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1. The Equilibrium Law:
- Keq is called the _________ _________.
eg. aA + bB + … ⇔ xX + yY + …
where
- a, b, x, y are _________ of the balanced equation.
- A,B,X & Y are the _________ and _________ in the equation
- The value of Keq does not change for a given rxn at a given _________.
- The factor representing any substance whose conc cannot change is always 1 and there fore not included (i.e. a solid or a pure liquid).
o i.e. if H 2 O is produced then [ X ]
x = 1 and therefore not written
2. The Meaning of K eq:
- A small Keq value means that the rxn is mostly in the form of _________
- A large “ “ “ “ “ “ “ “ “ “ “ “ _________.
Keq = [PRODUCTS] [REACTANTS]
Comparison
Small K (^) eq [1]/[10] = 0.1^ reactants > products Large K (^) eq [10]/[1] =10^ reactants < products K (^) eq = 1 [10]/[10] =1^ reactants = products
- The value of Keq changes when we… a) _________ rxns b) change the _____________ of a given rxn c) change the _________ of the equation.
3. Building K eq equations
The Equilibrium Constant (K eq)
Keq = [PRODUCTS] [REACTANTS]
Keq = [X]
x
[Y]
y
[A]
a
[B]
b
Reaction 1
N2(g) + 3H 2(g) ⇔ 2NH 3(g)
Reaction 2 (reaction 1 reversed)
2NH3(g) ⇔ N2(g) + 3H 2(g)
Reaction 3 (reaction 1 doubled)
2N2(g) + 6H 2(g) ⇔ 4NH 3(g) ,
4. Keq with Solids and Liquids
When we write the Keq expression for a rxn with solids or pure liquids , we simply leave them ___
E.g. CaCO3(s) CaO(s) + CO2(g)
Keq = [CaO(s) ] [CO2(g) ] therefore = [CO2(g) ] [CaCO3(s) ]
e.g. CaCO3(s) + 2HF(g) CaF2(s) + H 2 O(l) + CO2(g)
Keq = __________________
Keq = [ NH 3 ] 2
[ N 2 ] [ H 2 ] 3
Keq = [ N 2 ] [ H 2 ]
3
[ NH 3 ] 2
Keq = [ NH 3 ] 4
[ N 2 ] 2 [ H 2 ] 6
Therefore
- Whenever a rxn is reversed, its Keq value becomes the _________ of the original value
- “ “ “ “ doubled, “ “ “ “ “ _________ “ “ “ “
Definition p. 59
- Recall, a catalyst does NOT shift equilib
- Therefore No shift, Therefore NO change in Keq
8. Effect of ∆Pressure or/ ∆Volume on Keq
- if ∆Volume or ∆Pressure causes equilib shift, we see the same effect as ∆conc. on Keq
- if no ∆T, no ∆ Keq
- A change in total volume or total pressure does not change the value of the equilib constant Keq. The equilib will shift to keep the ratio equal to Keq
You will have to know how to calculate 5 different types of Keq questions
Type 1 - Find K eq , given equilib concs
Given the equilib system: PCl (^) 5(g) PCl3(g) + Cl (^) 2(g)
At equilib: [PCl 5 ] = 0.32 M, [PCl 3 ] = 0.40 M and the [Cl 2 ] = 0.40 M.
Keq = [PCl 3 ][Cl 2 ] = __________________ = _________ [PCl 5 ]
Type 2 – Find Equilib. Concs, given K eq
N2(g) + 3H 2(g) 2NH 3(g) Keq = 625
Calculate [H 2] if at equilib [N 2] = 0.030 M and [NH3] = 0.12 M
a) Write K eq expression:
b) Plug in known values and solve:
Keq = _________ ↔ ______ = _________ ↔
[H2] = _________
Note : Keq does not include any _______
K (^) eq CALCULATIONS
Type 3 – Find equilib concs or Keq , given initial concs and one equilib conc
The ICE method
[I] = Initial concentrations
[C] = Change in concentration
[E] = Equilib concentrations
e.g. Put only N 2 and H 2 into container…. [I] : [N2] = 0.32 M and [H 2] = 0.66 M.
wait for a while……. [E] : [H2] = 0.30 M. [C] : ∆ [H2] = -0.36 M
Question: a) Find the equilib [N 2] and [NH3].
b) Calculate K eq
a) use “ ICE ” table and write what you know
N 2 + 3H 2 2NH 3
Initial conc. [I]
(change in conc.) [C]
Equilibrium conc [E]
b) calculate change in conc using mole ratios
∆ [N2] = (_____________) x ________ = ________
∆ [NH3] = (_____________) x ________ = ________
c) Do the math:
d) Find value of K eq
Keq = ___________ = ________ = ___________
Type 4 – Find Equilib shift, given initial concs and K eq
N2(g) + 3H2(g) 2NH 3(g)
since [H 2 ]↓ and [N 2 ] ↓ , [NH 3 ] must ↑
[C]
[E] formula
[E]
Plug into K eq equation:
Keq =^ ________^ =^ ______
[H 2][I 2]
∴ x = ______
Therefore at eqb
[HI] = 2x = 2 (_________) = __________ M
[H2] = [I 2] = _____________ = _____________ = _____ M
Answer check :
Keq = ____ = _______ = _______ ≈ 55.
(close enough)
The Equilibrium Constant (K eq)
1. The Equilibrium Law:
- Keq is called the equilib constant.
eg. aA + bB + … ⇔ xX + yY + …
where
- a, b, x, y are coefficients of the balanced equation.
- A,B,X & Y are the products and reactants in the equation
- The value of Keq does not change for a given rxn at a given temperature.
- The factor representing any substance whose conc cannot change is always 1 and there fore not included (i.e. a solid or a pure liquid).
o i.e. if H 2 O is produced then [ X ]
x = 1 and therefore not written
2. The Meaning of K eq:
- A small Keq value means that the rxn is mostly in the form of reactants
- A large “ “ “ “ “ “ “ “ “ “ “ “ products.
Keq = [PRODUCTS] [REACTANTS]
Comparison
Small K (^) eq [1]/[10] = 0.1 reactants > products Large K (^) eq [10]/[1] =10 reactants < products K (^) eq = 1 [10]/[10] =1 reactants = products
- The value of Keq changes when we… d) switch rxns e) change the temperature of a given rxn f) change the form of the equation.
3. Building K eq equations
Keq = [PRODUCTS] [REACTANTS]
Keq = [X]
x
[Y]
y
[A]
a
[B]
b
- Rxn has shifted right because Keq increases.
- Fwd rxn is endo thermic, because it decreases
When the temperature changes, the value of K (^) eq also changes
- When the temperature increases in an endothermic rxn, the equilib will shift to the right and the value of Keq will increases.
- When the temperature decreases in an endothermic rxn, the equilib will shift to the left and the value of Keq will decreases.
- When the temperature increases in an exothermic rxn, the equilib will shift to the left and the value of Keq will decrease
- When the temperature decreases in an exothermic rxn, the equilib will shift to the right and the value Keq will increases
6. Effect of ∆ Concentration on Keq
If T remains constant, changing the conc of reactants or products does not change the value for K (^) eq Why?
Eg. (at eqilib) A + B C + D Keq = [C] [D] = 4. [A] [B]
- Now add C, so [C] is greater and now Keq ≠ 4
- but equilib will shift left
- new equilib also has Keq = 4
Therefore As long as T is not changed, the equilib will always shift just enough to keep the ratio equal to the value of the equilib constant (Keq )!
7. Catalyst effect on Keq
- Recall, a catalyst does NOT shift equilib
- Therefore No shift, Therefore NO change in Keq
8. Effect of ∆Pressure or/ ∆Volume on Keq
Recall: “ Whatever we do, nature tries to undo”
IMPORTANT: temperature is the only factor that affects K (^) eq
if ∆Volume or ∆Pressure causes equilib shift, we see the same effect as ∆conc. on Keq if no ∆T, no ∆ Keq
A change in total volume or total pressure does not change the value of the equilib constant Keq.
The equilib will shift to keep the ratio equal to Keq.
You will have to know how to calculate 5 different types of Keq questions
Type 1 - Find K eq , given equilib concs
Given the equilib system: PCl (^) 5(g) PCl3(g) + Cl (^) 2(g)
At equilib: [PCl 5 ] = 0.32 M, [PCl 3 ] = 0.40 M and the [Cl 2 ] = 0.40 M.
Keq = [PCl 3 ][Cl 2 ] = [0.40][0.40] = 0. [PCl 5 ] 0.
Type 2 – Find Equilib. Concs, given K eq
N2(g) + 3H 2(g) 2NH 3(g) Keq = 625
Calculate [H 2] if at equilib [N 2] = 0.030 M and [NH3] = 0.12 M
a) Write K eq expression:
b) Plug in known values and solve:
Keq = [NH3]
2
↔ 625 = [0.12]
2
[H 2]
3
[N2] [H 2]
3
[0.030]
[H2] = 0.092 M
Type 3 – Find equilib concs or Keq , given initial concs and one equilib conc
The ICE method
[I] = Initial concentrations
Note : Keq does not include any units
K (^) eq CALCULATIONS
eg:
CO (^) (g) + H 2 O (^) (g) CO (^) 2(g) + H (^) 2(g) Keq = 10.
How will equilib shift if the initial concentrations = 0.80M CO, 0.050M H 2 O, 0.50M CO 2 , and
0.40M H 2.?
Trial Keq = [CO2] [H2] = [0.50][0.40] = 5.
[CO] [H 2O] [0.80][0.050]
but actual Keq = 10.
therefore, Trial K eq < K eq and equilib shifts right
Type 5 – Find [E] of all species, given [I] of all species and K eq
H 2 + I 2 2HI The Keq = 55.
[I]: [H2] = 0.200 M and [I 2] = 0.200 M
Step 1. Which way will rxn shift? Æ Start with only H 2 and I 2 so
Trial Keq = HI] 2 = [0] 2 = 0 [H 2 ][I 2 ] [0.200][0.200]
Actual K (^) eq = 55.6. Therefore Keq increases , so rxn must shift right!
let ∆[H 2 ] = x and ([C] for H 2 = -x so…
Fill in first 3 rows on table on next page
H 2 + I 2 2HI
[I] 0.200 0.200 0
[C] -x -x +2x
[E] formula 0.200 - x 0.200 - x 0 + 2x = 2x
[E] 0.0423 0.0423 0.
Plug into K eq equation:
-x -x +2x H 2 + I 2 2HI
K (^) eq = HI] 2 = 55. [H 2 ][I 2 ]
= [2x] 2 = 55. [0.200-x][0.200-x]
= √ [2x] 2 = √ 55. √ ][0.200-x] 2
= 2x = 7. 0.200 - x
∴ x = 0.
Therefore at eqb
[HI] = 2x = 2 (0.1577) = 0.315 M
[H 2 ] = [I 2 ] = 0.200 – x = 0.200 – 0.1577 = 0.0423 M
Answer check :
K (^) eq = HI] 2 = [0.315] 2 = 55.45 ≈ 55. [H 2 ][I 2 ] [0.0423] 2
(close enough)
