*This preview shows
pages
1–2. Sign up to
view the full content.*

This ** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **f on those intervals? Exercise 4 The law of mass action states that the rate of a chemical reaction is proportional to the product of the concentrations of A and B , i.e., d @ C D d t = k @ A D@ B D , where @ A D , @ B D , and @ C D are all functions, and k is a constant. Let's assume the initial concentrations of A and B are equal, and write @ A D H L = @ B D H L = a . If we write c H t L = @ C D H t L for the concentration of C at time t , then the concentration of A (as well as that of B ) at time t is @ A D H t L = a-@ C D H t L . Therefore, in our case, the law of mass action can be stated as d c d t = k H a-c L 2 . Use Mathematica 's DSolve command to find c , the concentration of C , as a function of t . Hint: the reaction begins with an initial concentration @ C D H L = 0. Mathematica version 8.0, 3 April, 2012 2 Math 149 Lab 5 Exercises.nb...

View Full
Document