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Gr. 12 Chemical Systems and Equilibrium Cheat Sheet by

Introd­uction

Equili­brium = the point in a chemical reaction where the reactants and the products are formed and broken at the same rate
Dynamic equili­brium = a balance between the forward and backward rates that are occurring simult­ane­ously
Equili­brium law = mathem­atical descri­ption of a chemical system at equili­brium
Equili­brium constant (K or K
eq
) = the numerical value defining the equili­brium law for a system at a given temper­ature (changes with temper­ature)
Hetero­geneous equili­brium = products and reactants are in at least 2 different states; pure solids­/li­quids are not included in K
eq
formula

Equili­brium Constant (Keq)

K
eq
formula:
If a[A] + b[B] ⇌ c[C] + d[D]; then
K
eq
= ([C]c[D]d) ÷ ([A]a[B]b)
Magnitude of K
eq
: states whether the equili­brium position favours produc­ts/­rea­ctants
If K = 1
[products] = [react­ants]
If K1
[products] [react­ants]
If K1
[products] [react­ants]
K
forward
vs K
backward
If a[A] ⇌ b[B], then
K
forward
= ([B]b) ÷ ([A]a)
K
backward
= ([A]a) ÷ ([B]b)
So ∴ K
forward
= 1/
Kbackward
 @ equili­brium
Purpose of K
eq
: to determine equili­brium concen­tration of chemical entities given initial conditions (I.C.E. table)

Reaction Quotient (Q)

Helps to determine the position of the equili­brium of a system using the rate law for the system and comparing it with the K
eq
If Q < K
eq
               
[products] < [reactants]
Reaction has not reached ⇌ yet; reaction needs to shift right
If Q > K
eq
[products] > [reactants]
Reaction has not reached ⇌ yet; reaction needs to shift left
If Q = K
eq
[products] = [reactants]
Reaction has not reached equili­brium yet; no shift will occur

Variables Affecting Chemical Equilibria

Le Châtel­ier's Principle: When a chemical system at equili­brium is disturbed by a change in property, the system responds in a way that opposes the change
Concen­tra­tio­n/T­emp­erature
[conc]/T = shift to consume
[conc]/T = shift to replace
If you add more reacta­nt/heat to a system, the system will consume it to make more product, and vice versa
If you remove reacta­nt/heat from a system, the system will replace it from the existing product, and vice versa
Volume­/Pr­essure (gases only)
V =P = shift toward side with more gas entities (i.e. more mol of gas) (more space for particles)
V =P = shift toward side with fewer gas entities (i.e. less mol of gas) (less space for particles)
Cataly­sts­/Inert (noble) gases No effect
 

Equilibria of Slightly Soluble Compounds

Molar solubility: The amount of solute (in mol) that can be dissolved in 1 L of solvent at a certain temper­ature
T α solubi­lity: as temper­ature increases, so does molar solubility
All compounds have some solubility, even if they are considered "­ins­olu­ble­" (really very low molar solubility)
Very soluble compounds (high molar solubi­lity) = no ⇌ (complete disass­oci­ation into ions)
Slightly soluble compounds (low molar solubi­lity) = has ⇌ (incomp­let­e/p­artial disass­oci­ation into ions)
As a compound is placed in a solvent, some part of it will dissolve, but at the same time, the reverse reaction starts
Eventually the two reactions reach equili­brium, creating a saturated solution (at this point, [conc] of ions remains constant) solubility equili­brium
Equili­brium formula for solubi­lity:
If x[A
a
B
b
] 
(s)
⇌ a[Ab+] 
(aq)
+ b[Ba-] 
(aq)
, then
K
sp
= [Ab+]a[Ba-]b
K
sp
(Solub­ility product constant): the product of the [conc] of ions in a saturated solution
If K
sp
1
Amount (mol) of aqueous ionsamount (mol) of solid substance
If K
sp
1
Amount (mol) of aqueous ionsamount (mol) of solid substance
Molar solubility and the K
sp
describe the solubility of a substance in different ways, meaning you can use one to solve for the other (using an ICE table)
Using Q and K
sp
to predict precip­itate formation
If Q < K
sp
                             
Not enough ions in solution (unsat­ura­ted); no precip­itate; reaction will shift to the right
If Q > K
sp
Lots of ions in solution (satur­ated); precip­itate can form; reaction will shift to the left
If Q = K
sp
System is at ⇌; no precip­itate (saturated solution)

pH and pOH

pH/pOH: The measure of acidit­y/a­lka­linity of a solution
pH: The measure of [H+] in a solution
pH = -log[H+]
= -log[H
3
O+]
[H
3
O+] = [H+]
= 10-pH
pOH: The measure of [OH¯] in a solution
pH = -log[OH¯]
[OH¯] = 10-pOH
The pH and pOH values are related to the exponent of K
w
(14):
pH + pOH = 14
 

Acids and Bases

Arrhenius acid/base = solution that ionizes into H+ (acid)/OH¯ (base) ions
Bronst­ed-­Lowry acid/base = solution that donates (acid)/receives (base) H+ ions
Strong acid/base: solution that completely ionizes (acid)/disass­ociates (base) into ions
Weak acid/base: solution that partially ionizes (acid)/disass­ociates (base) into ions
Monoprotic acids: acids that donate one H+ ion
Polyprotic acids: acids that donate more than one H+ ion
(diprotic = 2 H+, triprotic = 3 H+, etc.)
Amphip­rotic substance: substance that can behave like an acid or as a base (i.e. can donate and receive H+ ions)
Neutra­liz­ation reactions
With strong acids/­bases:
acid + base salt + water
*Complete ioniza­tion, so no equili­brium analysis*
With weak acids/­bases:
acid + base ⇌ conjugate base + conjugate acid
*Partial ioniza­tion, so equili­brium has to be analyzed*
Acid/Base Constants (K
a
and K
b
)
K
a
/K
b
indicate the strength of an acid/base
If K
a
/K
b
1                
Strong acid/base (complete ioniza­tio­n/d­isa­sso­cia­tion)
If K
a
/K
b
1
Weak acid/base (partial ioniza­tio­n/d­isa­sso­cia­tion)
If one K is known, the other can be determined using K
w
through the formula:
K
w
= K
a
 (of acid) ⋅ K
b
 (of conjugate base)
K
w
= K
b
 (of base) ⋅ K
a
 (of conjugate acid)

Autoio­niz­ation of Water and Water Constant (Kw)

Water can dissociate into ions on its own:
H
2
O
(l)
⇌ H+
(aq)
+ OH¯
(aq)
But the H+ ion can also attack other H
2
O molecules:
H
2
O
(l)
+ H+
(aq)
⇌ H
3
O+
(aq)
1
Adding both reactions together:
2 H
2
O
(l)
⇌ H
3
O+
(aq)
+ OH¯
(aq)
All equilibria have a constant (K) value, therefore:
K
w
 = [H
3
O+][OH¯]
(H
2
O
(l)
not included because it is not 
(aq)
)
Since water is neutral (pH = 7):
[H+] = 1.0⋅10-7 [H
3
O+] = 1.0⋅10-7
pH + pOH = 14 pOH = 7
[OH¯] = 1.0⋅10-7
If [H
3
O+] = 1.0⋅10-7, and [OH¯] = 1.0⋅10-7, then:
K
w
 = (1.0⋅10-7)(1.0⋅10-7)
= 1.0⋅10-14*
[1]H
3
O+
(aq)
: Hydronium ion

* Value of K
w
is always 1.0⋅10-14
                                                   
 

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  • Published: 25th January, 2018
  • Last Updated: 27th January, 2018
  • Rated: 4 out of 5 stars based on 1 ratings

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