Question

Question #23afe

Answer 1

`K_p = 7.01 xx 10^-6`

Explanation:

We're asked to find the value of the `K_p` for a reaction at a certain temperature with given equilibrium partial pressures.

We do so the same way we would if we were solving for `K_c` (the equilibrium constant for concentration). The equilibrium-constant expression here is

`K_p = ([P_ ("H"_2)]^2[P_ ("O"_2)])/([P_ ("H"_2"O")]^2)`

Plugging in the partial pressure values for each species, we have

`K_p = ((0.00450"atm")^2(0.00250"atm"))/(0.0850"atm")^2 = color(blue)(7.01 xx10^-6`