Equilibrium, Redox Reactions, Hydrocarbons, and Functional Groups Chapters 18, 20, 22, and 23 Jennie L. Borders

Chapter 18 – Reaction Rates and Equilibrium  A rate is a measure of the speed of any change that occurs within an interval of time.  In chemistry, the rate of chemical change or the reaction rate is usually expressed as the amount of reactant changing per unit of time.

Section 18.1 – Rate of Reaction  According to collision theory, particles can react to form products when they collide with one another, provided that the colliding particles have enough kinetic energy.  The minimum energy that colliding particles must have in order to react is called the activation energy.

Activation Energy

Transition State  An activated complex is an unstable arrangement of atoms that forms momentarily at the peak of the activation-energy barrier.  The activated complex is sometimes called the transition state.

Factors Affecting Reaction Rates  The rate of a chemical reaction depends upon temperature, concentration, particle size, and use of a catalyst.

Temperature  Usually, raising the temperature speeds up reactions.  Increasing the temperature increases both the frequency of collisions and the number of particles that have enough energy to slip over the activation-energy barrier.

Concentration  The number of particles in a given volume affects the rate at which reactions occur.  More particles increase the concentration which increase the number of collisions leading to a higher reaction rate.

Particle Size  The surface area of a reactant affects the reaction rate.  The smaller the particle, the more surface area.  An increase in surface area increases the amount of the reactant exposed for reaction.

Catalysts  A catalyst increases the rate of reaction without being used up during the reaction.  Catalysts permit the reaction to proceed along a lower energy path.

Catalyst vs. No Catalyst catalyst

Catalysts  Since a catalyst is not consumed during a reaction, it does not appear as a reactant or a product.  Instead, the catalyst is written above the yield arrow.

Enzymes  Enzymes in your digestive tract act as catalysts for breaking down proteins.  It takes your body a few hours to digest proteins.  Without the enzymes, it would take many years for you to digest your food.

Section 18.1 Assessment 1. How is the rate of a chemical reaction expressed? 2. What are four factors that affect the rate of a chemical reaction? 3. Does every collision between reaction particles lead to products? Explain. 4. Why does refrigerated food stay fresh longer than room temperature food?

Section Equilibrium  A reversible reaction is one in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously.

Equilibrium  When the rates of the forward and reverse reactions are equal, the reaction has reached chemical equilibrium.  At chemical equilibrium, no net change occurs in the actual amounts of the components of the system.

Position  The relative concentrations of the reactants and products at equilibrium constitute the equilibrium position of a reaction.  The equilibrium position indicates whether the reactants or products are favored in a reversible reaction.

Le Chatelier’s Principle  Le Chatelier’s Principle states that if a stress is applied to a system in equilibrium, the system changes in a way that relieves the stress.  Stresses that upset the equilibrium include changes in the concentration of reactants or products, change in temperature, or changes in pressure.

Concentration  Changing the concentration of any reactant or product at equilibrium disturbs the equilibrium.  The system adjusts to minimize the effects of the change by shifting the equilibrium.

Temperature  Increasing the temperature causes the equilibrium position to shift to the direction that absorbs heat.  Think of heat as a reactant or product in the reaction.

Pressure  A change in pressure only affects gaseous equilibria that have an unequal number of moles of reactants and products.

Sample Problem  What effect do each of the following changes have on the equilibrium position for this following reaction? PCl 5(g) + heat  PCl 3(g) + Cl 2(g) a. Addition of Cl 2 b. Increase in pressure c. Removal of heat d. Removal of PCl 3 as it is formed equilibrium shifts to the left equilibrium shifts to the right equilibrium shifts to the left

Equilibrium Constant  The equilibrium constant (K eq ) is the ratio of the product concentration to reactant concentration at equilibrium, with each concentration raised to a power equal to the number of moles of that substance in the balanced chemical equation.

Equilibrium Constant Generic Reaction: aA + bB  cC + dD Equilibrium Constant: K eq = [C] c x [D] d [A] a x [B] b

Equilibrium Constant  K eq > 1, products favored at equilibrium  K eq < 1, reactants favored at equilibrium

Sample Problem  A liter of a gas mixture at equilibrium contains mol or N 2 O 4 and 0.030mol of NO 2. Calculate K eq. N 2 O 4(g)  2NO 2(g) K eq = [NO 2 ] 2 K eq = (0.030M) 2 = 0.20 [N 2 O 4 ] (0.0045M)

Practice Problem  At equilibrium, a 1L flask contains 0.15 mol H 2, 0.25 mol N 2, and 0.10 mol NH 3. Calculate K eq for the reaction. N 2 + 3H 2  2NH 3 K eq = [NH 3 ] 2 K eq = (0.10M) 2 = [N 2 ][H 2 ] 3 (0.25)(0.15) 3

Section 18.2 Assessment 1. How do the amounts of reactants and products change after a reaction has reached chemical equilibrium? 2. What are three stresses that can upset the equilibrium of a chemical system? 3. What does the value of the equilibrium constant tell you about the amounts of reactants and products present at equilibrium?

Section 18.2 Assessment 4. How can a balanced chemical equation be used to write an equilibrium-constant expression? 5. Can a pressure change shift the equilibrium position in every reversible reaction? Explain. 6. Using the following equilibrium constants, determine which reactions would favor the products. a. 1 x 10 2 b c. 3.5

Chapter 20 – Oxidation- Reduction Reactions  Oxidation-reduction reactions are also known as redox reactions.  Oxidation is the loss of electrons or the gain of oxygen.  Reduction is the gain of electrons or the loss of oxygen.

Section Oxidation vs. Reduction  The way to remember the difference in oxidation and reduction is OIL RIG. O = oxidation I = is L = loss of electrons R = reduction I = is G = gain of electrons

Oxidation-Reduction Reactions Mg (s) + S (s)  Mg +2 S -2 (s) Mg has a 0 charge and changes to Mg +2, so it loses electrons and is oxidized. S has a 0 charge and changes to S -2, so it gains electrons and is reduced.

Oxidizing and Reducing Agents  The substance that loses electrons is called the reducing agent.  The substance that accepts electrons is called the oxidizing agent.  In other words, the substance that is reduced is the oxidizing agent, and the substance that is oxidized is the reducing agent.

Sample Problem Determine what is oxidized and what is reduced. Identify the oxidizing agent and the reducing agent. 2AgNO 3(aq) + Cu (s)  Cu(NO 3 ) 2(aq) + 2Ag (s) 2Ag + NO 3 - (aq) + Cu 0 (s)  Cu +2 (NO 3 - ) 2(aq) + 2Ag 0 (s) Ag goes from +1 to 0, so it gains electrons. Cu goes from 0 to +2, so it loses electrons. Ag is reduced and is the oxidizing agent. Cu is oxidized and is the reducing agent.

Practice Problem Determine which substance is oxidized and which is reduced. Identify the oxidizing agent and the reducing agent. 4Al (s) + 3O 2(g)  2Al 2 O 3(s) 4Al 0 (s) + 3O 2 0 (g)  2Al 2 +3 O 3 -2 (s) Al goes from 0 to +3, so it loses electrons. O goes from 0 to -2, so it gains electrons. Al is oxidized and is the reducing agent. O is reduced and is the oxidizing agent.

Section 20.1 Assessment 1. Define oxidation and reduction in terms of loss or gain of oxygen. 2. Define oxidation and reduction in terms of loss or gain of electrons. 3. How do you identify the oxidizing agent and the reducing agent in a redox reaction?

Section 20.1 Assessment 4. Determine which substance is oxidized and which is reduced. Identify the oxidizing agent and reducing agent. Mg (s) + Cu(NO 3 ) 2(aq)  Mg(NO 3 ) 2(aq) + Cu (s) Mg 0 (s) + Cu +2 (NO 3 - ) 2(aq)  Mg +2 (NO 3 - ) 2(aq) + Cu 0 (s) Mg goes from 0 to +2, so it loses electrons. Cu goes from +2 to 0, so it gains electrons. Mg is oxidized and is the reducing agent. Cu is reduced and is the oxidizing agent.

Section 20.2 – Oxidation Numbers  An oxidation number is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction.  A bonded atom’s oxidation number is the charge that it would have if the electrons in the bonded were assigned to the atom of the more electronegative element.

Rules for Oxidation Numbers 1. The oxidation number of a monatomic ion is equal to its ionic charge. Ex: Br - = -1 Fe +3 = The oxidation number for H is +1 except in metal hydrides where it is -1. Ex: HCl, H = +1 NaH, H = The oxidation number for oxygen is -2 except in peroxides where it is -1. Ex: MgO, O = -2 H 2 O 2, O = -1

Rules for Oxidation Numbers 4. The oxidation number of a nonbonded element is 0. Ex: Ag = 0 N 2 = 0 5. For a neutral compound, the sum of the oxidation numbers must equal 0. Ex: NaCl = +1 and -1 equal 0. H 2 O = 2(+1) and -2 equal For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion. Ex: CO 3 -2, O = -2, so C = (3 x -2) = -2

Oxidation Numbers

Sample Problem What is the oxidation number of each element in Na 2 SO 4 ? Na 2 SO 4 = Na 2 + SO 4 -2 so Na is +1 SO 4 -2 = O is -2, so S has to be +6 Na=+1, O=-2, and S=+6

Practice Problems 1. Assign the oxidation numbers for each element in SO Assign the oxidation number for each element in (NH 4 ) 2 S. SO 2, O = -2, so S would be +4 O=-2 and S=+4 (NH 4 ) 2 S = (NH 4 ) 2 + S -2, so S = -2 NH 4 +, H = +1, so N would be -3 N=-3, H=+1, and S = -2

Oxidation Numbers  In some redox reactions, it is necessary to look at oxidation numbers instead of just using charges.

Sample Problem Identify which atoms are oxidized and which are reduced in the reaction. 2KNO 3(s)  2KNO 2(s) + O 2(g) KNO 3(s)  2KNO 2(s) + O 2(g) Since N went from +5 to +3, it is reduced. Since O went from -2 to 0, it is oxidized.

Practice Problem Identify which atoms are oxidized and which are reduced in the reaction. 2HNO 3(aq) + 6HI (aq)  2NO (g) + 3I 2(s) + 4H 2 O (l) HNO 3(aq) + 6HI (aq)  2NO (g) + 3I 2(s) + 4H 2 O (l) Since N goes form +5 to +2, it is reduced. Since I goes form -1 to 0, it is oxidized.

Section 20.2 Assessment 1. What is the general rule for assigning oxidation numbers? 2. Identify which atoms are oxidized and which are reduced. 2Na (s) + Cl 2(g)  2NaCl (s) Na (s) + Cl 2(g)  2NaCl (s) Since Na goes from 0 to +1, it is oxidized. Since Cl goes from 0 to -1, it is reduced.

Chapter 22 – Hydrocarbon Compounds  Compounds that contain carbon are classified as organic compounds.  The simplest organic compounds contain only carbon and hydrogen and are called hydrocarbons.

Section Hydrocarbons  Because carbon has four valence electrons, a carbon atom always forms four covalent bonds.

Alkanes  An alkane is a hydrocarbon in which there are only single covalent bonds.  The carbon atoms in an alkane can be arranged in a straight chain or in a chain that has branches.

Naming Alkanes  The following chart shows the names for straight- chain alkanes containing 1 to 10 carbons. # CarbonName Molecular Formula 1 Methane CH 4 2 Ethane C2H6 C2H6 3 Propane C3H8 C3H8 4 Butane C 4 H 10 5 Pentane C 5 H 12 6 Hexane C 6 H 14 7 Heptane C 7 H 16 8 Octane C 8 H 18 9 Nonane C 9 H Decane C 10 H 22

Writing the Formulas for Hydrocarbons  Molecular FormulaC 6 H 14

Branched-Chain Alkanes  An atom or group of atoms that can take the place of a hydrogen atom on a parent hydrocarbon is called a substituent.  The longest continuous carbon chain of a branched-chain hydrocarbon is called the parent alkane.

Alkyl Group  A hydrocarbon substituent is called an alkyl group.  Alkyl groups are named by removing the –ane ending from the parent hydrocarbon and adding –yl.  Ex: methane = methyl ethane = ethyl

Naming Branched-Chain Alkanes  The IUPAC rules for naming branched- chain alkanes are based on the name of the longest continuous carbon chain.  Each alkyl substituent is named according to the length of the chain and numbered according to its position on the main chain.

IUPAC Rules 1. The longest chain is the parent chain. 2. Number the carbons in the parent chain in order so that the groups attached to the chain will have the smallest numbers. 3. Add numbers to the names of the substituent groups to identify their positions.

IUPAC Rules 4. Use prefixes to indicate the appearance of the same group more than once in the structural formula. 5. List the names of the alkyl substituents in alphabetical order, but ignore any prefixes. 6. Use proper punctuation. Commas are used to separate numbers, and hyphens are used to separate numbers and words.

Sample Problems  Name the following compound. 4-ethyl-2,3-dimethylheptane

Sample Problems  Name the following compound ,3,4-trimethylhexane

Practice Problems 1. Name the following compound ethyl-3,3,4-trimethylheptane

Practice Problems 1. Name the following compound ,4-diethylheptane

Drawing a Structural Formula 1. Write the parent chain (the ending of the name). 2. Number the carbons. 3. Attach the substituent groups. 4. Add hydrogens as needed until all carbons have 4 bonds.

Sample Problem 1. Draw the structural formula for 2,2,4- trimethylpentane. 1 st draw pentane (5 carbons). C - C - C - C - C CH 3 I CH 3 I CH 3 2 nd add a methyl (-CH 3 ) group to the 2, 2, and 4. 3 rd fill in H’s until each C has 4 bonds. H 3 H 2 H H 3

Practice Problem 1. Draw the structural formula for 4-ethyl- 2,3,4-trimethyloctane. 1 st draw octane (8 carbons). 2 nd add an ethyl (CH 3 CH 2 -) to 4 and a methyl (CH 3 -) to 2, 3, and 4. 3 rd add H’s until each C has 4 bonds. C - C - C - C - C - C - C - C I CH 3 I CH 3 I CH 3 CH 2 CH 3 I H 3 H H H 2 H 2 H 2 H 3

Section 22.1 Assessment 1. Explain why carbon atoms form four covalent bonds. 2. Write the structural formula for the following alkanes. a. propaneb. pentane 3. Draw the structural formula for 2,2- dimethylbutane. CH 3 I CH 3 -C-CH 2 -CH 3 I CH 3

Section 22.2 – Unsaturated Hydrocarbons  Organic molecules that contain the maximum amount of hydrogen atoms per carbon atoms are called saturated compounds.  Compounds that contain double or triple bonds are called unsaturated compounds. AlkaneR–CH 2 –CH 2 –R AlkeneR–CH=CH–R AlkyneR–C≡C–R

Alkenes  Alkenes are hydrocarbons that contain one or more carbon-carbon double covalent bonds.  When naming alkenes, find the longest chain that includes the double bond and change the ending of the alkane name to –ene. Ex: butane = butene

Naming Alkenes  The carbons should be numbered so that the double bonded carbons have the lowest number.  Any substituents should follow the rules that we have already covered.

Alkynes  Hydrocarbons that contain one or more carbon-carbon triple bonds are called alkynes.  Alkynes are named like alkenes, except that the alkane name ending changes to –yne. Ex: propane = propyne

Sample Problem  Name the following compound methyl-2-heptene

Practice Problem  Name the following compound. CH 3 I CH 3 -CH 2 -C C-CH-CH-CH 2 -CH 3 I CH 2 CH ethyl-6-methyl-3-octyne

Section 22.2 Assessment  Describe the bonding between atoms in an alkene.  What types of bonds are present in an alkyne?  What is the difference between saturated and unsaturated hydrocarbons?

Chapter 23 – Functional Groups  The substituents of organic molecules often contain oxygen, nitrogen, sulfur, and/or phosphorus.  A functional group is a specific arrangement of atoms in an organic compound that is capable of characteristic chemical reactions.

Section 23.1 – Functional Groups  The symbol R represents any carbon chains or rings attached to the functional group.  Double bonds and triple bonds are also considered functional groups.  A halocarbon is a carbon containing compound with a halogen substituent.

Functional Groups phenyl

Functional Groups

Section 23.1 Assessment 1. What is a halocarbon? 2. Identify the functional group in each structure. a. CH 3 -OH b. CH 3 -CH 2 -NH 2 c. C - OH II O d. CH 3 -CH 2 -CH 2 -Br e. CH 3 -CH 2 -O-CH 2 -CH 3