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```Module 5: Equilibrium
Static and Dynamic Equilibrium:
Students:
● conduct practical investigations to analyse the reversibility of chemical reactions, for example:
– cobalt(II) chloride hydrated and dehydrated
– iron(III) nitrate and potassium thiocyanate
– burning magnesium
– burning steel wool
● model static and dynamic equilibrium and analyse the differences between open and closed systems
● analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy, for example:
– combustion reactions
– photosynthesis
● investigate the relationship between collision theory and reaction rate in order to analyse chemical
equilibrium reactions
Reversible vs irreversible reactions:
Reversible:
Irreversible:
● model static and dynamic equilibrium and analyse the differences between open and closed
systems
Static equilibrium:
- The rate of the forward and reverse reactions are zero
- irreversible
Dynamic equilibrium:
- The rate of both the forward and reverse reactions are NOT zero
- Reversible
Open and closed systems
Closed systems:
- The system cannot exchange material with the
surroundings
- If a combustion reaction occurs within a closed
system, any products formed (such as H2O) must
stay within the system, however, the energy
released may transfer out of the system.
Open systems:
- The system can exchange materials or energy with
the surroundings.
If a combustion reaction occurs within an open
system, the heat will quickly flow into the
surroundings and water vapour will exit the reaction
vessel into the surroundings.
Dynamic equilibria:
Le chatelier’s principle:
The rate of the forward and reverse reactions are influenced by:
- The temperature of the system (via enthalpy of reaction)
- The pressure of the system (via number of moles)
- The concentration of reactants and products
Le chatelier’s principle states that a system in equilibrium will act in a way
to minimise changes to the system.
The Haber-Bosch Process
-∆H
● conduct practical investigations to analyse the reversibility of chemical reactions, for example:
cobalt(II) chloride hydrated and dehydrated:
Crimson coloured
Pink in solution
Blue in solution
Endothermic
iron(III) nitrate and potassium thiocyanate:
Yellow in solution
Crimson in solution
Exothermic
Burning magnesium:
Burning steel wool:
Entropy
• Refers to the number of ways energy can be distributed throughout a
system
• Denoted by letter ‘S’
• Gases have higher entropy than liquids and solids
• The more molecules the higher the entropy
∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = ∆𝑆𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 − ∆𝑆𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠
∆𝐻𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 =
𝑇
∆𝑆𝑡𝑜𝑡𝑎𝑙 = ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 ≥ 0
Enthalpy
• A measure of the internal energy of the system
• Standard enthalpy refers to the enthalpy of one mole of a substance
• Reactions result in a change in enthalpy, meaning that energy is
either:
- taken into the system from the surroundings
(positive ∆𝐻, endothermic)
- Or lost from the system and given to surroundings
(negative ∆𝐻, exothermic)
Combustion reactions
• These are highly exothermic reactions that involve oxygen
• Hence enthalpy change is negative
• If oxygen is limited then incomplete combustion can occur
• Entropy change for reactions bellow is positive as moles of gas
increases
Photosynthesis
• This is the process utilised by plants to synthesise glucose from carbon dioxide
and water, using light as an energy source.
• This reaction has a negative entropy change
• This reaction cannot occur spontaneously in a closed system as the entropy would
decrease. Plants however, are an open system that use a series of complex
mechanisms to allow this process to occur with an overall positive entropy change
(overall closed system includes the plant, the atmosphere, the soil and the sun)
• 2nd law of thermodynamics: ∆𝑆𝑢𝑛𝑖𝑣𝑒𝑟𝑠𝑒 ≥ 0
Collision Theory
• Collision theory is the theory that reactions occur via reactant molecules randomly
colliding with each other.
• Reactions only occur if reactant molecules collide and also have sufficient energy to
react
• This results in a dependence of reaction rate on:
- Rate of collisions (concentration and temperature influence this)
- Effectiveness of collisions (energy of molecules  temperature)
- Orientation of reactants (some molecules might have reactive sites blocked)
• Reactions can be unimolecular (rate dependent of one molecule): rate = k [A]
• Can be bimolecular (rate dependent on two molecules): rate = k [A]*[B] or rate = k[𝐴]2
Factors that effect equilibrium:
Students:
● investigate the effects of temperature, concentration, volume and/or
pressure on a system at equilibrium and explain how Le Chatelier’s principle
can be used to predict such effects, for example:
– heating cobalt(II) chloride hydrate
– interaction between nitrogen dioxide and dinitrogen tetroxide
– iron(III) thiocyanate and varying concentration of ions
● explain the overall observations about equilibrium in terms of the collision
theory
● examine how activation energy and heat of reaction affect the position of
equilibrium
Le Chatelier’s Principle
• Le chatelier’s principle states that a system in equilibrium will act in a
way to minimise changes to the system.
𝐶𝑜𝑙𝑜𝑢𝑟𝑙𝑒𝑠𝑠
𝐵𝑟𝑜𝑤𝑛
positive ∆𝐻, endothermic
Increased pressure will favour reverse reaction
Increased concentration of reactants will favour forward reaction
Increased temperature will favour forward reaction
How would changing conditions
affect these reactions?
cobalt(II) chloride hydrated and dehydrated:
Crimson coloured
Pink in solution
Blue in solution
Endothermic
iron(III) nitrate and potassium thiocyanate:
Yellow in solution
Crimson in solution
Exothermic
Activation energy and heat of reaction
• Activation energy is the energy required to initiate a reaction
∆𝐺
The lower the activation energy, the
greater the rate of reaction
 greater rate of reaction pushes
equilibrium
∆𝐺
∆𝐺 = ∆𝐻 − 𝑇∆𝑆
• Change in Gibbs energy is proportional to
the change in enthalpy and inversely
proportional to the change in entropy
• A reaction is spontaneous if the gibbs
energy is negative
• The energy released or absorbed is proportional to the heat of reaction (q) and enthalpy change of
reaction
• When energy is released the heat of reaction is positive and enthalpy change is negative (vice versa)
Calculating the equilibrium constant
Students:
● deduce the equilibrium expression (in terms of Keq) for homogeneous reactions occurring
in solution
● perform calculations to find the value of Keq and concentrations of substances within an
equilibrium system, and use these values to make predictions on the direction in which a
reaction may proceed
● qualitatively analyse the effect of temperature on the value of Keq
● conduct an investigation to determine Keq of a chemical equilibrium system, for example:
– Keq of the iron(III) thiocyanate equilibrium
● explore the use of Keq for different types of chemical reactions, including but not limited
to:
– dissociation of ionic solutions
– dissociation of acids and bases
The equilibrium expression:
For,
𝛼𝐴 + 𝛽𝐵
γ𝐶 + 𝛿𝐷
𝐾𝑒𝑞
[𝐶]𝑦 [𝐷]𝛿
=
[𝐴]𝛼 [𝐵]𝛽
𝐾𝑒𝑞
[𝑁𝐻3 ]2
=
[𝑁2 ][𝐻2 ]3
For example:
• Keq is unitless
Calculations of Keq
• For a Haber-Bosch process, the concentration of ammonia is 10.0 mol L-1
and the concentration for Nitrogen and hydrogen is 2.00 M and 3.00 M,
respectively. Find the equilibrium constant.
[𝑁𝐻3 ]2
𝐾𝑒𝑞 =
[𝑁2 ][𝐻2 ]3
𝐾𝑒𝑞
2
=
= 1.85 (3 sf)
3
[2.0][3.0]
Calculations of Keq: determining rxn direction
Lets say a reaction starts with an initial concentration of 0.500 M of H2O(g) and 0.500 M CO(g) with
no products yet formed. If the equilibrium constant is 5.90 at 25° C then what will be the equilibrium
concentration of the products (H2 and CO2) and the reactants at 25 ° C for this reaction?
The ICE table:
H2O(g)
+
CO(g)
H2(g)
+
CO2(g)
Initial conc.
0.5
0.5
0
0
Change in conc.
-x
-x
+x
+x
Equilibrium conc.
0.5-x
0.5-x
x
x
𝑥 [𝑥]
𝑥2
𝐾𝑒𝑞 =
= 5.9 =
0.5 − 𝑥 [0.5 − 𝑥]
[0.25 − 𝑥 + 𝑥 2 ]
𝑥 2 = 5.9 0.25 − 𝑥 + 𝑥 2 = 1.475 − 5.9𝑥 + 5.9𝑥 2
4.9𝑥 2 − 5.9𝑥 + 1.475 = 0
−𝑏 ± 𝑏2 − 4𝑎𝑐 −(−5.9) ± (−5.92 −4 4.9 ∗ 1.475
𝑥=
=
= 0.850 𝑜𝑟 0.354
2𝑎
2(4.9)
As it cannot change by a negative number and cannot change more than the initial concentration, x = 0.354
Effect of temperature on Keq
• Increased temperature will increase the equilibrium constant if the
reaction is endothermic (positive ∆𝐻)
• Increased temperature will decrease the equilibrium constant if the
reaction is exothermic (negative ∆𝐻)
EXTRA INFO:
The qualitative analysis above can be explained quantitatively
by the below equation:
∆𝐻° ∆𝑆°
ln 𝐾 = −
+
𝑅𝑇
𝑅
Iron (III) thiosulfate equilibrium experiment
Step 1: Create five reference solutions made up of excess Fe3+ (from 0.2M Fe(NO3)3) and known concentrations of SCN–
(from 0.002M KSCN), making up a total volume of 10ml.
Step 2: Calculate the concentration of SCN– and therefore FeSCN2+ ions after dilution by adding iron (III) nitrate solution.
(using C1V1 = C2V2 dilution formula)
Step 3: Pour one reference solution into a cuvette (small rectangular tubular container) to place in the
spectrophotometer to calculate absorbance. Rinse cuvette thoroughly with water and dry it before measuring the
absorbance of the next reference solution.
Step 4: Draw a calibration curve using the absorbance of each reference solution against the concentration of
FeSCN2+ ions.
Step 5: Create test solutions by adding a known quantity of Fe(NO3)3 solution and varying quantity of KSCN solution and
water for a total of 10ml. Adding water helps ensure that the absorbance reading will be within the scale of your
calibration curve so you avoid extrapolation! (This will help improve accuracy of results)
Step 6: Calculate the concentration of Fe3+ and SCN– ions in each test solution.
Step 7: Pour one test solution into a cuvette and into the spectrophotometer to calculate absorbance of test solution.
Rinse the cuvette thoroughly with water and dry it before measuring the absorbance of the next test solution.
Step 8: Determine the concentration of FeSCN2+ for each test solution using the absorbance results and the calibration
curve.
Step 9: Calculate the equilibrium constant for each of the test solution, average the Keq for each test solution to find the
average Keq of the iron (III) thiocyanate solution and average deviation of the average Keq value. This is done by using the
results obtained in Step 6 (INITIAL concentrations of Fe3+ and SCN–) and Step 8 (EQUILIBRIUM concentration of FeSCN2+)
and the ICE table to compute the EQUILIBRIUM concentrations of Fe3+ and SCN– ions.
Dissociation of acids/bases in solution
The dissociation an acid into H+ and its conjugate base is an equilibrium reaction
(except for strong acids that fully dissociate).
This dissociation has an associated equivalent of the equilibrium constant called
the acid dissociation constant (Ka)
E.g. for acetic acid:
𝐻 + [𝐶𝐻3 𝐶𝑂𝑂− ]
𝐾𝑎 =
[𝐶𝐻3 𝐶𝑂𝑂𝐻]
The same process occurs for bases, except –OH ions and the conjugate acid are
released into solution. The constant is called the base dissociation constant (Kb)
E.g. for NaOH
𝐻𝑂− [𝑁𝑎+ ]
𝐾𝑏 =
[𝑁𝑎𝑂𝐻]
Dissociation of ionic compounds in solution
• The dissociation of ionic compounds is an equilibrium process with a
constant called the solubility product equilibrium constant (Ksp). Only
the product concentrations are included as the solid reactant does
not have a concentration.
e.g. for AgCl (s):
𝐾𝑠𝑝 = 𝐴𝑔+ [𝐶𝑙 − ]
• If the compound is completely soluble it will not follow an equilibrium
as the compound fully dissociates.
Solution Equilibria
Students:
● describe and analyse the processes involved in the dissolution of ionic compounds in water
● investigate the use of solubility equilibria by Aboriginal and Torres Strait Islander Peoples when
removing toxicity from foods, for example:
● conduct an investigation to determine solubility rules, and predict and analyse the composition of
substances when two ionic solutions are mixed, for example:
– potassium chloride and silver nitrate
– potassium iodide and lead nitrate
– sodium sulfate and barium nitrate
● derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility
of an ionic substance from its Ksp value
● predict the formation of a precipitate given the standard reference values for Ksp
Dissolution of ionic compounds in water
• Dipole-Dipole interactions of
water (H bonding) overcome by
electrostatic attraction to ions
• Electrostatic interaction between
Na+ & Cl- overcome by waters
electrostatic attraction to the
ions.
● investigate the use of solubility equilibria by Aboriginal and Torres Strait Islander Peoples when
removing toxicity from foods, for example:
Cycad fruit seeds contain toxins such as cycasin
and macrozamin. These toxins form equilibrium
between the solid and soluble state.
Aboriginal peoples used to eat the seeds of
cycad fruits such as Macrozamia. These seeds
are surrounded by a fleshy layer called
sarcotesta.
A process called leeching involved soaking the
seeds in the water of shallow lakes. As the
toxins solubility is a dynamic equilibrium and
the leeching occurs in an open system the toxins
fully dissolve.
Solubility Rules
N – Nitrates always soluble
A – Acetates always soluble
G – Group 1 always soluble
S – Sulfates always soluble**
A – Ammonium always soluble
G – Group 17 always soluble*
Silver
CastroBear* – EXCEPT Calcium,
Strontium, Barium
Predicting and analysing compositions of
reactions with two ionic compounds
KCl(aq) + AgNO3 (aq)  AgCl (s) + KNO3 (aq)
2KI (aq) + Pb(NO3)2 (aq)  PbI2 (s) + 2KNO3 (aq)
Na2SO4 + Ba(NO3)2  2NaNO3 (aq) + BaSO4 (s)
Deriving equilibrium expressions for Ksp
For A + B  C, forward rate = k1[A][B] and reverse rate = k2[C]
At equilibrium both rates are equal. Hence,
k1[A][B] = k2[C] and
𝑘1
𝑘2
=
[𝐶]
𝐴 [𝐵]
= 𝐾𝑒𝑞
𝐸
𝑎
−𝑅𝑇
Also, 𝑘 = 𝐴𝑒
Deriving Ksp for: AgCl(s) + H2O(l)
k1[AgCl][H2O] =
k2[Ag+][Cl-][H2O]
and
𝑘1
𝑘2
Ag+(aq) + Cl– (aq) + H2O(l)
[Ag+][Cl−][H2O]
=
= [Ag+][Cl−] = 𝐾𝑠𝑝
[AgCl][H2O]
Calculating molar solubility from Ksp
• Molar solubility is the number of moles of sparingly soluble compound
that can dissolve before the solution is saturated (Qsp=Ksp)
E.g. if 0.3 g of AgCl is dissolved in 50 mL of water and is fully saturated then
the molar solubility of AgCl is:
0.3 g/143.32 gmol-1 = 0.002903 mol,
Molar solubility = 0.002903 mol/0.05 L = 0.04 M
ICE is then used to find concentrations of Ag+ and Cl- (0.04 as mole ratio is
the same)
Ksp = [Ag+][Cl-] = 0.042 = 0.0016
```