In cis- and trans- isomers there are two parts that are on the same side (cis-) or across (trans-) a double bond.  For example these are cis- and trans-2-butene (C₄H₈).  To help you see the relationship of the parts, the hydrogens on the same (cis) side, or across (trans) the double bond are highlighted in blue.
 

Because the bonds can break apart and recombine, there is an equilibrium between the two forms which can be represented by this equation: cis-2-butene  trans-2-butene
 

The rate at which the cis molecules turn into trans (the forward reaction) is  Rf = kf [cis] where: Rf  the rate of the forward reaction kf  the rate constant for the forward reaction [cis] the concentration of cis molecules

The rate at which the trans molecules turn into cis (the reverse reaction) is  Rr = kr [trans] where: Rr the rate of the reverse reaction kr  the rate constant for the reverse reaction [trans] the concentration of trans molecules

 
If you started with a container in which there were no trans-molecules, the rate of the reverse reaction would initially be zero (since [trans] = 0).  However, over time, there will be a buildup of the trans- molecules ([trans] increases).  This means the reverse rate will increase.  At the same time, the [cis] must be going down.  This means the forward rate decreases. At some point in time, the R_f = R_r, and equilibrium is achieved.

You can see this in this Excel spreadsheet², which you should download and save on your computer.  Then open the spreadsheet, and use it to do the following:

1. Observe the spreadsheet graph for these default initial values¹:
 

Temperature 25 oC Activation energy (forward) 80 kJ Activation energy (reverse) 60 kJ [cis] 1.0 mol/L [trans] 0.0 mol/L

This is an endothermic reaction

 
At what point does the reaction reach equilibrium?  Explain why the spreadsheet graph has the shape it does.

2. Change the temperature (make it both higher and lower) and observe what happens to the equilibrium.  Explain what you see.

3. Change the size of the activation energies.  Make Activation energy (forward) = 60 and Activation Energy (reverse) = 80 (note: this will now be an exothermic reaction).  Does K_eq depend on the value of k_f and kr?  Since k_f and k_r depend on temperature, what can you conclude about the effect of a temperature change on the value of K_eq?  Is the change the same for an exothermic and endothermic reaction?

4. Reset the temperature and activation energies to their default values.  Now double the initial [cis] molecules.  What do you observe happens to the graph? Explain what you see.

5. What did you observe happen to the value of K_eq in step 4?  What can you conclude about the effect of concentration on K_eq's value?

6. Make the initial [trans] the same as the initial [cis].  What do you observe happens to the graph? Explain what you see.

7. Reset the default values for temperature and activation energy.  Make the initial [trans] = 1.0 mol/L , and set the [cis] = 0.  What do you observe happens to the graph?  Explain what you see.

8. Compare the results from steps 1 and 7.  What can you conclude about a reaction that has reached equilibrium, if you start from the products, as compared to starting with reactants?

9. Does the rate of a reaction become zero when it has reached equilibrium? Explain your answer. 

² This spreadsheet is a modification of one created by Larry Brown, Texas A&M University, Department of Chemistry,  P. O. Box 300012, College Station, TX 77842-3012