For the reaction `H_2 (g) +I_2(g) harr2HI_(g)` at `K_p =794` at 298K and `K_p=54` at 700K. Is  the formation of HI favored more at the higher or lower temperature?

Expert Answers
jeew-m eNotes educator| Certified Educator

`H_2+I_2 harr 2HI`

Equilibrium constant of the above reaction can be written as;

`K_p = ((P_(HI))^2)/(P_(H_2)xxP_(I_2))`


`K_p ` = equilibrium constant

`P_(HI)` = Partial pressure of `HI`

`P_(H_2)` = Partial pressure of` H_2`

`P_(I_2)` = Partial pressure of `I_2`


It is given that  `K_p` at 298k is 798 and `K_p` at 700k is 54.

`K_p = ((P_(HI))^2)/(P_(H_2)xxP_(I_2))`

`(K_p)_298 = 798`

`(K_p)_700 = 54`


Here when temperature increases we can see `K_p` decreases. This means the partial pressure of reactants will increase when temperature increases. This reveals that when temperature increases the reverse reaction encourages.

So at higher temperatures formation of HI will not be favoured. Hence reduction in temperature than 298K will enhance the formation of HI.