Wednesday, February 07, 2007
AP Le Chat chat
UPDATE: the Equilibrium Practice Tests that were posted on 020507 contained several ERRORS in the answer key; thanks to DM, we corrected the answer key at extra help; I just uploaded the corrected file which is still linked where it was originally.
I wrote this last year.
I think that this will be a good Le Chatelier review of the temperature, pressure, and inert substance stresses: Naturally, accompanying pictures and diagrams would greatly clarify and enhance the explanation and, perhaps, make the written part more concise.
inc P (or dec V), at constant T and n, is the same as increasing the concentrations of any gases. This occurs because, if you increase the external pressure on a system of gas(es) the piston of the container will be forced down because nothing initially occurs to the collision frequency of the gases (i.e. the internal pressure is initially the same). So there is a net force pushing the piston down which decreases the container volume which gradually increases the collision frequency of the gas particles with the container (the gas particles have the same average KE but they are moving in a smaller volume, which causes more frequent collisions). 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 GASEOUS 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), at constant n and T, is the same as a decrease in concentration (see above reasons) 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 GASEOUS 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 so a greater fraction of molecules meet or exceed the activation energy requirement for reaction) 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 (NET energy REQUIRING) 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 temporarily binding the reactant(s) in such a way that bonds are strained (and thus require less energy to break) or inter-particle 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. The number of effective collisions per second of both reactant and product particles remains the same because the inert gas both interferes with and CAUSES collisions to the SAME extent so that there is NO NET change in the effective collision rate.
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.
I wrote this last year.
I think that this will be a good Le Chatelier review of the temperature, pressure, and inert substance stresses: Naturally, accompanying pictures and diagrams would greatly clarify and enhance the explanation and, perhaps, make the written part more concise.
inc P (or dec V), at constant T and n, is the same as increasing the concentrations of any gases. This occurs because, if you increase the external pressure on a system of gas(es) the piston of the container will be forced down because nothing initially occurs to the collision frequency of the gases (i.e. the internal pressure is initially the same). So there is a net force pushing the piston down which decreases the container volume which gradually increases the collision frequency of the gas particles with the container (the gas particles have the same average KE but they are moving in a smaller volume, which causes more frequent collisions). 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 GASEOUS 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), at constant n and T, is the same as a decrease in concentration (see above reasons) 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 GASEOUS 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 so a greater fraction of molecules meet or exceed the activation energy requirement for reaction) 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 (NET energy REQUIRING) 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 temporarily binding the reactant(s) in such a way that bonds are strained (and thus require less energy to break) or inter-particle 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. The number of effective collisions per second of both reactant and product particles remains the same because the inert gas both interferes with and CAUSES collisions to the SAME extent so that there is NO NET change in the effective collision rate.
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 @ 10:25 AM 
