Chemistry 2720
Fall 1995 midterm examination

All questions are weighted equally.

1. 1. 2% of the maple sap is sugar so we want

      

      The mass of sap required is therefore 5kg.

   2. Sap is mostly water. We need to raise the temperature of the water from to and then boil it away. Note that there is 4.9kg of water (98% of 5kg) but, guessing that the heat capacity of sugar is similar to that of water, we could reasonably use 5kg as the mass of water to heat.

      

      The total heat which must be provided is therefore

      

      This heat comes from burning decane so

      

2. 1. Lavoisier's proposal was essentially that cells are heat engines. The maximum efficiency of a heat engine operating between and is

      

      The heat to run this engine is provided by burning glucose. The heat of combustion of glucose is just for the reaction

      

      Looking at the data sheet, we see that we have limited information at our disposal. We don't have the heat capacity of glucose so we can't work out the enthalpy of combustion at , which is what we really need. (The fire would be burning in the hot organelle in this model.) We also only have data gaseous oxygen and carbon dioxide whereas the reactions in a cell would certainly involve the dissolved forms. Our numbers are therefore going to give only a rough approximation, but of course the maximal efficiency calculated above can only give us wildly optimistic estimates anyway. (Obviously, I don't expect you to tell me all this on an exam.)

      

      The efficiency is the work performed divided by the heat input. Therefore

      

   2. Isothermal free energy machines are far more efficient than heat engines. The maximum work for an isothermal free energy machine is calculated by computing the change in free energy:

      

      The maximum work is therefore , or over five times as much as is available from the equivalent heat engine.

3. The overall process is

   

   where G represents a glucose molecule. for this process is -31.3kJ/mol since the standard free energy of formation of glucose is the same inside as outside the cell. The maximum ratio of glucose in to glucose out is maintained when . (This means that all of the free energy from the reaction is used to transport glucose.) Therefore

   

   assuming that the activity of water is 1 and with the understanding that each concentration term should be divided by 1mol/L to convert it to an activity. Therefore

   

   so that

   

4. 1. The vapour pressure is related to the evaporation process:

      

      The equilibrium constant for this process is

      

      so the equilibrium constant is the vapour pressure. The standard change in free energy is

      

      Therefore

      

      (I have cheated slightly and converted from the dimensionless activity to pressure directly.)

   2. There are at least two ways to solve this problem:

      Method 1: At the temperature at which a phase change occurs, . This implies that . For this process

      

      We have at 298K and so

      

      If you prefer the Celcius scale, that's about .

      Method 2: Since the vapour pressure is just an equilibrium constant, we can use the equation which relates equilibrium constants to temperature. At the boiling point, the vapour pressure is 1atm.

      

      In our case, let the label . Calculate as above and solve for :

      

5. The formation reaction for steam is

   

   Since both enthalpy and free energy are state functions, we imagine converting hydrogen and oxygen at 500K to steam at 500K by the following path:

   1. Cool the hydrogen and oxygen from 500K to 298K. For this step,

      

   2. React the hydrogen and oxygen to form steam at 298K:

      

   3. Heat the steam from 298K to 500K.

      

   For the complete process at 500K,