Chemistry 2720, Fall 96, Assignment 2 Solutions

1. We first need to determine the of the acid. is the equilibrium constant for the reaction

   

   

   However

   

   The activity of acetate is equal to that of the protons because the two are produced in equal quantities by dissociation. Furthermore, since is quite small, we can take . Solving the equilibrium condition for , we get

   

   (Note that the small size of this number confirms the validity of the approximation made with respect to the degree of dissociation.)

   

2. The overall flow-coupled synthesis process is

   

   The maximum PG synthesis will occur when (no free energy is being wasted), i.e. when

   

   But

   

   We use the usual approximation that . Then

   

3. We are interested in melting, i.e. in the process . For a phase transition between condensed phases,

   

   is just the inverse of the density:

   

   To get , we note that at the normal melting point, solid and liquid are in equilibrium. Therefore which implies that

   

   Thus we have

   

   We know the melting point at 1atm and we want to know it at 0atm so consider a change in pressure of -1atm (-101325Pa):

   

   The melting point of tin in a vacuum is therefore virtually identical to its melting point at 1atm, i.e. we may expect tin to melt very near in a vacuum.

4. Boiling refers to the process for a pure liquid. For this process, the equilibrium constant is just . A liquid boils when its partial pressure P equals the atmospheric pressure. Thus, in our case, we want to calculate the temperature at which knowing that when . If we rearrange the equation relating equilibrium constants and temperature to solve for , we get

   

   In degrees Celcius, this is .

5. Let us first consider an exothermic reaction. Le Chatelier's principle predicts that the equilibrium constant of an exothermic reaction decreases as the temperature increases. (The reactive system attempts to ``soak up'' the extra thermal energy by favoring the direction which absorbs heat.) Take . For an exothermic reaction, so the right-hand side of the equation

   

   is negative. Thus, (at the higher temperature ) is lower than which is precisely as predicted by Le Chatelier's principle. Conversely, for an endothermic reaction, the right-hand side of the equation is positive implying that which agrees with Le Chatelier's principle since in this case heat is absorbed by forming more products.