Karl's Calculus Tutor - Solution to Exercise 6.3-7

Solution to Exercise 6.3-7

I'll not repeat the entire problem here. If you need to recall anything about it, then just go back and reread it. We had a formula for entropy of

                                          Uc_B
   S  =  -R c_A ln(c_A)  -  R c_B ln(c_B)  +     
                                           T

We also had an equation of

   c_A + c_B  =  1 mole/liter

In the general class of problems to which this belongs, the above equation is called a constraint, because it limits the freedom of the variables to vary. They can vary, but only to the extent that they still obey the constraint.

The problem instructed you to substitute c_A with (1 mole/liter - c_B). And for brevity sake, I will abbreviate 1 mole/liter with simply the number 1. So the entropy equation becomes

                                                Uc_B
   S  =  -R (1 - c_B) ln(1 - c_B) - R c_B ln(c_B) +    
                                                 T

Now you get to the fun part. The problem tells you to take the derivative of this with respect to c_B. You will need the chain rule to find the derivative of ln(1 - c_B). You will also need the product rule to take the derivatives where the logs are multiplied by other functions of c_B. After applying them both you get (and you must pay close attention to the sign of each term):

   dS        1 - c_B                    c_B               U
        =  R        + R ln(1 - c_B) - R    - R ln(c_B) +   
   dc_B       1 - c_B                    c_B               T

Observe that the two ratios involving c_B are both 1. Each is multiplied by the same constant, R, but with opposite sign, so those terms cancel. So rewriting the equation after the cancellation, and setting the derivative to zero, which the problem instructs you to do, you get

                                U
   0  =  R(ln(1 - c_B) - ln(c_B)) +  
                                T

or equivalently

                          U
   ln(c_B) - ln(1 - c_B)  =    
                         RT

If you substitute back the c_A wherever you see 1 - c_B and then apply the product identity, this becomes

                         æ  c_B   ö        U
   ln(c_B) - ln(c_A)  =  ln ç        ÷   =    
                         è  c_A   ø       RT

Now you can eliminate the log by raising e to the power on both sides of the equation

   c_B
       =  e^U/RT
   c_A

This is the answer. What does it mean? If U is large, then the chemical reaction of going from A to B gives off a lot of heat. And when that is the case, the exponential will be a large number. What effect does that have on the ratio of c_B:c_A? It means that the concentration product of the reaction, B, will, when the reaction is complete, predominate over the concentration of A. Almost all the starting material will be consumed. If you raise the temperature, T, then the exponent will become smaller, and likewise the exponential function. And that means that B's predominance over A will become less. More of the starting material will be left over.

If the reaction absorbs heat rather than giving it off, then U is negative. I'll let you discover for yourself what effect that has.

Chemists are fond of representing the concentration of, say A and B, with the symbology of [A] and [B]. The value of the exponential on the right they like to call K. So with that nomenclature, the equation becomes

   [B]
        =  K
   [A]

If you've taken a chemistry class, you've certainly seen that kind of equation before. And you will remember that the K is always given with a temperature as a subscript. From the exponential we derived here, you can see why.

If you feel especially ambitious, you might try doing this problem again, except that this time have it be that every 2 moles of A changes into 1 mole of B. If you start out with c_totalmoles/liter of A, (and c_total remains constant) the constraint equation is

   c_A + 2c_B  =  c_total

Again, each mole of B formed liberates U units of heat.

The problem is a little harder, but it has a more interesting result.

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