Equilibrium effects Flashcards
What will increasing the concentration of H+ do?
2 CrO4 2− + 2 H+ ⇌ Cr2O7 2− H2O
The addition of H+ increases it’s concentration; this increases the collision frequency between the reactant particles, in turn immediately increasing forwards ROR. There is no change in the initial reverse rate. [IMMEDIATE spike for H+, GRADUAL decrease in CrO4 2-]
As products, CrO72- and H2O, are now produced faster, the reverse rate of reaction gradully increases. [GRADUAL increase in CrO72- and H2O]
Because the forwards rate of reaction increased before the reverse rate did, the forwards rate overtook the reverse, shifting equilibrium right; this makes the solution appear more orange.
NOTE: Solution does not appear more yellow because it wasn’t chromate that was originally added, only H+ which had no affect on colour. After equilibrium is re-established, CrO7 2- is the only ion that has seen a net increase in concentration; CrO42- has in fact seen a decrease.
What will adding some NaOH do?
2 CrO4 2− + 2 H+ ⇌ Cr2O7 2− H2O
NaOH ionises completely in water to produce OH- ions, which neutralise the H+ ions in the solution, thereby decreasing their concentration. This decreases the rate of collisions between the reactant particles H+ and CrO42-, which in turn immediately decreases the forwards rate of reaction. There is no change in initial reverse rate. [IMMEDIATE drop in H+, GRADUAL drop in CrO42-]
As products are produced at a slower rate the reverse rate of reaction gradually decreases. [GRADUAL decrease in CrO72- and H2O].
Because the forwards rate of reaction decreased before the reverse rate did, the reverse rate overtook the forwards, shifting equilibrium left; this makes the solution appear more yellow.
What would happen if more N2 gas was added to this system?
N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g)
The addition of N2 gas increases the gases partial pressure, meaning there are a greater number of N2 particles per unit volume; this in turn increases the rate of collisions between reactant particles, immediately increasing the forwards rate of reaction. There is no change in the initial reverse rate of reaction. [IMMEDIATE spike in N2, gradual decrease for H2]
As the product is produced at a faster rate, the reverse rate of reaction gradually increases. [GRADUAL increase in NH3]
Because the forwards rate of reaction increased before the reverse rate did, the forwards rate of reaction overtook the reverse, shifting equilibrium right.
NOTE: Don’t have to account for the initial spikes in both rates like you do with total pressure. This is the same as concentration, but for gaseous reactants. Just like dilutions are volume changes for aqueous reactants.
What would be the effect of adding more solid sulphur
S2O32- (aq) + 2 H+ (aq) ⇄ S (s) + H2SO3 (aq)
There would be no effect. Sulphur is a solid with a fixed concentration. Adding it has no effect on the species which do have variable concentration. There is already unreacted suplhur since this is an equilibrium system. Now there is just more sitting on the bottom. Adding more will not affect the reaction rate in either direction. Therefore equilibrium experiences no change.
What would be the effect of adding O2 gas
N2 (g) + 3 H2 (g) ⇄ 2 NH3 (g)
O2 is an inert gas because it is not part of the system that doesn’t participate in the system, so the concentrations of gases that do participate in the reaction remain unchanged. Therefore it has no effect on the reaction rate in either direction. Equilibrium experiences no change.
What would be the effect of decreasing the volume
2 NO2 (g) ⇆ N2O4 (g)
Decreasing the volume would increase the pressure of the system. This would increase collision frequency of all particles because there are more particles per volume; in turn, this increases the forwards and reverse reaction rate instantly. [IMMEDIATE spike in partial pressure of both gases, but one more than the other]
However, the forwards rate of reaction is increased to a greater extent, because there are more molecules colliding, seen by the greater molar coefficient on this side.
This results in the forwards rate overtaking the reverse, shifting equilibrium right. This makes the solution appear more colourless.
What would be the effect of increasing the volume
2 NO2 (g) ⇆ N2O4 (g)
Increasing the volume would decrease the pressure of the system. This would decrease collision frequency of all particles because there are less particles per volume; in turn, this decreases the forwards and reverse reaction rate instantly. [IMMEDIATE drop in partial pressure of both gases].
However, the forwards rate of reaction is decreased to a greater extent, because it relies on more molecules colliding, seen by the greater molar coefficient on this side.
This results in the reverse rate overtaking the forwards, shifting equilibrium left. This makes the solution browner.
What would be the effect of adding water to this system?
2 CrO42- (aq) + 2 H+ (aq) ⇆ Cr2O72- (aq) + H2O (l)
NOTE: Water is a pure subtance with a fixed concentration. Adding more is a dilution, i.e a change in volume, rather than a concentration change.
Adding water would increase the total volume of the solution, decreasing the concentration of all ions present. This would decrease the collision frequency of all particles because there are less particles per volume; in turn, this decreases the forwards and reverse reaction rates instanty. [IMMEDIATE drop in concentration of all aqueous species]
However, the forwards rate of reaction is decreased to a greater extent, because it relies on more molecules colliding, seen by the greater molar coefficient on this side.
This results in the reverse rate overtaking the forwards, shifting equilibrium left. This makes the solution more yellow.
What would happen if you increased the temperature
2 NO2 (g) ⇆ N2O4 (g) + 57 kJ
Increasing the temperature would increase the average kinetic energy of particles in the system, which increases the frequency and proportion of successful particle collisions. This, in turn, this increases the forwards and reverse reaction rates instantly. [IMMEDIATE spike in partial pressure, GRADUAL change in conc]
However, the reverse rate of reaction is increased to a greater extent, because it has a larger activation energy barrier, and is therefore proportionally affected more.
This results in the reverse rate overtaking the forwards, shifting equilibrium left. This makes the solution browner.
What would happen if you decreased the temperature
2 NO2 (g) ⇆ N2O4 (g) + 57 kJ
Decreasing the temperature would decrease the average kinetic energy of particles in the system, which decreases the frequency and proportion of successful collisions. This in turn, this decreases the forwards and reverse reaction rates instantly. [IMMEDIATE drop in partial pressure, GRADUAL change in conc]
However, the reverse rate of reaction is decreased to a greater extent, because it has a larger activation energy barrier, and is therefore proportionally affected more.
This results in the forwards rate overtaking the reverse, shifting equilibrium right. This decreases the intensity of the brown color in the solution.
Determine whether reaciton was endo or exothermic
thicyanite reaction
Given equilibrium shifted left when temperature was increased
An increase in temperature increases the average kinetic energy of particles. This increases the frequency and proportion of successful particle collisions. This, in turn, immediately increases both the forwards and reverse rate of reaction. The endothermic rate is increased more proportionally because it has the larger activation energy barrier. Therefore, the reverse rate must be the endothermic reaction because based on our observations, the system shifted left. Therefore, the forwards reaction is exothermic.
Which effects give instant spikes/drops in both rates
Volume and temperature
The forwards and reverse rate of reaction instantly increase/drop, but one does so more
Which effects give gradual gain/drop in one rate
Concentration and partial pressure
The side with the species that is added will immediately increase. There is no change in initial reverse rate.
