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Solutions. – Solution , A homogeneous mixture in which all of the material is in the same state. – S ubstances present in lesser amounts, called solutes , are dispersed uniformly throughout the substance in the greater amount, the solvent

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Solutions
Solutions

– Solution, A homogeneous mixture in which all of the material is in the same state.

– Substances present in lesser amounts, called solutes, are dispersed uniformly throughout the substance in the greater amount, the solvent

– Aqueous solution — a solution in which the solvent is water

– Nonaqueous solution — any substance other than water is the solvent

– Many of the chemical reactions that are essential for life depend on the interaction of water molecules with dissolved compounds.


Polar substances

Aqueous Solutions

Polar Substances

  • An individual water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure.

  • The oxygen atom in each O–H covalent bond attracts the electrons more strongly than the hydrogen atom.

  • O and H nuclei do not share the electrons equally.

  • – Hydrogen atoms are electron-poor compared with a neutral hydrogen atom and have a partial positive charge, indicated by the symbol δ+.

  • – The oxygen atom is more electron-rich than a neutral oxygen atom and has a partial negative charge, indicated by the symbol 2δ-.

  • Unequal distribution of charge creates a polar bond, which makes them good solvents for ionic compounds.

  • Individual cations and anions are called hydrated ions.


  • Aqueous solutions
    Aqueous Solutions

    • Electrolyte — any compound that can form ions when it dissolves in water

    – When strong electrolytes dissolve, constituent ions dissociate completely, producing aqueous solutions that conduct electricity very well.

    – When weak electrolytes dissolve, they produce relatively few ions in solution. Aqueous solutions, of weak electrolytes do not conduct electricity as well as solutions of strong electrolytes.

    – Nonelectrolytes dissolve in water as neutral molecules and have no effect on conductivity.

    CH3CO2H(aq) → CH3CO2-(aq) +H+(aq)


    Molarity

    Aqueous Solutions

    Molarity

    • Most common unit of concentration

    • Most useful for calculations involving the stoichiometry of reactions in solution

    • Molarity of a solution is the number of moles of solute present in exactly 1 L of solution:

    • Units of molarity — moles per liter of solution (mol/L),

      abbreviated as M

    • Relationship among volume, molarity, and moles is expressed as:



    Calculating moles from volume
    Calculating Moles from Volume

    Calculating Volume from Mass


    Preparation of Solutions

    Problem: What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution?


    Preparation of Solutions

    0.250 L of water was used to make 0.250 L of solution. Notice the water left over.


    Preparation of Solutions

    (Vs) (Ms) = moles of solute = (Vd) (Md).


    Problem you have 50 0 ml of 3 0 m naoh and you want 0 50 m naoh what do you do

    Preparation of Solutions

    PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?


    Ion concentrations in solution
    Ion Concentrations in Solution

    • Classify each compound as either a strong electrolyte or a nonelectrolyte.

    • If a compound is a nonelectrolyte, the concentration is the same as the molarity of the solution.

    • If a compound is a strong electrolyte, determine the number of each ion contained in one formula unit and find the concentration of each species by multiplying the number of each ion by the molarity of the solution.


    Preparation of Solutions

    PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate molarity of the solution and the concentration of each of the ions.


    Zinc reacts with acids to produce H2 gas.

    Have 10.0 g of Zn

    What volume of 2.50 M HCl is needed to convert the Zn completely?

    SOLUTION STOICHIOMETRY


    Zinc reacts with acids to produce H2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely?

    SOLUTION STOICHIOMETRY


    Limiting reactants in solutions
    Limiting Reactants in Solutions

    • The concept of limiting reactants applies to reactions that are carried out in solution and reactions that involve pure substances.

    • If all the reactants but one are present in excess, then the amount of the limiting reactant can be calculated.

    • When the limiting reactant is not known, one can determine which reactant is limiting by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation.

    • Use volumes and concentrations of solutions of reactants to calculate the number of moles of reactants.


    Ionic equations
    Ionic Equations

    • Chemical equation for a reaction in solution can be written in three ways:

      1. Overall equation — shows all of the substances present in their undissociated form

      2. Complete ionic equation — shows all of the substances present in the form in which they actually exist in solution

      3. Net ionic equation

      – Derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equation with the same coefficients

      – Demonstrate that many different combinations of reactants can give the same net chemical reaction


    Ionic Equations

    • There are three ways to write reactions in aqueous solutions.

    • Molecular equation:

      • Show all reactants & products in molecular or ionic form

    • Total ionic equation:

      • Show the ions and molecules as they exist in solution

    • Net ionic equation:

      • Shows ions that participate in reaction and removes spectator ions. Spectator ions do not participate in the reaction.


    Mg(s)+ 2HCl(aq)→ H2(g) + MgCl2(aq)

    The molecular formula above can be written as the total ionic formula

    Mg(s)+ 2H+(aq)+ 2Cl-(aq)→ H2(g)+ Mg2+(aq)+ 2Cl-(aq)

    The two Cl- ions are SPECTATOR IONS — they do not participate. Could have used NO3- for the spectator ion as salts of nitrates are all soluble.

    By leaving out the spectator ions out you get the net ionic reaction

    Mg(s) + 2 H+(aq) ---> H2(g) + Mg2+(aq)

    Net Ionic Equations


    Classifying Chemical Reactions

    • Exchange reactions

      • Single Displacement Reactions – one element displaces another from a compound – AB + C  AC + B

      • Metathesis Reactions - Exchange Reactions – AB + CD  AD + CB

        • Precipitation: products include an insoluble substance which precipitates from solution as a solid

        • Acid-base neutralization: product is a salt and water

        • Gas formation – primarily the reaction of metal carbonates

  • Condensation reactions (and the reverse, cleavage reactions)

    • Combination Reactions (Condensation) – More than one reactant, one product. Some condensation reactions are redox reactions. – A + B  AB

    • Decomposition Reactions (Cleavage) – Single reactant, more than one product – AB  A + B

  • Redox (Oxidation Reduction Reactions) – Electrons are transferred between reactants. Oxidation numbers of some elements change; at least one element must increase and one must decrease in oxidation number.

    • Single Displacement Reactions are always Redox reactions as well.

    • oxidant + reductant  reduced oxidant + oxidized reductant


  • Aqueous chemical reactions metathesis
    Aqueous Chemical Reactions: Metathesis

    EXCHANGE REACTIONS

    The anions exchange places between cations.

    Precipitation

    Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2KNO3(aq)

    Pb2+(aq) + 2 I-(aq) → PbI2(s)

    Neutralization:

    NaOH(aq) + HCl(aq)→ NaCl(aq)+ H2O(l)

    OH-(aq) + H+(aq) → H2O(lq)

    Gas Formation

    MgCO3(s)+ 2HCl(aq) → 2Mg(Cl)2(aq)+ H2O(l) + CO2(g)


    Precipitation reactions
    Precipitation Reactions

    • A reaction that yields an insoluble product, a precipitate, when two solutions are mixed

    • Are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble

    • Used to isolate metals that have been extracted from their ores and to recover precious metals for recycling


    Water solubility of ionic compounds
    Water Solubility of Ionic Compounds

    If one ion from the “Soluble Compound” list is present in a compound, the compound is water soluble.


    Precipitation reactions1
    Precipitation Reactions

    In addition to understanding solubility. It is equally important to know if a reaction will occur.

    K2Cr2O7(aq) +

    AgNO3(aq) →

    2KNO3(aq) +

    Ag2Cr2O7(s)

    KBr(aq) + NaCl(aq) → KCl(aq) + NaBr(aq)


    Neutralization Reactions

    • A reaction in which an acid and a base react in stoichiometric amounts to produce water and a salt

    • Strengths of the acid and base determine whether the reaction goes to completion

      1. Reactions that go to completion

      • Reaction of any strong acid with any strong base

      • Reaction of a strong acid with a weak base

      • Reaction of weak acid with a weak base

        2. Reaction that does not go to completion is a reaction of a weak acid or a weak base with water


    Neutralization Reactions

    A brief history of

    Acid-Base

    Identification Systems


    Acids

    StrongBrönsted-Lowry acids are strong electrolytes

    HCl hydrochloric

    H2SO4 sulfuric

    HClO4perchloric

    HNO3 nitric

    ACIDS

    Acetic acid

    HNO3

    A Brönsted-Lowry Acid → H+ in water

    Weak Brönsted-Lowry acids are weak electrolytes

    CH3CO2H acetic acid (CH3COOH)

    H2CO3 carbonic acid

    H3PO4 phosphoric acid

    HF hydrofluoric acid

    Carbonic Acid


    Polyprotic Acids

    • Acids differ in the number of hydrogen ions they can donate.

      – Monoprotic acids are compounds capable of donating a single proton per molecule.

      – Polyprotic acids can donate more than one hydrogen ion per molecule.


    Ammonia nh 3 a n important weak base
    Ammonia, NH3 an Important weak Base

    NaOH(aq) → Na+(aq) + OH-(aq)

    BASES

    Brönsted-LowryBase → OH- in water

    NaOH is a strong base


    Acids1
    ACIDS

    Nonmetal oxides can be acids

    CO2(aq) + H2O(l) → H2CO3(aq)

    SO3(aq) + H2O(l) → H2SO4(aq)

    NO2(aq) + H2O(l) → HNO3(aq)

    Acid Rain is an example of nonmetal oxides behaving as acids. This process can result from burning coal and oil.

    BASES

    • Metal oxides can be bases

    • CaO(s)+H2O(l) → Ca(OH)2(aq)

    CaO in water. Phenolphthalein indicator shows a of calcium oxide solution is basic.


    Ph a concentration scale
    pH, a Concentration Scale

    pH: a way to express acidity -- the concentration of H+ in solution.

    Low pH: high [H+]

    High pH: low [H+]

    Acidic solution pH < 7

    Neutral pH = 7

    Basic solution pH > 7


    The ph scale
    The pH Scale

    pH = log (1/ [H+]) = - log [H+]

    In a neutral solution, [H+] = [OH-] = 1.00 x 10-7 M at 25 oC

    pH = - log [H+] =

    If the [H+] of soda is 1.6 x 10-3 M, the pH is ____.

    If the pH of Coke is 3.12, it is _____.


    Acid-Base Strength

    Identification

    You should know the strong acids & bases


    Oxidation reduction reactions in solution
    Oxidation-Reduction Reactions in Solution

    • Oxidation-reduction reactions — electrons are transferred from one substance or atom to another.

    • Oxidation-reduction reactions that occur in aqueous solution are complex, and their equations are very difficult to balance.

    • Two methods for balancing oxidation-reduction reactions in aqueous solution are:

      • Oxidation states — overall reaction is separated into an oxidation equation and a reduction equation

      • Half-reaction


    Oxidation-Reduction Reactions

    • The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides.

      – Metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule.

      – Oxygen atoms acquire a negative charge and form oxide ions (O2-).

      – Metals lose electrons to oxygen and have been oxidized—oxidation is the loss of electrons.

      – Oxygen atoms have gained electrons and have been reduced—reduction is the gain of electrons.


    Oxidation-Reduction Reactions

    • Oxidation and reduction reactions are now characterized by a change in the oxidation states of one or more elements in the reactants.

    • Oxidation states of each atom in a compound is the charge that atom would have if all of its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form are assigned an oxidation state of zero.

    • Oxidation-reduction reactions are called redox reactions, in which there is a net transfer of electrons from one reactant to another. The total number of electrons lost must equal the total number of electronsgained.

    • Oxidants and reductants

      • Oxidants – Compounds that are capable of accepting electrons are called oxidants, or oxidizing reagents, because they can oxidize other compounds.

        • An oxidant is reduced in the process of accepting electrons.

      • Reductants – Compounds that are capable of donating electrons are called reductants, or reducing agents, because they can cause the reduction of another compound.

        • A reductant is oxidized in the process of donating electrons.


    Oxidation numbers
    OXIDATION NUMBERS

    NH3

    ClO-

    H3PO4

    MnO4-

    Cr2O72-


    Recognizing a redox reaction
    Recognizing a Redox Reaction

    2 Ag(s) + Cu2+(aq) → Al+(aq) + Cu(s)

    Hydrogen Fuel Cell

    2 H2(g) + O2(g) → 2H2O(l)

    Thermite reaction

    Fe2O3(s) + 2Al(s) → 2 Fe(s) + Al2O3(s)



    Quantitative analysis titrations
    Quantitative Analysis: Titrations

    • Quantitative analysis — used to determine the amounts or concentrations of substances present in a sample by using a combination of chemical reactions and stoichiometric calculations

    • Titration

      – A method in which a measured volume of a solution of known concentration, called the titrant, is added to a measured volume of a solution containing a compound whose concentration is to be determined (the unknown)

      – Reaction must be fast, complete, and specific (only the compound of interest should react with the titrant)

      – Equivalence point — point at which exactly enough reactant has been added for the reaction to go to completion (computed mathematically)


    Acid base titrations
    Acid-Base Titrations

    • Most common acids and bases are not intensely colored – Rely on an indicator

    • Endpoint — point at which a color change is observed, which is close to the equivalence point in an acid-base titration


    Standard solutions
    Standard Solutions

    • A solution of a primary standard whose concentration is known precisely

    • A primary standard is non-hygroscopic, has a high mass, is fairly inexpensive compound, is of known reactive ability that can be accurately weighed for use as a titrant.

    • A standard solution is used to determine the concentration of the titrant.

    • Accuracy of any titration analysis depends on accurate knowledge of the concentration of the titrant.

    • Most titrants are first standardized—their concentration is measured by titration with a standard solution.


    Titration
    Titration

    1. Add titrant solution from the buret.

    2. Reagent (base) reacts with compound (acid) in solution in the flask.

    3. Indicator shows when exact stoichiometric reaction has occurred.

    4. Net ionic equation

    H++ OH- → H2O

    5. At equivalence point

    moles H+= moles OH-


    Acid base reactions titrations
    ACID-BASE REACTIONSTitrations

    H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)

    acidbase

    Carry out this reaction using a TITRATION.

    Oxalic acid,

    H2C2O4


    PROBLEM: Standardize a solution of NaOH — i.e., accurately determine its concentration. 1.065 g of H2C2O4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH?

    Quantitative Analysis: Titrations


    PROBLEM : Use standardized NaOH to determine the amount of an acid in an unknown. Apples contain malic acid, C4H6O5. 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

    Quantitative Analysis: Titrations

    HOOCCH2COHCOOH(aq) + 2NaOH(aq) → Na2C4H4O5(aq) + 2H2O(l)


    Energy chemistry
    Energy & Chemistry an acid in an unknown

    2H2(g) + O2(g) → 2H2O(g) + heat and light

    This can be set up to provide

    ELECTRIC ENERGY in a fuel cell.

    Oxidation:

    2 H2 → 4 H+ + 4 e-

    Reduction:

    4 e- + O2 + 2 H2O → 4 OH-

    H2/O2 Fuel Cell Energy


    Energy & Chemistry an acid in an unknown

    ENERGYis the capacity to do work (w) or transfer heat (q).

    HEATis the thermal energy that can be transferred from an object at one temperature to an object at another temperature

    – Net transfer of thermal energy stops when the two objects reach the same temperature.

    Other forms of energy —

    • Radiant (light)— energy in light, microwaves,

      and radio waves

    • Thermal (kinetic and potential) — results from

      atomic and molecular motion – Temperature of an

      object is a measure of the thermal energy content

    • Chemical — results from the particular arrangement

      of atoms in a chemical compound; radiant and thermal

      energy produced in this reaction due to energy released

      during the breaking and reforming of chemical bonds

    • Nuclear — radiant and thermal energy released when

      particles in the nucleus of the atoms are rearranged

    • Electrical— due to the flow of electrically charged

      particles


    Potential kinetic energy
    Potential & Kinetic Energy an acid in an unknown

    Kinetic energy

    • Potential energy – Energy storedin an object because of the relative positions or orientations of its components –

    • PE = Fd = mad = mgh = work

    • Kinetic energy – Energy due to the motion of an object –

    • KE = ½ mv2


    Potential & Kinetic Energy an acid in an unknown


    Internal energy e
    Internal Energy (E) an acid in an unknown

    • PE + KE = Internal energy (E or U)

    • Internal Energy of a chemical system depends on

      • number of particles

      • type of particles

      • temperature

    • The higher the T the higher the internal energy

    • So, use changes in T (∆T) to monitor changes in E (∆E).


    Internal energy e1

    heat transfer in an acid in an unknown

    (endothermic), +q

    heat transfer out

    (exothermic), -q

    w transfer in

    (+w)

    w transfer out

    (-w)

    Internal Energy (E)

    SYSTEM

    ∆E = q + w


    Energy chemistry1
    Energy & Chemistry an acid in an unknown

    All of thermodynamics depends on the law of

    CONSERVATION OF ENERGY.

    • The total energy is unchanged in a chemical reaction.

    • If PE of products is less than reactants, the difference must be released as KE.

    Energy Change in Chemical Processes

    Potential Energy of system dropped. Kinetic energy increased. Therefore, you often feel a Temperature increase.


    Thermodynamics enthalpy
    Thermodynamics -Enthalpy an acid in an unknown

    • Thermodynamics – the science of heat (energy) transfer.

    Heat transfers until thermal equilibrium is established. ∆T measures energy transferred.

    • SYSTEM – The object under study

      • 1. Open system — can exchange both matter and energy with its surroundings

      • 2. Closed system — can exchange energy but not matter with its surroundings

      • 3. Isolated system — exchanges neither energy nor matter with the surroundings; total energy of the system plus the surroundings is constant

    • SURROUNDINGS – Everything outside the system


    First law of thermodynamics

    heat energy transferred an acid in an unknown

    work done by the system

    energy

    change

    FIRST LAW OF THERMODYNAMICS

    ∆E = q + w

    Energy is conserved!


    The first law of thermodynamics
    The First Law of Thermodynamics an acid in an unknown

    • Exothermic reactions generate specific amounts of heat.

    • This is because the potential energies of the products are lower than the potential energies of the reactants.


    The first law of thermodynamics1
    The First Law of Thermodynamics an acid in an unknown

    • There are two basic ideas of importance for thermodynamic systems.

    • Chemical systems tend toward a state of minimum potential energy.

    • Chemical systems tend toward a state of maximum disorder.

    • The first law is also known as the Law of Conservation of Energy.

      • Energy is neither created nor destroyed in chemical reactions and physical changes.


    State of a System an acid in an unknown

    • The state of the system is a complete description of the system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter Some examples of state functions are:

      • T (temperature), P (pressure), V (volume), E (change in energy), H (change in enthalpy – the transfer of heat), and S (entropy)

    • Examples of non-state functions are:

      • n (moles), q (heat), w (work)

    ∆H along one path = ∆H along another path

    • This equation is valid because ∆H is a STATE FUNCTION

    • These depend only on the state of the system and not how it got there.

    • V, T, P, energy — and your bank account!

    • Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H.


    Standard states and standard enthalpy changes
    Standard States and Standard Enthalpy Changes an acid in an unknown

    • Thermochemical standard state conditions

      • The thermochemical standard T = 298.15 K.

      • The thermochemical standard P = 1.0000 atm.

        • Be careful not to confuse these values with STP.

    • Thermochemical standard states of matter

      • For pure substances in their liquid or solid phase the standard state is the pure liquid or solid.

      • For gases the standard state is the gas at 1.00 atm of pressure.

        • For gaseous mixtures the partial pressure must be 1.00 atm.

      • For aqueous solutions the standard state is 1.00 M concentration.

    • ∆Hfo = standard molar enthalpy of formation

      • the enthalpy change when 1 mol of compound is formed from elements under standard conditions.


    State function
    State Function an acid in an unknown

    • The properties of a system that depend only on the state of the system are called state functions.

      • State functions are always written using capital letters.

    • The value of a state function is independent of pathway.

    • An analog to a state function is the energy required to climb a mountain taking two different paths.

      • E1 = energy on the first floor of Heldenfels

      • E1 = mgh1

      • E2 = energy on the fourth floor of Heldenfels

      • E2 = mgh2

      • E = E2-E1 = mgh2 – mgh1 = mg(h)


    Enthalpy
    ENTHALPY an acid in an unknown

    Most chemical reactions occur at constant P, so

    Heat transferred at constant P = qp

    qp = ∆H where H = enthalpy

    and so ∆E = ∆H + w (and w is usually small)

    ∆H = heat transferred at constant P ≈ ∆E

    ∆H = change in heat content of the system

    ∆H = Hfinal - Hinitial


    Directionality of heat transfer

    T(system) goes down an acid in an unknown

    T(surr) goes up

    Directionality of Heat Transfer

    • Heat always transfer from hotter object to cooler one.

    • EXOthermic: heat transfers from SYSTEM to SURROUNDINGS.


    Directionality of heat transfer1

    T(system) goes up an acid in an unknown

    T (surr) goes down

    Directionality of Heat Transfer

    • Heat always transfers from hotter object to cooler one.

    • ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM.


    Enthalpy1
    ENTHALPY an acid in an unknown

    ∆H = Hfinal - Hinitial

    If Hfinal > Hinitial then ∆H is positive

    Process is ENDOTHERMIC

    If Hfinal < Hinitial then ∆H is negative

    Process is EXOTHERMIC


    USING ENTHALPY an acid in an unknown

    Consider the formation of water

    H2(g) + 1/2 O2(g) → H2O(g) + 241.8 kJ

    Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ


    USING ENTHALPY an acid in an unknown

    Example of HESS’S LAW— Making liquid H2O from H2 + O2 involves twoexothermic steps.

    Liquid H2O

    H2O vapor

    H2 + O2 gas

    Making liquid H2O from H2 involves two steps.

    H2(g) + 1/2 O2(g) → H2O(g) + 242 kJ

    H2O(g) → H2O(l) + 44 kJ

    H2(g) + 1/2 O2(g) → H2O(l) + 286 kJ

    If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.


    Enthalpy values
    Enthalpy Values an acid in an unknown

    H2(g) + 1/2 O2(g) → H2O(g) ∆H˚ = -242 kJ

    2H2(g) + O2(g) → 2H2O(g) ∆H˚ = -484 kJ

    H2O(g) → H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ

    H2(g) + 1/2 O2(g) → H2O(l) ∆H˚ = -286 kJ

    Depend on how the reaction is written and on phases of reactants and products


    Standard molar enthalpies of formation h f o
    Standard Molar Enthalpies of Formation, an acid in an unknownHfo

    • The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements.

      • The symbol for standard molar enthalpy of formation is Hfo.

    • The standard molar enthalpy of formation for MgCl2 is:


    Standard molar enthalpies of formation h f o1
    Standard Molar Enthalpies of Formation, an acid in an unknownHfo

    • Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 15-1 and Appendix K in the text.

    • Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero.

    • Example 15-4: The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for whichHorxn = -1281 kJ.

      P in standard state is P4

      Phosphoric acid in standard state is H3PO4(s)


    Standard molar enthalpies of formation h f o2
    Standard Molar Enthalpies of Formation, an acid in an unknownHfo

    H2(g) + ½ O2(g) → H2O(g) ∆Hf˚ (H2O, g)= -241.8 kJ/mol

    C(s) + ½ O2(g) → CO(g)∆Hf˚ of CO = - 111 kJ/mol

    By definition, ∆Hfo= 0 for elements in their standard states.

    Use ∆H˚’s to calculate enthalpy change for

    H2O(g) + C(graphite) → H2(g) + CO(g)


    Standard molar enthalpies of formation h f o3
    Standard Molar Enthalpies of Formation, an acid in an unknownHfo

    • Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere.


    Standard molar enthalpies of formation h f o4
    Standard Molar Enthalpies of Formation, an acid in an unknownHfo

    • Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere.


    Hess’s Law – Enthalpies of an acid in an unknown

    Formation and Reaction

    • Standard enthalpies of reaction (ΔH°xn)

      – The enthalpy that occurs when a reaction is carried out with all reactants and products in their standard states

      – General reaction: aA + bB  cC + dD where A, B, C, and D are chemical substances and a, b, c, and d are their stoichiometric coefficients

      – Magnitude of ΔH°rxn is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient , minus the sum of the standard enthalpies of formation of the reactants, multiplied by their coefficients

      ΔH°rxn= [cΔH°f(C) + dΔH°f(D)]  [aΔH°f(A) + bΔH°f(B)]

      or

      ΔH°rxn= mΔH°f(product)  nΔH°f(reactants)

      where  means “sum of” and m and n are the stoichiometric coefficients of each of the products and reactants, respectively


    Hess s law energy level diagrams
    Hess’s Law & Energy Level Diagrams an acid in an unknown

    Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.


    Enthalpies of Formation an acid in an unknown

    • Standard enthalpies of formation

      –Themagnitude of ΔH for a reaction depends on the physical states of the reactants and products, the pressure of any gases present, and the temperature at which the reaction is carried out.

      – A specific set of conditions under which enthalpy changes are measured are used to ensure uniformity of reaction conditions and data.

      – Standard conditions: a pressure of 1 atmosphere (atm) for gases and a concentration of 1 M for species in solution (1 mol/L), each pure substance must be in its standard state, its most stable form at a pressure of 1 atm at a specified temperature.

      – Enthalpies of formation measured under standard conditions are called standard enthalpies of formation (ΔH°f).

      – Standard enthalpy of formation of any element in its standard state is zero.


    Enthalpies of Reaction an acid in an unknown

     Hess’s law allows the calculation of the enthalpy change for any conceivable chemical reaction by using a relatively small set of tabulated data

    • Enthalpy of combustion, ΔHcomb—enthalpy change that occurs when a substance is burned in excess oxygen

    • Enthalpy of fusion, ΔH fus—enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance

    • Enthalpy of vaporization, ΔHvap—the enthalpy change that accompanies the vaporization of 1 mol of a substance

    • Enthalpy of solution, ΔHsoln—enthalpy change when a specified amount of solute dissolves in a given quantity of solvent

    • Enthalpy of formation, ΔHf—enthalpy change for the formation of 1 mol of a compound from its component elements


    Hess s law
    Hess’s Law an acid in an unknown

    • Hess’s Law of Heat Summation, Hrxn = H1 +H2 +H3 + ..., states that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps.

      • Hess’s Law is true because H is a state function.

    • If we know the following Ho’s


    Hess’s Law an acid in an unknown

    • For example, we can calculate the Ho for reaction [1] by properly adding (or subtracting) the Ho’s for reactions [2] and [3].

    • Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a product.

      • Arrange reactions [2] and [3] so that they also have FeO and O2 as reactants and Fe2O3 as a product.

        • Each reaction can be doubled, tripled, or multiplied by half, etc.

        • The Ho values are also doubled, tripled, etc.

        • If a reaction is reversed the sign of the Ho is changed.


    Hess s law1
    Hess’s Law an acid in an unknown

    • Given the following equations and Hovalues

      calculate Ho for the reaction below.


    Hess s law2
    Hess’s Law an acid in an unknown

    • Calculate the H o298 forthe following reaction:


    Hess s law3
    Hess’s Law an acid in an unknown

    • Application of Hess’s Law and more algebra allows us to calculate the Hfofor a substance participating in a reaction for which we know Hrxno , if we also know Hfofor all other substances in the reaction.

    • Given the following information, calculate Hfo for H2S(g).


    Heat capacity
    HEAT CAPACITY an acid in an unknown

    The heat required to raise an object’s T by 1 ˚C.

    Which has the larger heat capacity?

    Thermal energy cannot be measured easily.

    Temperature change caused by the flow of thermal energy between objects or substances can be measured.

    Calorimetry is the set of techniques employed to measure enthalpy changes in chemical processes using calorimeters.


    Specific heat capacity

    Specific heat capacity an acid in an unknown

    =

    heat lost or gained by substance (J)

    (mass,

    g) (T change,

    K)

    Specific Heat Capacity

    How much energy is transferred due to Temperature difference?

    The heat (q) “lost” or “gained” is related to

    a) sample mass

    b) change in T and

    c) specific heat capacity


    Table of specific heat capacities an acid in an unknown

    • Heat capacity of an object depends on both its massand itscomposition.

    • Molar heat capacity (Cp) — amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; units of Cp are J/(mol•°C)

    • Specific heat (Cs) — amount of energy needed to increase the temperature of 1 g of a substance by 1°C, units are J/(g•°C)

    • The quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change are related in two ways:

      q = nCpΔT where n = number of moles of substance

      q = mCsΔT where m = mass of substance in grams

    Aluminum


    Specific heat capacity1
    Specific Heat Capacity an acid in an unknown

    If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?


    Heat energy transfer no change in state
    Heat/Energy Transfer an acid in an unknownNo Change in State

    q transferred = (sp. ht.)(mass)(∆T)


    Heat transfer
    Heat Transfer an acid in an unknown

    • Use heat transfer as a way to find specific heat capacity, Cp

    • 55.0 g Fe at 99.8 ˚C

    • Drop into 225 g water at 21.0 ˚C

    • Water and metal come to 23.1 ˚C

    • What is the specific heat capacity of the metal?


    Heating cooling curve for water
    Heating/Cooling Curve for Water an acid in an unknown

    Note that T is constant as ice melts or water boils


    Thermochemical equations
    Thermochemical Equations an acid in an unknown

    • Thermochemical equations are a balanced chemical reaction plus the H value for the reaction.

      • For example, this is a thermochemical equation.

    • The stoichiometric coefficients in thermochemical equations must be interpreted as numbers of moles.

    • 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to as one mole of reactions.


    Thermochemical equations1
    Thermochemical Equations an acid in an unknown

    • This is an equivalent method of writing thermochemical equations.

    • H < 0 designates an exothermic reaction.

    • H > 0 designates an endothermic reaction


    Thermochemical equations2
    Thermochemical Equations an acid in an unknown

    • Write the thermochemical equation for the reaction for

      CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

      50.0mL of 0.400 M CuSO4 at 23.35 oC Tfinal 25.23oC

      50.0mL of 0.600 M NaOH at 23.35 oC

      Density final solution = 1.02 g/mL CH2O = 4.184 J/goC


    Standard temperature and pressure

    Gas Laws an acid in an unknown

    Standard Temperature and Pressure

    • At the macroscopic level, a complete physical description of a sample of a gas requires four quantities:

      • 1. Temperature (expressed in K)

      • 2. Volume (expressed in liters)

      • 3. Amount (expressed in moles)

      • 4. Pressure (given in atmospheres)

  • • These variables are not independent — if the values of any three of these quantities are known, the fourth can be calculated.

    • Standard temperature and pressure is given the symbol STP.

      • It is a reference point for some gas calculations.

    • Standard P  1.00000 atm or 101.3 kPa

    • Standard T  273.15 K or 0.00oC

      • Gas laws must use the Kelvin scale to be correct.

    • Relationship between Kelvin and centigrade.


    Boyle s law the volume pressure relationship
    Boyle’s Law: The Volume-Pressure Relationship an acid in an unknown

    • V  1/P or V= k (1/P) or PV = k

    • P1V1 = k1 for one sample of a gas.

    • P2V2 = k2 for a second sample of a gas.

    • k1 = k2 for the same sample of a gas at the same T.

    • Mathematically we write Boyle’s Law as P1V1 = P2V2

    • This relationship between pressure and volume is known as Boyle’slaw which states that at constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure.

    Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.


    Charles law the volume temperature relationship the absolute temperature scale
    Charles’ Law: The Volume-Temperature Relationship; an acid in an unknownThe Absolute Temperature Scale

    Charles’s and Gay-Lussac’s findings can be stated as: At constant pressure, the volume of a fixed amount of a gas is directly proportional to its absolute temperature

    (in K).

    This relationship is referred to as Charles’s law and is stated mathematically as

    V = (constant) [T (in K)] or V  T (in K, at constant P).

    Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist.


    Charles law the volume temperature relationship the absolute temperature scale1
    Charles’ Law: The Volume-Temperature Relationship; an acid in an unknownThe Absolute Temperature Scale

    35

    30

    25

    20

    Volume (L)

    15

    Gases liquefy

    10

    before reaching 0K

    5

    • Gay-Lussac showed that a plot of V versus T was a straight line that could be extrapolated to –273.15ºC at zero volume, a theoretical state.

    • The slope of the plot of V versus T varies for the same gas at different pressures, but the intercept remains constant at –273.15ºC.

    • Significance of the invariant T intercept in plots of V versus T was recognized by Thomson (Lord Kelvin), who postulated that –273.15ºC was the lowest possibletemperature that could theoretically be achieved, and he called it absolute zero (0 K).

    0

    0

    50

    100

    150

    200

    250

    300

    350

    400

    Temperature (K)

    absolute zero = -273.15 0C


    The combined gas law equation
    The Combined Gas Law Equation an acid in an unknown

    • Boyle’s and Charles’ Laws combined into one statement is called the combined gas law equation.

      • Useful when the V, T, and P of a gas are changing.


    Avogadro s law and the standard molar volume
    Avogadro’s Law and an acid in an unknownthe Standard Molar Volume

    • Avogadro’s Law states that at the same temperature and pressure, equal volumes of two gases contain the same number of molecules (or moles) of gas.

    • If we set the temperature and pressure for any gas to be STP, then one mole of that gas has a volume called the standard molar volume.

    • Stated mathematically: V = (constant) (n) or V  n (at constant T and P)

    • The standard molar volume is 22.4 L at STP.

      • This is another way to measure moles.

      • For gases, the volume is proportional to the number of moles.

        11.2 L of a gas at STP = 0.500 mole

        44.8 L of a gas at STP = ? Moles


    Boyle s law the volume pressure relationship1
    Boyle’s Law: an acid in an unknownThe Volume-Pressure Relationship

    • At 25oC a sample of He has a volume of 4.00 x 102 mL under a pressure of 7.60 x 102 torr. What volume would it occupy under a pressure of 2.00 atm at the same T?


    Charles law the volume temperature relationship the absolute temperature scale2
    Charles’ Law: an acid in an unknownThe Volume-Temperature Relationship; The Absolute Temperature Scale

    • A sample of hydrogen, H2, occupies 1.00 x 102 mL at 25.0oC and 1.00 atm. What volume would it occupy at 50.0oC under the same pressure?


    The combined gas law equation1
    The Combined Gas Law Equation an acid in an unknown

    • A sample of nitrogen gas, N2, occupies 7.50 x 102 mL at 75.00C under a pressure of 8.10 x 102 torr. What volume would it occupy at STP?


    Avogadro s law and the standard molar volume1
    Avogadro’s Law and the an acid in an unknownStandard Molar Volume

    • One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given temperature and pressure. (a) What is its molar mass? (b) What is its density at STP?


    Summary of gas laws the ideal gas law
    Summary of Gas Laws: an acid in an unknownThe Ideal Gas Law

    • What volume would 50.0 g of ethane, C2H6, occupy at 1.40 x 102 oC under a pressure of 1.82 x 103 torr?


    Summary of gas laws the ideal gas law1
    Summary of Gas Laws: an acid in an unknownThe Ideal Gas Law

    • Calculate the number of moles in, and the mass of, an 8.96 L sample of methane, CH4, measured at standard conditions?


    Dalton s law of partial pressures
    Dalton’s Law of Partial Pressures an acid in an unknown

    The ideal gas law assumes that all gases behave identically. If V, T, and n for each gas in a mixture are known, the pressure of each gas, its partial pressure, can be calculated. Holding V and T constant, P is directly proportional to n:

    P = n(RT/V) = n(constant)

    Generally, for a mixture oficomponents, the total pressure is given by

    Pt = (n1 + n2 + n3 + - - - +ni) (RT/V).

    The above equation makes it clear that, at constant T and V, P depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of gaseous species. Nothing in the equation depends on the nature of the gas, only on the quantity.

    The total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law is known as Dalton’s law of partial pressures and can be written mathematically as

    Pt= P1+ P2+ P3- - - + Pi

    where Pt is the total pressure and the other terms are the partial pressures of the individual gases.

    John Dalton 1766-1844

    • Vapor Pressure is the pressure exerted by a substance’s vapor over the substance’s liquid at equilibrium.

    • Partial pressure is the pressure the gas would exert if it were the only one present (at the same temperature and volume).


    Dalton s law of partial pressures1
    Dalton’s Law of Partial Pressures an acid in an unknown

    • If 1.00 x 102 mL of hydrogen, measured at 25.0 oC and 3.00 atm pressure, and 1.00 x 102 mL of oxygen, measured at 25.0 oC and 2.00 atm pressure, were forced into one of the containers at 25.0 oC, what would be the pressure of the mixture of gases?


    Dalton s law of partial pressures2
    Dalton’s Law of Partial Pressures an acid in an unknown

    • A sample of hydrogen was collected by displacement of water at 25.0 oC. The atmospheric pressure was 748 torr. What pressure would the dry hydrogen exert in the same container?


    Mass volume relationships in reactions involving gases
    Mass-Volume Relationships an acid in an unknownin Reactions Involving Gases

    2 mol KClO3 yields 2 mol KCl and 3 mol O2

    2(122.6g) yields 2 (74.6g) and 3 (32.0g)

    Those 3 moles of O2 can also be thought of as:

    3(22.4L)

    or

    67.2 L at STP

    • In this section we are looking at reaction stoichiometry, like in Chapter 3, just including gases in the calculations.


    Mass volume relationships in reactions involving gases1
    Mass-Volume Relationships in Reactions Involving Gases an acid in an unknown

    • What volume of oxygen measured at STP, can be produced by the thermal decomposition of 120.0 g of KClO3?


    Real gases deviations from ideality
    Real Gases: Deviations from Ideality an acid in an unknown

    • The Ideal Gas Law ignores both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive.

    • In reality, all gases have a volume and the molecules of real gases interact with one another.

    • For an ideal gas, a plot of PV/nRT versus P gives a horizontal line with an intercept of 1 on the PV/nRT axis.

    • Real gases behave ideally at ordinary temperatures and pressures. At low temperatures and high pressures real gases do not behave ideally.

    • The reasons for the deviations from ideality are:

      • The molecules are very close to one another, thus their volume is important.

      • The molecular interactions also become important.

    J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910.


    Real gases deviations from ideality1
    Real Gases: Deviations from Ideality an acid in an unknown

    The van der Waals’ equation

    (P + an2/V2) (V – nb) = nRT

    accounts for the behavior of real gases at low temperatures and high pressures.

    • The van der Waals constants a and b are empirical constants that differ for each gas that take into account two things:

      • Pressure term, P + (an2/V2)

        • a corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law

        • For nonpolar gases the attractive forces are London Forces

        • For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds.

      • n2/V2 represents the concentration of the gas (n/V) squared because it takes two particles to engage in the pairwise intermolecular interactions

    • Volume term, V – nb, corrects for the volume occupied by the gaseous molecules

      • b accounts for volume of gas molecules

    At large volumes a and b are relatively small and the van der Waal’s equation reduces to the ideal gas law at high temperatures and low pressures.


    Real gases deviations from ideality2
    Real Gases: an acid in an unknownDeviations from Ideality

    • Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the ideal gas law.

    PV = nRT

    P = nRT/V n = 84.0g * 1mol/17 g T = 200 + 273

    P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K)

    (5 L)

    P = 38.3 atm


    Real gases deviations from ideality3
    Real Gases: Deviations from Ideality an acid in an unknown

    • Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the van der Waal’s equation. The van der Waal's constants for ammonia are: a = 4.17 atm L2 mol-2 b =3.71x10-2 Lmol-1

    n = 84.0g * 1mol/17 g T = 200 + 273

    P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K) (4.94 mol)2*4.17 atm L2 mol-2

    5 L – (4.94 mol*3.71E-2 L mol-1) (5 L)2

    P = 39.81 atm – 4.07 atm = 35.74

    P = 38.3 atm

    7% error


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