Date: ____________________

SCH 4U

Name: ____________________

EQUILIBRIUM

o In Gr.9, learned that chemical changes are irreversible changes o In Gr.11, learned that limiting reactants are all used up when the reaction is complete o In all studies, the deeper you investigate, the more nuanced the truth is, even in science, there are no

all-or-nothing rules Reversible Reaction = chemical reactions that can proceed in forward and reverse directions

Forward Reaction = reactants form products (read equation from left to right ) Reverse Reaction = products decompose to re-form reactants (read equation from right to left ) Both forward and reverse reaction can be happening at the same time in the chemical system

Equilibrium = a state of balance, opposing forces or processes are balanced

Nature naturally settles into a state of equilibrium Chemical Equilibrium = when forward reaction and reverse reaction are occurring at the same rate in a chemical system

If opposing reactions are occurring at the same rate, there is no net change in [reactants] or [products]

o Does not mean that the [reactants] and [products] are the same or equal Each reactant and product stabilizes at different concentrations

Equilibrium is achieved in a closed system where no matter can escape or be added Dynamic Equilibrium = chemical system is not changing on a macroscopic level but still changing on a microscopic / molecular level

On a macroscopic level, measurements / values of concentration, colour, temperature, pressure, pH of the reaction are not changing – reaction “looks” complete

But on a microscopic / molecular level, forward and reverse reactions are still occurring Liquid-Vapour Equilibrium (A) = when the rate of particles entering liquid and gaseous phase of a system is balanced

ratevap = ratecond Homogeneous Equilibrium = a chemical system in equilibrium where all components are in same physical state (ie/ all liquids, or all solids, or all aqueous etc.)

N2(g) +3 H2(g) ↔ 2 NH3(g) Heterogeneous Equilibrium = a chemical system in equilibrium its components are in different physical states

H2O(l) ↔ H2O(g) CaCO3(s) ↔ CaO(s) + CO2(g) (limestone ↔ quicklime + carbon dioxide)

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EQUILIBRIUM EXPRESSION

Law of Chemical Equilibrium = when a chemical system is at equilibrium, there is a constant ratio between [reactants] and [products]

aka Law of Mass Action or Guldberg-Waage Law o Cato Guldberg (1836-1902) o Peter Waage (1833-1900)

Equilibrium Constant (Keq) = ratio of concentrations for all components of a chemical system at equilibrium at a particular temperature Equilibrium Constant Expression:

aA + bB ↔ cC + dD

Keq = [products] Keq = [C]c [D]d [reactants] [A]a [B]b

[A], [B], [C], [D]… = concentrations of reactants and products in reaction a, b, c, d = stoichiometric coefficients from balanced chemical equation of reaction Keq > 1 means [reactants] < [products]

o Position of equilibrium lies to the right , in direction of forward reaction, more products o Keq > 1010 are said to proceed to completion

Keq = 1 means [reactants] = [products] Keq < 1 means [reactants] > [products]

o Position of equilibrium lies to the left , in direction of reverse reaction, more reactant formation

o Keq < 10-10 are believed to not proceed at all, reactants just sit there in reaction chamber (B) Practice Question for Homogeneous Equilibrium 1. Sulphur dioxide gas reacts with oxygen gas to produce sulphur trioxide in gaseous form. Write the

equilibrium constant expression for this reaction. Practice Question for Heterogeneous Equilibrium For Heterogeneous Equilibrium:

o Concentrations for solids and liquids are considered constant, they don’t change, so absorbed into Keq

o Concentrations for gases and aqueous solutions can vary, so they are the only concentrations expressed in the equilibrium constant expression

2. Calcium carbonate (solid limestone) decomposes to form calcium oxide (solid quicklime) and carbon

dioxide gas. Write the equilibrium constant expression for this reaction.

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LE CHÂTELIER’S PRINCIPLE

At equilibrium, a system is closed and the following remain constant:

• [reactants]

• [products]

• Volume of system

• Pressure of system

• Temperature of system Le Châtelier’s Principle = a system is no longer at equilibrium if there are any changes to concentration, pressure, volume or temperature to the system at equilibrium

• Rate of forward and reverse reactions will not be equal, forward and reverse reaction will begin to proceed at different rates

• Eventually, equilibrium will be reached again Haber-Bosch Process (C)

• N2(g) +3 H2(g) ↔ 2 NH3(g)

• Temperature can be manipulated in this chamber so product, NH3(g), is liquefied and pumped out of reaction vessel

• N2(g) and H2(g) can be continuously added to drive reaction to go forward, shift to the right How Does Concentration Affect Equilibrium

• ↑ [reactant], reaction shifts to the right , make more products, ↑ [product] Keq stays the same since ↑ [reactant] resulted in ↑ [product], reactant:product stays the

same

• ↓ [reactants], reaction shifts to the left , make more reactants, ↑ [reactant] Keq stays the same since ↓ [reactants] resulted in ↑ [reactant], no net change in [reactant]

or [product]

• ↑ [product], reaction shifts to the left , make more reactants, ↑ [reactant] Keq stays the same since ↑ [product] resulted in ↑ [reactant], reactant:product stays the

same

• ↓ [product], reaction shifts to the right , make more product, ↑ [product] Keq stays the same since ↓ [product] resulted in ↑ [product], no net change in [reactant] or

[product]

• H2CO3(aq) ↔ CO2(aq) + H2O(l) Carbonic acid in bloodstream is regulated by breathing out carbon dioxide

How Does Pressure & Volume Affect Equilibrium (D)

• No effect on solids & liquids, their volume is constant

• ↑ P, then V ↓, reaction shifts so that total number of particles ↓, so pressure would ↓

• ↓ P, then V ↑, reaction shifts so that total number of particles ↑, so pressure would ↑

• Addition of inert gas does not shift reaction in either direction, stays at equilibrium

• Keq stays the same whether P & V ↑ or ↓, because ratio of reactants to products stays the same

• 2 SO3(g) ↔ 2 SO2(g) + O2(g) How Does Temperature Affect Equilibrium

• ↑ T, add thermal energy, favours endothermic reactions that need energy to proceed For endothermic reaction:

• ↑ T, forward reaction proceeds, [product] ↑, then Keq ↑ since [product] > [reactant]

• ↓ T, reverse reaction proceeds, [product] ↓, then Keq ↓ since [product] < [reactant]

• ↓ T, remove thermal energy, favours exothermic reactions because energy will be released For exothermic reaction:

• ↑ T, reverse reaction proceeds, [product] ↓, then Keq ↓ since [product] < [reactant]

• ↓ T, forward reaction proceeds, [product] ↑, then Keq ↑ since [product] > [reactant]

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How Does a Catalyst Affect Equilibrium

• Catalyst increases rate of reaction in forward and reverse direction Reaction reaches equilibrium faster But Keq does not change

3. Phosphorus pentachloride gas is heated to produce phosphorus trichloride gas and chlorine gas. Using

Le Châtelier’s Principle, explain in which direction the reaction shifts and whether Keq is affected, when:

a) more phosphorus pentachloride gas was added b) chlorine gas is removed c) temperature is decreased d) pressure is increased by adding an inert gas (helium) e) a catalyst is used

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EQUILIBRIUM CALCULATIONS

How to Calculate Keq for Homogeneous Equilibrium 4. Nitrogen gas and chlorine gas react to product nitrogen trichloride gas in a 5.0 L flask. At equilibrium,

there was 0.0070 mol of nitrogen gas, 0.0022 mol of chlorine gas, and 0.95 mol of nitrogen trichloride gas. Calculate the equilibrium constant for this reaction.

How to Calculate [reactant/product] with Keq 5. Nitrogen gas and hydrogen gas react to form ammonia (nitrogen trihydride) gas in a closed 3500mL

flask. At equilibrium, there was 0.25 mol of ammonia gas and 0.080 mol of hydrogen gas. The equilibrium constant for this reaction is 5.81 x 105. Determine the amount (mol) of nitrogen gas that was present at equilibrium.

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How to Calculate Kp for Homogeneous Equilibrium Composed of Only Gases

Ideal Gas Law PV = nRT o P = pressure (kPa) o V = volume (L) o N = amount of gas (mol) o R = universal gas constant 8.314 (kPa·L/mol·K) o T = temperature (K)

If all reactants & products are gases, partial pressures (kPa) can be used in calculations instead of molar concentrations (mol / L) n = P

V RT

Kp = equilibrium constant expressed in pressure (kPa) aA + bB ↔ cC + dD

Kp = PC PD PA PB

Kp = PCc PD

d PA

a PBb

o Keq is often different from Kp even if they’re calculated for the same system o Can convert Keq to Kp via:

∆n = sum gas product coefficients – sum gas reactant coefficients Kp = Keq(RT)∆n

6. Methyl chloride (chloromethane) is used in the production of silicones and also used as a sealant.

It can be found in oceans naturally or made in industrial laboratories. Methane gas and chlorine gas can combine to form methyl chloride gas and hydrogen chloride gas. If at 1500 K, the mixture contained 0.13 atm of methane, 0.035 atm chlorine gas 0.24 atm of methyl chloride and 0.47 atm hydrogen chloride, what is the equilibrium constant Kp.

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How to Use ICE Tables Initial = initial [reactant / product] (mol/L) Change = amount of change in each reactant / product from initial to equilibrium conditions Equilibrium = [reactant / product] (mol/L) at equilibrium

7. Hydrogen iodide gas decomposes to make hydrogen gas and iodine gas in a 2.0 L flask. If 0.200

mol of hydrogen iodide at 453oC reacts to a concentration of 0.078 mol/L at equilibrium, calculate the concentrations of hydrogen and iodine gas and the equilibrium constant for the reaction.

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8. Hydrogen gas is a non-polluting and sustainable fuel source used in fuel cells and internal combustion engines. Carbon monoxide gas and water vapour can be combined to form hydrogen gas and carbon dioxide gas. If at 700K in a 5.0L container, 1.0 mol of carbon monoxide gas reacts with 1.0 mol of water vapour and the equilibrium constant is 0.83, how many moles of each substance is present in the container at equilibrium?

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How to Use Quadratic Formula in Equilibrium Problems Quadratic equation ax2 + bx + c = 0 Quadratic formula

• Can have up to 2 solutions 9. Hydrogen gas and iodine gas can react to form hydrogen iodide gas. At 1100 K in a 1.00 L

reaction vessel with an equilibrium constant of 25.0, 2.00 mol of hydrogen gas and 3.00 mol of iodine gas are combined. Determine the equilibrium concentration of each gas.

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How to Solve Problems with a Really Small Keq Recall: Keq = [products]

[reactants]

If Keq is very small, 1000x smaller than initial concentrations, only a very small amount of product was made, [products] is very small

o Therefore, the change (x) in the amount of initial concentrations is almost negligible

when considering significant digits 10. Nitrogen gas and oxygen gas in the atmosphere don’t usually react unless they are in high

temperatures. In a car engine, nitrogen gas and oxygen gas react to form nitrogen oxide in the exhaust, which is a serious pollutant. If in a 1.5 L cylinder 0.085 mol of nitrogen gas reacts with 0.038 mol of oxygen gas and the equilibrium constant is 4.2 x 10-8, what is the concentration of nitrogen monoxide gas at equilibrium?

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REACTION QUOTIENTS

What if a chemical reaction (chemical system) is not at equilibrium?

At equilibrium, Keq and Kp are constant, the value is a single number, it does not change because the [reactants / products] are not changing

Reaction Quotient Qeq and Qp are used to quantify [reactants / products] when reaction is not at equilibrium (basically same equation)

Qeq = [products] Qeq = [C]c [D]d

[reactants] [A]a [B]b

Qp = PC PD Qp = PCc PD

d PA PB PA

a PBb

If Qeq < Keq (Qp < Kp), lower [products]:[reactants], reaction shifts to the right to ↑ product

formation If Qeq = Keq (Qp = Kp), system is at equilibrium

If Qeq > Keq (Qp > Kp), higher [products]:[reactants], reaction shifts to the left to ↑ re-formation

of reactants 11. For the Haber-Bosch process, the equilibrium constant is 0.40 at 500oC. At some point in time

during the reaction, the concentrations of the chemicals were measured to be 0.10 M for nitrogen gas, 0.30 M for hydrogen gas and 0.20 M for ammonia. Determine whether the system is at equilibrium. Describe what will happen to the reaction system by comparing the reaction quotient and equilibrium constant.

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ACID & BASE THEORY Acid & Base Properties

PROPERTY ACID BASE Taste Sour Bitter Touch N/A Slippery

Conductivity Conducts electricity Reaction with Metal Corrosive

Litmus Paper Red Blue Phenolphthalein Clear, colourless Pink

Bromothymol Blue Yellow Blue Methylene Orange Orange Yellow

Acid & Base Theory

Arrhenius Theory

Svanté Arrhenius (1859 – 1927)

Dissociation = when a compound separates into its ions when in solution

Acid Definition = compound that dissociates H+ o HBr(aq), H2SO4(aq), HClO4(aq)

Base Definition = compound that dissociates OH-

o LiOH(aq), KOH(aq), Ba(OH)2(aq) This is not a complete definition, as there are exceptions to this description

Brønsted-Lowry Theory

Johannes Bronsted (1879 – 1947) Thomas Lowry (1874 – 1936) Acid Definition = compound that is H+ (proton) donor

o HCl(aq) H+

(aq) o Conjugate base of an acid is the particle that remains after the H+ (proton) was

donated

Base Definition = compound that is H+ acceptor o NH3(aq) + H+

(aq) NH4+

(aq) o Conjugate acid of a base is the resulting particle after the H+ (proton) was accepted

by the base

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A substance can only behave like an acid if another substance behaves like a base at the same time, and vice versa

o HCl(aq) + H2O(l) Cl-(aq) + H3O+

(aq)

o NH3(aq) + H2O(l) NH4+

(aq) + OH-(aq)

12. Draw and label the conjugate acid-base pairs for ethanoic acid. 13. Draw and label the conjugate acid-base pairs for hydrobromic acid.

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ION-PRODUCT CONSTANT OF WATER & DEFINITION OF ACID & BASE Electrolyte = can conduct electricity because it has ions

Amphiprotic = can behave like an acid or a base, can donate or accept H+ protons (ie/ water)

H2O + H+ H3O+

H2O H+ + OH-

Ion-Product Constant of Water (Kw) = the equilibrium constant for water when it produces its ions in water

Kw = [H3O

+][OH-] [H3O+] = Kw [OH-] = Kw

[OH-] [H3O+]

Kw

= [H3O+][OH-] = 1.0 x 10-14 (at 25oC)

o [H3O+] = 1.0 x 10-7 M (at 25oC)

o [OH-] = 1.0 x 10-7 M (at 25oC)

Autoionization of water means that in all aqueous solutions (solute dissolved in water) there will be some concentration of H3O

+ and OH- o If [H3O

+] > [OH-], acidic solution o If [H3O

+] = [OH-], neutral solution

o [H3O+] < [OH-], basic solution

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14. Determine whether a window cleaner solution is acidic, neutral or basic if the concentration of hydroxide ions in the solution is 0.0020 M at 25oC.

pH = power of Hydrogen, a measure of [H3O

+] in solution to describe its acidity (E) [H3O

+] and [OH-] are often really small numbers, need a simpler number to quantify acids & bases pH

pH = – log [H3O+]

pKw = pH + pOH = 14.00 (at 25oC) Rules for logarithm significant digits

o If concentration has 2 sig dig, pH has 2 sig dig to the right of the decimal o If pH has 2 digits to the right of the decimal, concentration has 2 sig dig

15. Determine the pH, pOH and [OH-] of a solution if it has a hydronium concentration of 0.50 M.

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ACID-BASE STRENGTH & DISSOCIATION

Strength and Extent of Ionization

The more a substance ionizes into H3O+ or OH-, the stronger it is as an acid or base

o More dissociation, more H3O

+ or OH- ions in solution to behave like an acid or base Strong Acid = completely dissociates into ions in water (F)

High levels of [H3O+], pH very low, more acidic

HCl, HBr, HI, HClO4, HNO3, H2SO4

Weak Acid = dissociates slightly into ions in water (G)

Less [H3O+], pH won’t be as low, less acidic

Strong Base = completely dissociates into ions in water

High levels of ions to accept H+ (protons), pH very high, more basic LiOH, NaOH, KOH, Ca(OH)2, Ba(OH)2

Weak Base = dissociates slightly into ions in water

Fewer ions to accept H+, pH not as high, less basic 16. Predict the direction of the reaction if the sulphate ion reacts with ethanoic acid (acetic acid).

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Monoprotic Acid = an acid that dissociates 1 H+ (proton) HF, HCl, HBr, HI…

Diprotic Acid = an acid that dissociates 2 H+ (protons)

H2SO4, H2CO3, H2S… Triprotic Acid = an acid that dissociates 3 H+ (protons)

H3PO4… Acid-Dissociation Constant (Ka) = equilibrium constant for weak acids (aka acid-ionization constant)

Strong acids dissociate to completion, don’t need a special equilibrium constant because at equilibrium the strong acid has completely dissociated

The stronger the acid, ↑ [H3O+], ↑ Ka

The weaker the acid, ↓ [H3O+], ↓ Ka

Effects of autoionization of water is negligible Difference in [HA] before and after dissociation is negligible, since weak acids dissociate only

slightly For a monoprotic acid:

HA(aq) + H2O(l) ↔ H3O+

(aq) + A-(aq)

Ka = [H3O

+] [A-] [HA] Percent Dissociation = ratio of [ionized acid @ equilibrium] to [acid before ionization]

For weak acids only percent dissociation = [HA]dissociated x 100%

[HA]initial Acid-Dissociation Constant - Polyprotic Acids (Ka1, Ka2, Ka3)

H3A(aq) + H2O(l) ↔ H3O+

(aq) + H2A-(aq)

H2A

-(aq) + H2O(l) ↔ H3O

+(aq) + HA2-

(aq) HA2-

(aq) + H2O(l) ↔ H3O+

(aq) + A3-(aq)

Ka1 = [H3O

+] [H2A-]

[H3A] Ka2 = [H3O

+] [HA2-] [H2A

-] Ka3 = [H3O

+] [A3-] [HA2-]

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17. Determine the acid-dissociation constant and the percent dissociation of the weak acid propanoic acid, which inhibits mould formation in bread, if its initial concentration is 0.10 M and the pH is 2.96.

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18. Determine the pH of a 0.050 M solution of nitrous acid.

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19. Determine the concentration of hydronium of a 0.10 M solution of methanoic acid (formic acid) if the acid-dissociation constant is 1.8 x 10-4. Methanoic acid is a toxin found in stinging ants!

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BASE IONIZATION

Base-Dissociation Constant (Kb) = equilibrium constant for ionization of a base (aka base-ionization constant)

B(aq) + H2O(l) ↔ BH+(aq) + OH-

(aq)

Kb = [BH+] [OH-] [B]

20. Aniline (C6H5NH2), a weak base, is used in manufacturing dyes. Determine the base-dissociation

constant for 5.0 g/L of aniline with a pH of 8.68.

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21. Quinine (C20H24N2H2) is a naturally occurring white crystal that can be used to treat malaria and accounts of the bitter taste in tonic water. Determine the concentration of hydroxide ions and pH of a 3.6 x 10-3 solution of quinine if its base dissociation constant is 3.3 x 10-6.

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EQUILIBRIUM CONSTANT RELATIONSHIPS

For a reaction that is the sum of 2 or more reactions, the overall equilibrium constant is the product of the individual equilibrium constants.

KaKb = Kw 22. Determine the base-dissociation constant for the conjugate base of benzoic acid. 23. Determine the acid-dissociation constant for the conjugate acid of ethylamine.

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BUFFERS

Buffer Solution = resists changes in pH when a limited amount of acid / base is added Made of weak acid + conjugate base as a salt OR Made of weak base + conjugate acid as a salt ex/ CH3COOH(aq) & CH3COONa(s)

o If add base, CH3COOH would donate H+ (protons) for base to accept, base would be neutralized

o If add acid, CH3COONa dissociates into CH3COO- which can accept H+ (protons), acid is neutralized

TITRATIONS

Neutralization Reaction

Acid + base salt (ionic compound) + water HA + BOH AB + H2O

Titration = an experiment to find the unknown concentration of an acid with a known base, or the unknown concentration of a base with a known acid

Endpoint = the point when the indicator changes colour, signifying that the addition of the known acid / base has neutralized the unknown base / acid

Equivalence Point = the exact point where the amount of moles of the known acid / base has neutralized the exact proportional amount of moles of the unknown base / acid

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3 Types of Titrations Strong Acid–Strong Base Titration (H)

Strong acid / strong base dissociate completely Strong acid is titrated by a strong base

o Strong acid in flask (initial pH << 7) o Strong base in buret

pH = 7.00 at equivalence point 24. In a strong acid–strong base titration, 40.00 mL of 0.1000 M of hydrochloric acid is titrated with

0.1000 M of sodium hydroxide. Calculate the pH of: ) original hydrochloric acid ) titration solution after adding 20.00 mL of base ) titration solution at equivalence point ) titration solution after adding 50.00 mL of base

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Weak Acid–Strong Base Titration (I) Weak acid is titrated by a strong base

Weak acid in flask (initial pH < 7) Strong base in buret

Weak acid dissociates partially, less H3O+ than a strong acid, initial pH will be higher than

strong acid – strong base titration Buffer Region

As strong base neutralizes weak acid (HA), conjugate base (A-) of the weak acid forms Acid-conjugate base buffer is created in the titration mixture

o Conjugate base (A-) of the weak acid can accept H+ (proton) from H2O, leaving behind OH- in titration mixture

o ↑ [OH-], ↑ pH of titration mixture, equivalence point will be higher than 7 At midpoint of buffer reaction (buffer region on graph), half of weak acid has been

neutralized, [HA] = [A-] 25. In a weak acid–strong base titration, 25.00 mL of 0.1000 M of ethanoic acid (acetic acid) is titrated

with 0.1000 M of sodium hydroxide. Calculate the pH after: ) 10.00 mL of sodium hydroxide was added ) 25.00 mL of sodium hydroxide was added

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Weak Base–Strong Acid Titration (J) Weak base is titrated by a strong acid

o Weak base in flask (initial pH > 7) o Strong acid in buret

Buffer Region o As weak base (NH3) is neutralized by strong acid, conjugate acid (NH4

+) of weak base is formed

o Acid-conjugate base buffer is created in the titration mixture

• Conjugate acid (NH4+) of weak base donates H+ to H2O, creating H3O

+ in titration mixture

• ↑ [H3O+], ↓ pH of titration mixture, equivalence point will be lower than 7

o At midpoint of buffer reaction (buffer region on graph), pH = pKa of conjugate acid 26. In a weak base-strong acid titration, 40.00 mL of 0.1000 M ammonia is titrated with 0.1000 M

hydrochloric acid. Find the pH at the equivalence point.

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SLIGHTLY SOLUBLE IONIC COMPOUNDS

Solubility = maximum amount of solute that can dissolve in a given amount of solvent at a particular temperature

• Ionic compounds can be highly soluble, slightly soluble, or insoluble Ion-Product Expression Constant (Qsp) = dissociation constant of slightly soluble ionic compound in solution, not at equilibrium Solubility-Product Constant (Ksp) = equilibrium constant for slightly soluble ionic compounds

• Bigger Ksp, more ions produced, ↑ [Mn+] & [ Xz-], ionic compound is more soluble

• Small Ksp, less ions produced, ↓ [Mn+] & [ Xz-], more stays as MpXq, ionic compound is less soluble

• Qsp < Ksp, solution unsaturated, no precipitate forms

• Qsp = Ksp, solution saturated, no change occurs

• Qsp > Ksp, precipitate forms until solution is saturated

MpXq ↔ pMn+ + qXz-

Qsp = [Mn+]p [ Xz-]q = Ksp 27. Calculate the solubility-product constant for 1.3 x 10-4 M of silver carbonate. 28. Find the molar solubility (mol/L) of barium fluoride if its Ksp is 1.7 x 10-6.

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29. Determine the Ksp for 0.67 g/L of calcium sulphate. 30. Determine if a precipitate will form if 200 mL of 0.0040M barium chloride solution is mixed with

600 mL of 0.0080 M potassium sulphate solution. Assume the only precipitate that can form is barium sulphate.

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EQUILIBRIUM IN THE REAL WORLD

Scuba Diving

• Pressure pressing onto a diver ↑ 1 atm / 10m

• More gas in breath dissolved in blood

• Nitrogen Narcosis = when too much nitrogen gas is in blood Starts at 30m below surface, impair electrical signals in nerves Each + 15m, diver feels more and more “drunk” 60m, confusion, drowsiness, impaired judgement 90m, hallucination, unconsciousness

• Resurfacing Problems: Need to resurface slowly, gases slowly come out of blood into breath, slowly

breathe gases out The Bends (decompression sickness) = gases coming out of blood too quickly,

bubbles in body damage tissue / organs

• We’re not meant to breathe in pure oxygen, too much oxygen is toxic seizures Carbon Monoxide Poisoning

• Incomplete combustion CO

• Equilibrium constant for CO and hemoglobin 250 X greater than O2 and hemoglobin, not enough O2 in blood, body “suffocates” inside

• Hyperbaric Chamber = high pressure high oxygen chamber, allows equilibrium to shift, CO comes out of blood (used for nitrogen narcosis, the bends etc)

Coral Reefs

• Built by polyps using carbonate (CO32-

(aq)) to make calcium carbonate (CaCO3(s)) Ca2+

(aq) + CO32-

(aq) ↔ CaCO3(s)

• One would think the more CO2(g) in oceans, the more CO32-

(aq) to build coral reefs, but no, the equilibrium between these compounds is:

CO2(g) + CO32-

(aq) + H2O(l) ↔ 2 HCO3-(aq)

• ↑ CO2(g), ↓ CO32-

(aq), ↑ HCO3-(aq), fewer coral reefs, and ocean is more acidic too

Haber-Bosch Process & Ammonia (K)

• NH3 is used to make 500 million tons (453 billion kg) fertilizers every year, feeds 40% world’s population

• A complicated process to choose optimal temperature, pressure, catalyst, rate of removal to maximize yield, while not having extreme conditions that drive up operating costs or accidents

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Sulphuric Acid Production

• Similar factors considered like ammonia

• Used in so many industries (L)

• The health of a country’s economy can be directly proportional to its sulphuric acid production

Methanol Production

• Methane from natural gas is converted into a mixture of carbon monoxide, carbon dioxide & hydrogen gas (called syngas)

• Methanol used by ancient Egyptians to make embalming solution to preserve mummies, now used in many industries (M)