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?

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`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))`

Where;

`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.*

**Sources:**

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