 2HI
The reaction is product-favored.
Work Step by Step
H$_2$ + I$_2$---> 2HI
(H$_2$)=6.50*10$^{-5}$
(I$_2$)= 1.06*10$^{-3}$
(HI)= 1.87*10$^{-3}$
K= (HI)$^{2}$ / (H) (I)
k= (1.87*10$^{-3}$) / (6.5*10$^{-5}$) (1.06*10$^{3}$)
k= 50.75
50.75> 1
Therefore, the reaction is product favored.
