R2.3 How Far?

S2.3 How Far?

​Review of Physical Changes

​Are changes of size, shape or state of matter (including dissolving). Physical changes tend to be reversible, but most importantly no new chemicals are produced.

​Review of Chemical Changes

​Also known as chemical reactions. They are not easily reversible and result in the production of new chemicals.

Categorize

Options (9)

Example: Evaporation

Involve breaking and forming of bonds between atoms.

No new chemicals are made.

Involves forming new chemicals.

Example: Combustion

Involve overcoming intermolecular forces

State Changes

Dissolving

Cooking

Organize these options into the right categories

Physical Changes

Chemical Changes

​Dynamic Equilibrium

​When the forward and reverse rates of reaction (either for a chemical or physical process) are equal. It is dynamic because the reaction has not stopped, both the forward and reverse reactions continue to occur.

​Requirements of Dynamic Equilibrium

​As we watch the video, note down the conditions needed:

​Requirements of Dynamic Equilibrium

​As we watch the video, note down the conditions needed:

​Closed System

Reversible Process

​Systems

​Dynamic equilibrium can only occur in closed systems because if the products leave the system, they can no longer reform the reactants in the reverse reaction.

Let's look at a real world example of physical equilibrium.

​Arrows in Chemical Equations

​We use double half-arrows when:

• talking about equilibrium

• reactions in which there are significant amounts of reactants and products at equilibrium (reaction does not go to completion).

​Example: N₂ + H₂ ⇌ NH₃

​Observing Systems: Physical System

Activity: Dissolving Salt* & Simulation

​✔ EQUILIBRIUM

✘ EQUILIBRIUM

✘ EQUILIBRIUM

​Observing Systems: Chemical System

​Activity: Blue Bottle Equilibrium & Simulation

​✘ EQUILIBRIUM

​✘ EQUILIBRIUM

​​✔ EQUILIBRIUM

​​✔ EQUILIBRIUM

​​✔ EQUILIBRIUM

​Characteristic 1

​At equilibrium, there is no visible change (no change in macroscopic properties) occurring (ie, it looks like nothing is happening).

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​​​✔ EQUILIBRIUM

​✘ EQUILIBRIUM

​​​✔ EQUILIBRIUM

​✘ EQUILIBRIUM

​✘ EQUILIBRIUM

​✘ EQUILIBRIUM

​Characteristic 2

​At equilibrium, there is no visible change (no change in macroscopic properties) occurring (ie, it looks like nothing is happening).

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​​​✔ EQUILIBRIUM

​✘ EQUILIBRIUM

​​​✔ EQUILIBRIUM

​✘ EQUILIBRIUM

​✘ EQUILIBRIUM

​✘ EQUILIBRIUM

Reason: Color is related to concentration. When concentrations remain constant, color does too.

​Characteristic 2

​At equilibrium, the concentrations of reactants and products remain constant.

​Characteristic 3

​At equilibrium, the rates of the forward and reverse reactions are equal.

​Characteristic 4

​The same equilibrium is reached regardless of starting concentrations of reactants/products.

​If you start with only products:

​If you start with only reactants:

​2SO₂ + O₂ ⇌ 2SO₃

​Misconception Equil ≠ Equal

​A common error students make is thinking concentrations of reactants and products will be equal at equilibrium. Rather it is the rates that are equal, not necessarily the concentrations (see Characteristic 4’s graphs for evidence).

​Summary

​For the reaction:
N₂O₄ ⇌ 2NO₂

​Equilibrium Law

​Tells us at equilibrium, there is a constant ratio (the equilibrium constant, K) based on concentrations and the stoichiometry of a chemical system:

​Equilibrium Constant Expression

STEP 1: The numerator is the equilibrium concentrations of the products multiplied together.

STEP 2: The denominator is the equilibrium concentrations of the reactants multiplied together.

STEP 3: Raise the concentration of each reactant and product concentration to their coefficient in the balanced chemical equation.

STEP 4: Remove (l) and (s) states from the equation as they do not have concentrations.

Multiple Choice

What is the equilibrium constant expression, Kc, for the following reaction?

2NH₃ + 2O₂ ⇌ N₂O(g) + 3H₂O

• 2[NH3]2[0₂]

• [NH3]²[0₂]²

Multiple Choice

What is the equilibrium constant expression, Kc, for the formation of hydrogen iodide from its elements? H₂(g) + I₂(g) → 2HI(g)

​Magnitude of K

These fact follow mathematically from the equilibrium constant expression:

Multiple Choice

The equilibrium constant for the reaction is 4.31 x 10¹². The position of equilibrium would be...

• To the left

• To the right

• Neither side would be favored.

Multiple Choice

The equilibrium constant for the reaction is 2.67 x 10^-7. At equilibrium, the mixture would be primarily:

• Reactants

• Products

• An approximately equal mix of reactants and products

Temperature Dependence

The value of K determined at equilibrium is only constant at one given temperature. If the temperature changes, K will change. This will be explored later with Le Chatelier’s Principle.

​Reverse Reaction Values of K

​The equilibrium constant (K) of the reverse reaction is simply the inverse of the K of the forward reaction.

​Reverse Reaction Values of K

Fill in the Blanks

If K for the reaction N₂O₄ ⇌ 2NO₂ is 4.66 x 10^-3, then what is K of the reverse reaction?

​Stresses

​When a system is at equilibrium, a stress can be added that disturbs the equilibrium. The systems will then respond to establish a new equilibrium:

​Stresses can be:

• Changing concentration by adding/removing chemicals

• Changing pressure (for gaseous reaction)

• Changing temperature

​Le Chatelier's Principle

​A closed system at dynamic equilibrium responds by counteracting stresses (changes). Let’s explore this in the following videos...

​​Le Chatelier's Principle SUmm

Multiple Choice

What will happen if the pressure is increased in the following reaction mixture at equilibrium? CO₂ (g) + H₂O ⇒ H⁺ (aq) + HCO₃^- (aq)

• The equilibrium will shift to the right.

• The equilibrium will shift to the left.

• There will be no shift.

Multiple Choice

N₂O₄ (g) ⇌ 2NO₂ (g) ΔH = +58 kJ mol^-1

Which changes shift the position of equilibrium to the right?

I. Increasing the temperature

II. Decreasing the pressure

III. Adding a catalyst

• I and II only

• I and III only

• II and III only

• I, II and III

Multiple Choice

What is the effect of increasing temperature on the equilibrium?

• A.

• B.

• C.

• D.

Multiple Choice

What happens when the temperature of the following equilibrium system is increased?

CO(g) + 2H₂(g) ⇌ CH₃OH(g) ΔΗ = -91kJ

• A

• B

• C

• D

Multiple Choice

What would have happened at time = t₁ for the reaction:

2HI ⇌ H₂ + I₂

• Addition of H₂

• Addition of HI

• C. addition of a catalyst

• D. a decrease in volume

Le Chatelier's Principle
Experiment

​Graphical Representations

Flat line =
Previous
Equilibrium

Flat line =
New Equilibrium

Sharp rise of 1 chemical = conc. change

​SO₂ + O₂ ⇌ SO₃

​When a stress is introduced, it can be visualized by changes in concentration in a graph.

Slopes = system countering change

​Graphical Representations

Flat line =
Previous
Equilibrium

Flat line =
New Equilibrium

Sharp change of all chemicals = pressure change

​SO₂ + O₂ ⇌ SO₃

Slopes = system countering change

​Graphical Representations

Flat line =
Previous
Equilibrium

Flat line =
New Equilibrium

Gradual change of all chemicals = temp change

​SO₂ + O₂ ⇌ SO₃ ​ΔH = -197 kJ/mol

Slopes = system countering change

Multiple Choice

What stress was introduced at t₁ for this system at equilibrium?

CH₃COOH + H₂O ⇌ CH₃COO^- + H₃O⁺ + heat

• Addition of H₃O⁺

• Decreasing temperature

• Addition of CH₃COO^-

• increasing the volume of the container.

Multiple Choice

What stress was introduced to the system:

N₂O₄(g) ⇌ 2NO₂(g)

• Catalyst was added.

• Pressure was changed.

• Temperature was changed.

• NO₂ was added

​Increase reactant (to shift to right), remove products

​Increase pressure (to shift to right)

Decrease temperature (shift to right)

No effect on yield

​Increase concentration of reactants (faster)

​Increase pressure (faster)

​Increase temperature (faster)

​Use a catalyst (faster)

Equilibrium says to decrease temp, while kinetics says to use high temp.
Solution: Use a moderate temperature.

​Real Life Application: Ocean Acidification

​Ocean Acidification exemplifies a heterogeneous equilibrium of gaseous CO₂ from the atmosphere and aqueous CO₂ dissolved in the oceans. We will cover this more in the acids and bases unit.

​Reaction Quotient (Q or K_Trial)

​Tells us which way the reaction needs to proceed in order to establish equilibrium. It is calculated using non-equilibrium concentrations into the equilibrium constant expression.

​Answer: K = 49.4, Q = 4, forward reaction is favored.

​Link to Example Video

​Worked Example Video 1 | Worked Example Video 2

​Answer: 1.84

​Ans: 0.00240 mol dm^-3

​Answer: 33

​Connecting to Enthalpy

​Lower (minimum) enthalpy corresponds to more stable products, which is favorable naturally.

​Entropy

​Measures the disorder of a system. According to the 2nd Law of Thermodynamics (Law of Entropy), the entropy of the universe is always increasing. Therefore reactions that increase entropy are favorable.

​Qualitatively Describing Entropy

​State Changes: If you see state changes in the chemical equation, you can deduce the entropy change:

​Qualitatively Describing Entropy

Moles of Gas: Total the moles of gas on both sides: if products > reactants, then entropy increased (gases have much more entropy compared to solids/liquids).

​Qualitatively Describing Entropy

Number of Particles: If no moles of gas change, but the number of particles changes: more particles = higher entropy.

Multiple Choice

In which reaction is the enthalpy of the reactants greater than the enthalpy of the products?

• H₂O(s) → H₂O(l)

• H₂O(s) → H₂O(g)

• H₂O(l) → H₂O(s)

• H₂O(l) → H₂O(g)

Multiple Choice

In which reaction is entropy decreasing?

• H₂O(l) → H₂O(g)

• N₂O₄(g) → 2NO₂(g)

• CaCO₃(s) → CaO(s) + CO₂(g)

• Fe³⁺(aq) + SCN^-(aq) → FeSCN²⁺(aq)

Multiple Choice

Which of the following describes the changes in

enthalpy and entropy as the reaction proceeds?

N₂(g) + 3H₂(g) → 2NH₃(g) + heat

• A

• B

• C

• D

Multiple Choice

Consider the following reaction:

Na2CO3(s) + 2HCl → NaCl(aq) + CO2(g) + H2O(l) ΔH = -27.7 kJ

• minimum enthalpy and maximum entropy both favour products.

• minimum enthalpy and maximum entropy both favour reactants.

• minimum enthalpy favours products and maximum entropy favours reactants.

• minimum enthalpy favours reactants and maximum entropy favours products.

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