In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (c) What is the value of Kp for any liquid in equilibrium with its vapor at the normal boiling point of the liquid?

Video transcript

Hey, everyone, we're told that the vapor pressure of any liquid and the pressure exerted by its vapor are equal when the two are in equilibrium, determine the value of our equilibrium constant expression calculated from our partial pressures at the normal boiling point of a liquid in equilibrium with its vapor. So a key thing here to remember is that the vapor pressure of a liquid at its normal boiling point is equal to the atmospheric pressure. So in this case, it will be one atmospheric pressure. So let's go ahead and write out our reaction. We were told that we have a liquid which we will label as X and this is an equilibrium with its vapor. So in its gaseous state, and as we said, the vapor pressure of our X is going to be equal to one atmospheric pressure. So let's go ahead and calculate our equilibrium constant expression. And this is going to be the partial pressure of our products divided by the partial pressure of our reactant. Now, since the reactant here is a liquid, it will not be included in our equilibrium expression. So in this case, our equilibrium constant expression is going to be equal to the vapor pressure of our x in its gaseous state. So as we said, this is one atmospheric pressure and this will be our final answer. Now, I hope that made sense and let us know if you have any questions.