# Question 23afe

Jul 2, 2017

${K}_{p} = 7.01 \times {10}^{-} 6$

#### 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

${K}_{p} = \left({\left[{P}_{\text{H"_2)]^2[P_ ("O"_2)])/([P_ ("H"_2"O}}\right]}^{2}\right)$

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#