In Section 11.5, we defined the vapor pressure of a liquid in
ter...

Question

In Section 11.5, we defined the vapor pressure of a liquid in
terms of an equilibrium. 
(b) By using data
in Appendix B, give the value of Kp for this reaction at 30 C.

Accepted Answer

Hey, everyone, we're asked to consider the following equilibrium. We have our water in its liquid state At equilibrium with our water in our gaseous state at 70°C. What is the value of our equilibrium constant for our partial pressures for this reaction? So, key thing to note here is that the vapor pressure of any liquid and the pressure exerted by its vapor are equal when the two are in equilibrium. So when we look up the vapor pressure of water At 70°C, we find that it is 233. tour. Now we want to go ahead and convert this into atmospheric pressure. So we have the vapor pressure of water And this is, as we said, equal to 233.7 tour, Using our conversion factor, we know that we have 760 tour per one atmospheric pressure. So when we calculate this out, we end up with a vapor pressure of 0.03075 atmospheric pressure. Now to calculate our equilibrium constant, calculated from our partial pressures, this is going to be equal to the partial pressure of our products over the partial pressure of our reactant. But since the reactant is a liquid, it will not be included in the equilibrium constant expression. So this means our equilibrium constant expression is going to be equal to the partial pressure of water, Which is 0.03075 atmospheric pressure. And this is going to be our final answer. Now, I hope this made sense and let us know if you have any questions.

In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (b) By using data in Appendix B, give the value of Kp for this reaction at 30 C.

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