Le Chatelier's principle: what are the effects of decreasing volume? Which of the following is increased by decreasing the volume of the
  reaction system in the following reaction: $\ce{2H_2S(g) +3O_2(g)&lt;=&gt;
 2SO_2(g) +2H_2O(g)} + {\text{heat}}$ 
  
  I. Rate of Reaction 
  
  II. Equilibrium concentration of reactants 
  
  III. Value of $\ce{K_{eq}}$


My attempt: By the gas laws, decreasing the volume of the container will increase the pressure, so equilibrium should shift to the right, as there are less gases on the right. Thus III is true. Additionally, heat is also generated by the shifting of the equilibrium, and the increased pressure on the system due to the decreasing of volume would also speed up the reaction, so I is also true. II is not true because the Equilibrium concentration of the products increases, not the reactants. So my final answer is I and III. Yet the answer key says I only. Why is that?
I think a good place to start is with the idea that that $K_{eq}$ will not change in value unless the temperature changes (see the second answer here for a good explanation). With this in mind we can say that statement III is false because the temperature is not being changed. 

Statement II is shown to be false via Le Chatelier's principle, as the equilibrium concentration of reactants should decrease rather than increase due to the decrease in volume because the products have fewer gaseous moles and so would exert less pressure on the container.

Statement I can be shown to be true via Le Chatelier's principle by the same logic as Statement II: for the reaction to shift towards the product, the rate of formation of product must increase. 

Therefore, only Statement I is true. 

You might wonder how $K_{eq}$ can stay constant, but the rate of forward reaction can increase. Its important to remember that $K_{eq}$ is the ratio of the forward and reverse rate constants, not the forward and reverse rates. The rate constants have no dependence on volume (or pressure) so they won't change.