JCHS Honors Chemistry Spring Final S.G (Part 2) CH Williams

CHEMICAL EQUILIBRIUM Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action

Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical Equilibrium: When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged 2HgO(s)  2Hg(l) + O 2 (g) Arrows going both directions (  ) indicates equilibrium in a chemical equation

2NO 2 (g)  2NO(g) + O 2 (g)

Law of Mass Action For the reaction For the reaction: Where K is the equilibrium constant, and is unitless jA + kB  lC + mD

Product Favored Equilibrium Large values for K signify the reaction is “product favored” When equilibrium is achieved, most reactant has been converted to product

Writing an Equilibrium Expression 2NO 2 (g)  2NO(g) + O 2 (g) K = ??? Write the equilibrium expression for the reaction:

Conclusions about Equilibrium Expressions  The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse 2NO 2 (g)  2NO(g) + O 2 (g ) 2NO(g) + O 2 (g)  2NO 2 (g)

Conclusions about Equilibrium Expressions  When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. 2NO 2 (g)  2NO(g) + O 2 (g ) NO 2 (g)  NO(g) + ½O 2 (g )

Significance of the Reaction Quotient  If Q = K, the system is at equilibrium  If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved  If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

Solving for Equilibrium Concentration Consider this reaction at some temperature: H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) K = 2.0 Assume you start with 8 molecules of H 2 O and 6 molecules of CO. How many molecules of H 2 O, CO, H 2, and CO 2 are present at equilibrium? “ICE” Here, we learn about “ICE” – the most important problem solving technique in the second semester. You will use it for the next 4 chapters!

Solving for Equilibrium Concentration H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) K = 2.0 Step #1: We write the law of mass action for the reaction:

Solving for Equilibrium Concentration H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) Step #2: We “ICE” the problem, beginning with the Initial concentrations x +x 8-x6-x x x

Solving for Equilibrium Concentration Step #3: We plug equilibrium concentrations into our equilibrium expression, and solve for x H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) x = 4

Solving for Equilibrium Concentration Step #4: Substitute x into our equilibrium concentrations to find the actual concentrations H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) x = 4

Le Chatelier’s Principle

When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress and restore a state of equilibrium. Henry Le Chatelier

When you take something away from a system at equilibrium, the system shifts in such a way as to replace some what you’ve taken away. Le Chatelier Translated: When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

LeChatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water The system temporarily shifts to the _______ to restore equilibrium. right

LeChatelier Example #2 A closed container of N 2 O 4 and NO 2 is at equilibrium. NO 2 is added to the container. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium. left

LeChatelier Example #3 A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor The system temporarily shifts to the _______ to restore equilibrium. right

LeChatelier Example #4 A closed container of N 2 O 4 and NO 2 is at equilibrium. The pressure is increased. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. left

Moles and Formula Mass

The Mole 1 dozen = 1 gross = 1 ream = 1 mole = x There are exactly 12 grams of carbon-12 in one mole of carbon-12.

Avogadro’s Number x is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ). Amadeo Avogadro I didn’t discover it. Its just named after me!

Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li = g Li 1 mol Li 6.94 g Li 45.1

Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li = mol Li 6.94 g Li 1 mol Li 2.62

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol = atoms 1 mol 6.02 x atoms 2.07 x 10 24

Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li = atoms Li 1 mol Li x atoms Li 1.58 x g Li1 mol Li (18.2)(6.022 x )/6.94

Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO g g + 3(16.00 g) = g

Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. From previous slide: g g + 3(16.00 g) = g

Formulas  molecular formula = (empirical formula) n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 O C 12 H 22 O 11

Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x Empirical formula: C3H5O2C3H5O2

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = g

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g 2. Divide the molecular mass by the mass given by the emipirical formula.

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

Solution Formation

Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

Steps in Solution Formation  H 2 Step 2 -  H 2 Step 2 - Expanding the solvent Overcoming intermolecular forces of the solvent molecules

Steps in Solution Formation  H 3 Step 3 -  H 3 Step 3 - Interaction of solute and solvent to form the solution

Predicting Solution Formation

Solubility Chart

Ionic Bonding  Electrons are transferred  Electronegativity differences are generally greater than 1.7  The formation of ionic bonds is always exothermic!

Determination of Ionic Character Compounds are ionic if they conduct electricity in their molten state Electronegativity difference is not the final determination of ionic character

Coulomb’s Law “The energy of interaction between a pair of ions is proportional to the product of their charges, divided by the distance between their centers”

Table of Ion Sizes

Sodium Chloride Crystal Lattice Ionic compounds form solid crystals at ordinary temperatures. Ionic compounds organize in a characteristic crystal lattice of alternating positive and negative ions. All salts are ionic compounds and form crystals.

Properties of Ionic Compounds

Polar-Covalent bonds Nonpolar-Covalent bonds Covalent Bonds  Electrons are unequally shared  Electronegativity difference between.3 and 1.7  Electrons are equally shared  Electronegativity difference of 0 to 0.3

Bond Length and Energy Bonds between elements become shorter and stronger as multiplicity increases.

Bond Energy and Enthalpy D D = Bond energy per mole of bonds Energy requiredEnergy released Breaking bonds always requires energy Breaking = endothermic Forming bonds always releases energy Forming = exothermic

The Octet Rule and Covalent Compounds  Covalent compounds tend to form so that each atom, by sharing electrons, has an octet of electrons in its highest occupied energy level.  Covalent compounds involve atoms of nonmetals only.  The term “molecule” is used exclusively for covalent bonding

Solids Image:Wikimedia Commons User Alchemistry-hp

Types of Solids  Crystalline Solids: highly regular arrangement of their components  Amorphous solids: considerable disorder in their structures (glass, plastic).

Representation of Components in a Crystalline Solid Lattice: A 3-dimensional system of points designating the centers of components (atoms, ions, or molecules) that make up the substance.

Bragg’s Law xy + yz = n and xy + yz = 2d sin   n = 2d sin 

Crystal Structures - Cubic SimpleFace-CenteredBody-Centered **Knowledge of specific types of crystal structures and the study of the specific varieties of crystal lattices for ionic compounds is beyond the scope of this course and the AP Exam.

Crystal Structures - Monoclinic Simple End Face-Centered

Crystal Structures - Tetragonal Simple Body-Centered

Crystal Structures - Orthorhombic SimpleEndFace-CenteredBodyCenteredFaceCentered

Crystal Structures – Other Shapes Rhombohedral Triclinic Hexagonal

Metal Alloys  Substitutional Alloy: some metal atoms replaced by others of similar size. brass = Cu/Zn brass = Cu/Zn

Metal Alloys (continued)  Interstitial Alloy: Interstices (holes) in closest packed metal structure are occupied by small atoms. steel = iron + carbon steel = iron + carbon

Network Atomic Solids Some covalently bonded substances DO NOT form discrete molecules. Diamond, a network of covalently bonded carbon atoms Graphite, a network of covalently bonded carbon atoms

Semiconductors Pure silicon is structurally the same as diamond, but is a semiconductor rather than an insulator.  The conductivity increases at higher temperature.  Conductivity of silicon can be improved by doping with other elements.

Molecular Solids Strong covalent forces within molecules Weak covalent forces between molecules Sulfur, S 8 Phosphorus, P 4

Ionic Solids Sodium Fluoride Ionic compounds at room conditions are generally crystal lattices of alternating cations and anions. Sodium chloride and sodium fluoride form simple cubic crystals. NaCl Unit Cell