Wednesday, February 08, 2006
Le Chat chat
AP: okay, here's a transcript of what I wanted to say today. I think that this will be a good Le Chatelier review of the temperature and pressure stresses:
inc P (or dec V) is the same as increasing the concentrations of any gases, so there is an increase in collision frequency and therefore a greater forward and reverse reaction rate (due to the greater number of molecules available for collision per unit of volume = inc concentration); an inc in conc of all gases on the side with MORE molecules causes a disproportionately greater inc in that side's reaction rate so there is a NET shift toward the side with FEWER molecules as the new equilibrium is reached.
dec P (or inc V) is the same as a decrease in concentration of all gases so there is a decrease in collision frequency and therefore a decreased forward and reverse reaction rate (due to the lower number of molecules available for collision per unit of volume = dec concentration); so, a dec in conc of all molecules causes a disproportionately greater decrease in reaction rate on the side with more molecules; thus, there is a NET shift towards the side with more molecules as the new equilibrium is reached.
inc T speeds up both forward and reverse reaction rates because there is increased collision frequency and a greater fraction of effective collisions (due to the higher T = higher avg. KE of the molecules in the system) but there will be a disproportionately larger increase in the NET "energy consuming", endothermic, direction.
dec T slows down forward and reverse reaction rates because there is decreased collision frequency and a decreased fraction of effective collisions due to the lower T = lower avg. KE of the molecules in the system) but there is disproportionately greater decrease in the endothermic direction so that there is a net shift to the EXOTHERMIC direction.
Of course, giving specific examples with made up numbers for the forward and reverse rates is the BEST thing that you can do because you can then quantitatively show towards which side (reactants or products) a NET shift occurs as the system proceeds towards the new equilibrium.
more stresses...
addition of a catalyst: a catalyst affects the orientation of the colliding reactant(s) by temporatily binding the reactant(s) in such a way that bonds are strained (and thus require less energy to break) or interparticle attractions are weakened (so the less energy is needed to overcome the attractions); thus, a catalyst lowers the activation energy for both the forward and reverse reaction. Catalysts lower the activation energy of both forward and reverse reactions EQUALLY. Therefore, though both the forward and reverse reaction rates increase (because, at the same temperature, a greater fraction of reactant particles have enough kinetic energy for an effective collision due to the activation energy- lowering effect of the catalyst), there is NO NET shift towards the reactants or products because both rates are increased EQUALLY.
addition of an INERT gas:
addition of any non-reacting substance will NOT affect the equilibrium concentrations of the gases (or anything) as long as the system is at CONSTANT VOLUME!!! This is because, at constant volume, even if you add many many moles of an inert gas, the PARTIAL PRESSURES and, thus, the CONCENTRATIONS of the reactant and product gases remain CONSTANT because the number of MOLES per LITER of the reactant gases is NOT CHANGING as the inert gas is added.
BUT, if an inert gas is added at constant TOTAL pressure i.e. the VOLUME expands, the effect is to LOWER the PARTIAL PRESSURES/concentrations of the reactant and product gases (initially). So, just treat that situation as an INCREASED VOLUME or DECREASED PRESSURE stress. Thus, the equilibrium will shift to the side with a greater number of moles of gaseous molecules (in the balanced equation) due to the decrease in BOTH the forward and reverse reaction rates but a disproportionately greater decrease in the reaction involving the side with a greater number of moles of gaseous molecules.
inc P (or dec V) is the same as increasing the concentrations of any gases, so there is an increase in collision frequency and therefore a greater forward and reverse reaction rate (due to the greater number of molecules available for collision per unit of volume = inc concentration); an inc in conc of all gases on the side with MORE molecules causes a disproportionately greater inc in that side's reaction rate so there is a NET shift toward the side with FEWER molecules as the new equilibrium is reached.
dec P (or inc V) is the same as a decrease in concentration of all gases so there is a decrease in collision frequency and therefore a decreased forward and reverse reaction rate (due to the lower number of molecules available for collision per unit of volume = dec concentration); so, a dec in conc of all molecules causes a disproportionately greater decrease in reaction rate on the side with more molecules; thus, there is a NET shift towards the side with more molecules as the new equilibrium is reached.
inc T speeds up both forward and reverse reaction rates because there is increased collision frequency and a greater fraction of effective collisions (due to the higher T = higher avg. KE of the molecules in the system) but there will be a disproportionately larger increase in the NET "energy consuming", endothermic, direction.
dec T slows down forward and reverse reaction rates because there is decreased collision frequency and a decreased fraction of effective collisions due to the lower T = lower avg. KE of the molecules in the system) but there is disproportionately greater decrease in the endothermic direction so that there is a net shift to the EXOTHERMIC direction.
Of course, giving specific examples with made up numbers for the forward and reverse rates is the BEST thing that you can do because you can then quantitatively show towards which side (reactants or products) a NET shift occurs as the system proceeds towards the new equilibrium.
more stresses...
addition of a catalyst: a catalyst affects the orientation of the colliding reactant(s) by temporatily binding the reactant(s) in such a way that bonds are strained (and thus require less energy to break) or interparticle attractions are weakened (so the less energy is needed to overcome the attractions); thus, a catalyst lowers the activation energy for both the forward and reverse reaction. Catalysts lower the activation energy of both forward and reverse reactions EQUALLY. Therefore, though both the forward and reverse reaction rates increase (because, at the same temperature, a greater fraction of reactant particles have enough kinetic energy for an effective collision due to the activation energy- lowering effect of the catalyst), there is NO NET shift towards the reactants or products because both rates are increased EQUALLY.
addition of an INERT gas:
addition of any non-reacting substance will NOT affect the equilibrium concentrations of the gases (or anything) as long as the system is at CONSTANT VOLUME!!! This is because, at constant volume, even if you add many many moles of an inert gas, the PARTIAL PRESSURES and, thus, the CONCENTRATIONS of the reactant and product gases remain CONSTANT because the number of MOLES per LITER of the reactant gases is NOT CHANGING as the inert gas is added.
BUT, if an inert gas is added at constant TOTAL pressure i.e. the VOLUME expands, the effect is to LOWER the PARTIAL PRESSURES/concentrations of the reactant and product gases (initially). So, just treat that situation as an INCREASED VOLUME or DECREASED PRESSURE stress. Thus, the equilibrium will shift to the side with a greater number of moles of gaseous molecules (in the balanced equation) due to the decrease in BOTH the forward and reverse reaction rates but a disproportionately greater decrease in the reaction involving the side with a greater number of moles of gaseous molecules.
# posted by C @ 12:56 AM 
