Chemistry 2720 Fall 1998 Test 2 Solutions

1. There are at least two alternatives:
   1. If the reaction is exothermic, reduce the temperature. If the reaction is endothermic, increase the temperature.
   2. Use coupled reactions: Add reactants which combine to form either or such that the free energy change for the overall reaction is more negative than the free energy change for the original reaction.
2. 1. Possible: n=8, ,
   2. and n=4 but should be less than n so this is not a possible hydrogenic state.
   3. and but can't be larger than so this is not a possible hydrogenic state.
3. The hydrogenic atom energies are proportional to so all energy differences are also proportional to . The photon energies are therefore proportional to so the wavelengths ( ) are inversely proportional to . Accordingly, the 434nm line of the hydrogen spectrum appears at 108.5nm ( ) in the spectrum.
4. The radius of a Bohr orbit is proportional to . Therefore, atoms in the imaginary universe described would be larger by a factor of .
5. 1. The reaction is

      

      For this reaction,

      

      To calculate the maximum work, we need :

      

      The activity of the solvent (water) is normally taken to be 1 and the activity of solid sulfur is exactly 1.

      

      A cell could therefore extract up to 152kJ/mol of work from the oxidation of glucose by sulfur under the stated conditions.

   2. 

      

      A cell could therefore extract up to 2903.52kJ/mol of work from the oxidation of glucose by oxygen under the stated conditions, which is much more than is obtained from oxidation by sulfur under similar conditions. There is therefore a huge metabolic advantage to using oxygen rather than sulfur as an oxidizing agent.

6. We need to calculate the equilibrium constant, so we need :

   

   This equilibrium ratio is extremely small and suggests that very little of the reactants will in fact be consumed. Accordingly, we guess that the final pressures of and HF will be close to the initial pressures. By stoichiometry, . Our equilibrium condition then reads

   

   Note: The exact answer is , , , so our approximation gives pretty good results.

7. We will use the formula

   

   For the process , the equilibrium constant is where X is the mole fraction of the liquid. We will take to be the normal freezing point of pure water (273.15K, X=1) while will be the freezing point of the solution. For the water in the solution,

   

   The equilibrium constant for the solution at its freezing point is therefore K=1/0.993=1.007.