Chemistry 2720 Spring 1998 Assignment 2 Solutions

1. The reaction is

   

   To save typing, I will call these two compounds G6P and F6P from here on. The reaction is spontaneous if

   

   For this reaction, . To make negative, Q must be sufficiently small, i.e. there must be enough reactant relative to the product concentration. You can solve this problem by trial and error with your calculator. I find that a ratio of reactant to product of 2:1 (i.e. ) is adequate to get a negative so, for example, we could have a concentration of F6P of 1mM and a concentration of G6P of 2mM.

2. Maximum work for an isothermal reaction is computed from the free energy change. For this we need the activities of the various substances appearing in the reaction. The activities of acetaldehyde, NADH, ethanol and are easy enough to get since the standard concentrations for these species is 1mol/L. Thus , , and . Since we are using the biochemists' standard state, the standard concentration of protons however is so

   

   Therefore

   

   Calculating is easy after that:

   

   The maximum work available is therefore 38.4kJ/mol of ethanol produced.

3. The three ionization reactions are

   

   Using the free energies of formation of the phosphates, we find

   

   We now find the 's of the three steps using the formula :

   

   It is convenient (but not essential) to convert these acid dissociation constants into 's:

   

   At the , an acid and its conjugate base are present in equal amounts. At pH's far above , the acid form is present in only very small amounts. It follows that at pH 11.5, phosphoric acid exists almost exclusively as the phosphate and hydrogen phosphate anions. Thus, to a very good approximation, we need only consider the last equilibrium. For this equilibrium,

    

   Furthermore

    

   (the total concentration of phosphates is 0.01mol/L and we know that the other phosphate species are present only in tiny amounts) and . We now have two equations in two unknowns. Solving eq 1 for , we get

    

   Substituting this relation into eq 2, we get a simple equation in one unknown with solution

   

   We substitute this result back into eq 3:

   

   Once we have found the concentrations of the phosphate and hydrogen phosphate anions, we can work backwards through the other equilibria. For the second equilibrium,

   

   Since we know everything that goes into this formula except for , we can solve for this quantity:

   

   Similarly,

   

   Since the standard concentration is 1mol/L, the concentrations of phosphates in the buffer are

   

4. HCl is a strong acid so it dissociates completely in water. However, at this low concentration, we have to consider the water autoionization equilibrium:

   

   The amount of in solution is a combination of protons released by water and by HCl:

    

   This relation and the ionization constant equation

    

   provide us with the two equations we need to find our two unknowns. ( , which you can look up in any first-year textbook, has the value at 298K.) If we solve eq 4 for and substitute this into eq 5, we get

   

   We solve this equation using the quadratic equation:

   

   The pH is therefore 6.96.

5. The free energy of mixing obeys the relationship

   

   For this particular mixture at (973K), we have

   

   It is impossible to tell whether these two substances will mix at this temperature. We know that the free energy of mixing must be larger than this small negative number, but we don't know how much larger. It could be slightly negative (in which case the two substances will mix) or it could be positive (in which case they won't).