The question is: The rate equation for the reaction A2 + B2 -> 2C (all gases) is rate = k[A2]2[B2]. If the gaseous reaction misture is compressed to half its original volume, by what factor will the reaction rate change. Assume temperature is constant. My thoughts are that the reaction rate wouldn't change. I know the rate changes if the concentrations of any of the reactants do, which is quite simple to calculate, but I'm not totally sure on pressure's effect on reaction rate. Any pushes in the right direction would be great.