dynamic equilibrium …going back and forth… …at the same time… …at the same rate…
TRANSCRIPT
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Dynamic Equilibrium
…going back and forth……at the same time…
…at the same rate…
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What happens in a reversible reaction?
Consider: aW + bX ↔ cY + dZ
Where
W reacts with X to produce Y and Za, b, c, and d are the coefficients of the
balanced equation
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What happens in a reversible reaction?
Consider: aW + bX ↔ cY + dZ
1) W is mixed with X and begins to react quickly
W and X are at maximum concentrations Y and Z are not present at the beginning
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What happens in a reversible reaction?
Consider: aW + bX ↔ cY + dZ
2) The “forward” reaction (W + X) continues, but is slowing
3) When enough product is formed, the “reverse” reaction begins
W and X concentrations are decreasing Y and Z concentrations are increasing
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What happens in a reversible reaction?
Consider: aW + bX ↔ cY + dZ
When the rate of the forward reaction is equal to the rate of the reverse reaction, the system has reached “dynamic equilibrium”
RateFWD = RateREV
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Dynamic Equilibrium
The reaction does not stop! Products are still being formed Products are still combining to
reform the reactants Nothing “appears” to be happening
Concentrations stop changing Color changes cease, etc…
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Graphing R P
time
[conc.]
Red = reactants Blue = products
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Dynamic Equilibrium
The only thing “equal” about equilibrium are the rates of the forward and reverse reactions
Equilibrium might be reached when there is mostly products, mostly reactants, or maybe a 50/50 mix of both
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Dynamic Equilibrium
Note: the initial amounts of reactants and products do not matter
Example: N2O4(g) ↔ 2 NO2(g)
Start with only [N2O4] …..
at eq., [N2O4] = 0.83M and [NO2] = 0.33M
Start with only [NO2]….
at eq., [N2O4] = 0.83M and [NO2] = 0.33M
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Dynamic Equilibrium
Quantitatively the situation at equilibrium can be expressed using the reactant and product concentrations in the “mass-action expression”
Consider: aW + bX ↔ cY + dZ
ba
dc
XW
ZYQ][][
][][
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Dynamic Equilibrium
Consider: aW + bX ↔ cY + dZ
At equilibrium, because the concentrations stop changing, Q becomes constant and is replaced by K – the equilibrium constant.
ba
dc
XW
ZYK][][
][][
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Dynamic Equilibrium
Note : 1) [products] on top and [reactants]
on bottom2) Coefficients become exponents
ba
dc
XW
ZYK][][
][][
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Dynamic Equilibrium
K = the “equilibrium constant” K is unitless
K is temperature dependentas long as the temperature is constant, so is K for a reversible reaction
ba
dc
XW
ZYK][][
][][
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Dynamic Equilibrium
Note : aW(aq) + bX(s) ↔ cY(aq) + dZ(l)
Pure substances (solids, liquids) do not have a changeable [molarity] and so drop out of the equilibrium expression
a
c
W
YK][
][
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K – the equilibrium constant
The size of K tells us something about the equilibrium “position” i.e. what are the concentrations of the
reactants and products at equilibrium?Because the products are in the
numerator, the larger the K value, the more products that are present at equilibrium and the fewer reactants that remain.
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K – the equilibrium constant
General rule: K<10-4
the equilibrium mixture is mostly reactants
The reaction does not “proceed” very far forward in order to reach equilibrium
The smaller K is, the fewer products formed
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K – the equilibrium constant
General rule: 10-4<K<104
the equilibrium mixture has substantial amounts of reactants and products
Not necessarily a “50/50” mix, but reasonably similar amounts of reactants and products
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K – the equilibrium constant
General rule: K>104
the equilibrium mixture is mostly products
The larger K gets, the more the forward reaction “goes to completion”
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Dynamic Equilibrium
…going back and forth……at the same time…
…at the same rate…
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Dynamic Equilibrium
…going back and forth……at the same time…
…at the same rate…
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LeChatelier’s Principle
If a system at equilibrium is disturbed it will respond in the direction that counteracts the disturbance and re-establishes equilibrium
Disturbed(?) 1. add/remove a chemical2. change temperature3. change pressure (gases) 4. add/remove catalyst
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1. adding/removing a chemical
If you add a chemical, the system tries to “remove” it
This is done by reacting it awayThis uses up the chemicals on its “side”
of the equation and making more of the chemicals on the other “side”
Equilibrium is re-established (Q =K), but the individual concentrations are different
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1. adding/removing a chemical
Consider: A + B ↔ C + D If you add more A… The system tries to remove it by reacting it
away, which makes more products [C] [D] [B] ↓ It is said the equilibrium has “shifted to the
right” or “shifted towards the products”
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1. adding/removing a chemical
Consider: A + B ↔ C + D If you add more C… The system tries to remove it by reacting it
away, which makes more reactants [A] [B] [D] ↓ It is said the equilibrium has “shifted to the left”
or “shifted towards the reactants”
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1. adding/removing a chemical
Consider: A + B ↔ C + D If you remove some B… The system tries to replace it by reacting to
make more of it (and whatever else is on its side of the equation)
[A] [C] ↓ [D] ↓ It is said the equilibrium has “shifted to the left”
or “shifted towards the reactants”
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1. adding/removing a chemical
Consider: A + B ↔ C + D If you remove some D… The system tries to replace it by reacting to
make more of it (and whatever else is on its side of the equation)
[C] [A], [B] ↓ The reaction is driven forward in this case, or
towards the products
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Just how does one “remove” a chemical?
Note: [D] is the concentration of DSolids do not have a molarity because
they are not dissolved into anythingIf one product in an aqueous system is a
solid, the solid is called a “precipitate” and is not in the equilibrium mixture
This “drives” the reaction forwardDouble replacements
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Just how does one “remove” a chemical?
Same for gases in an open containerThey can bubble out of the mixture
(leave)Ex: opening a soda bottle
H2CO3(aq) ↔ H2O(l) + CO2(g)
Ex: Mg(s) + 2 HCl(aq) ↔ H2(g) + MgCl2(aq)
If the container is open, the reaction just keeps going forward
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2. Changing the volume
Remember Boyle’s LawChanging the volume of a container of
gases changes their pressure as well Inverse relationship If V↓, P If V , P↓
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2. Changing the volume
If V↓, PThe equilibrium will shift to try to make
the P↓How is this done?Shift to whichever side has less gas
Fewer moles of a gasCoefficients in the balanced equation
Less gas means lower pressure
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2. Changing the volume
Ex: N2(g) + 3 H2(g) ↔ 2 NH3(g)
4 moles of gas in the reactants, 2 in products
If V↓, P…the system will try to make P↓ by shifting to the products (less gas)
Every time the reaction proceeds forward, 4 moles of gas becomes 2…which means the P↓
Vice versa if V
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3. Changing the temperature
Consider : A + B ↔ C + D + Heat For this system…
The forward reaction is exothermic The reverse reaction is endothermic
Treat heat as if it were a substance being added or removed
Add heat, equilibrium shifts away from the side with heat [A],[B] [C],[D]↓
Remove heat, equilibrium shifts toward the side with heat [A],[B]↓ [C],[D]
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4. Catalytic effect
Adding or removing a catalyst has no effect on the value of K
The activation energy is lowered for the forward and the reverse reaction, and they both speed up by the same amount, so RateFWD still = RateREV
If not at equilibrium, it will be reached quicker is a catalyst is used.
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LeChatelier’s Principle
If a system at equilibrium is disturbed it will respond in the direction that counteracts the disturbance and re-establishes equilibrium
Disturbed(?) 1. add/remove a chemical2. change temperature3. change pressure (gases) 4. add/remove catalyst
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Now…lets try it out…
Ex: PCl3(g) + Cl2(g) ↔ PCl5(g) ΔH= -88kJ
How will [Cl2] at equilibrium be changed by…
Adding some PCl3?
System will try to react the PCl3 away
More products are formedTo do this, more Cl2 is consumed
[Cl2] ↓
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Now…lets try it out…
Ex: PCl3(g) + Cl2(g) ↔ PCl5(g) ΔH= -88kJ
How will [Cl2] at equilibrium be changed by…
Adding some PCl5?
The system will try to react it awayMore reactants are formed
[Cl2]
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Now…lets try it out…
Ex: PCl3(g) + Cl2(g) ↔ PCl5(g) ΔH= -88kJHow will [Cl2] at equilibrium be changed by…Increasing the temperature?Heat is added, the system shifts away
from the side with heatΔH = negative, so heat is a productSystem shifts towards the reactants
[Cl2]
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Now…lets try it out…
Ex: PCl3(g) + Cl2(g) ↔ PCl5(g) ΔH= -88kJHow will [Cl2] at equilibrium be changed by…Decreasing the volume?If V↓, P System shifts towards the side with less
gas to make P↓Product side has fewer moles of gas
[Cl2] ↓