Chemistry  111          Lecture Notes                                               

Week 6    September 27-Oct.1, 2004

 

Oxidation-Reduction Reactions (continued)

 

Assign oxidation numbers to all atoms in all the materials listed in the reactions below.  Show where there has been a loss of electrons and where there has simultaneously been a gain of electrons.

 

     Mg(s)  +  2 HCl (aq) ---->  H2(g)   +    MgCl2(aq)

 

Practice.  Examples:   Assign oxidation numbers to all atoms in all the materials listed in the reactions below.  Show where there has been a loss of electrons and where there has simultaneously been a gain of electrons.

 

1.   2 Ca(s)  +  O2(g) ------>  2 CaO(s)   

2.   Mg(s)  + 2 HCl (aq)  ------>   MgCl2(aq)  +   H2(g)

3.  Zn(s)  + 2 NO 3- (aq)   +    4 H+(aq)  ---->  Zn+2(aq)   +   2 NO2   +   2 H2O(l)

 

 

Types of Redox(Oxidation/Reduction) Reactions:

  1. Combination Reactions:

Elements combine to form a compound

S(s)  +  O2(g) à  SO2(g)

 

  1. Decomposition Reactions:

A compound breaks down into two or more components of which at least one is an element.

            2 KClO3(s) à  2 KCl(s)   +  3 O2(g)

 

  1. Displacement Reactions:
    1.  Hydrogen Displacements:

Some metals can displace hydrogen from a compound:

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

The Activity Series lists metals by their ability to displace hydrogen from

an acid, cold water, or hot water.

 

 

 

 

 

 

    1.  Metal Displacements:  Metals can displace other metals in a compound.

    The Activity Series:  A metal can displace metals below it in the series

   Cu(s) +  AgNO3(aq) à  Ag(s)  +  Cu(NO3)2 (aq)

 

 

    1. Halogen Displacements:  F2>Cl2>Br2>I2 reactivity

                                     Cl2  + 2 NaBr(aq)  à Br2 (aq)    +  2 NaCl(aq)

 

 

 

 

 

 


           

4.      Disproportionation Reactions:  An element in the reaction undergoes both

oxidation and reduction.   Cl2  + 2 OH-  à  ClO-  +   Cl-  +  H2O

 

 

 

Solutions:

           

Expressing Concentration of a solution: There are many ways to express the concentration of a solution. 

 

M= Molarity= #moles of solute/ #Liters of solution   

                       

m= molality= #moles of solute/# kg of solvent

 

mass % = (mass of X/Total mass of the sample) x100%

 

volume % = volume of X/Total volume of the sample)  x 100%

 

X = mole fraction=  # moles of A / (Total # moles of everything in the sample)

 

Ppm= parts per million= #grams of solute/ 1 million grams of solute

            (in water solutions that is = #mg solute/Liter of water)

 

Ppb = parts per billion = #g solute/ 1 billion grams of solute

 

 

 

 

 

Preparing Solutions

A.  By mixing solute with solvent

                        Example:  Make 250.00 mL of a 1.00 M solution of NaOH.

                                    M= # moles solute/# Liters of solution

 

1.00 M = x moles/ 0.25000 Liters

 

x = 0.250 moles NaOH

                       

0.250    moles NaOH x 40.0 grams/1 mole NaOH = 10.0 grams NaOH

                       

Add 10.0 g of NaOH to a 250.00 mL volumetric flask, add about 100 mL of water, swirl to dissolve the NaOH.  When all of the solid has dissolved, dilute to the mark on the flask.  Stopper tightly and invert 20 times to mix the solute and the solvent.

                                  

           

B.  By Diluting a more concentrated solution

Example:  Make 500.00 mL of a 1.00 M HCl solution using commercial HCl solution that is 12.00 M HCl.

 

Useful formula = Dilution Formula            Derive this one line formula that you can use for

C1V1 = C2V2                                   dilution determinations:                                                                                           

 

Using Concentration Information in “Stoichiometry” problems:

  1. Gravimetric Analyses:
    1.  CaCO3 (s) + 2 HCl(aq)  à CaCl2(aq)  + H2O(l)  +  CO2(g)

            What  mass of calcium carbonate will react with 25.0 mL of 0.750 M HCl?

 

 

 

                

 

 

 

 

 

 b.  A 0.5662 grams sample of an ionic salt of an unknown cation and chloride ion is

                        dissolved in water and treated with excess AgNO3.   If 1.0882 grams of AgCl

                        precipitate from the reaction, what is the % of chloride in the original sample of

                        compound?

 

2.  The careful, controlled addition of an acid to a base or a base to an acid is called “titration”.   If you know the concentration of either the acid or the base, this method can be used to determine the concentration of the other. 

a.      Write the balanced equation for the reaction of NaOH and H2SO4 to yield Na2SO4 and water.

b.      Determine the concentration of a NaOH solution if 25.00 mL of the NaOH solution require 33.20 mL of 2.000M H2SO4 to completely react, i.e. to reach the equivalence point (equivalence point = the point where the base and acid have completely  reacted to form salt and water)

 

           

  

Gases

 

Kinetic Molecular Theory

            Gas particles are in constant, random motion.

            Gas particles collide with each other and with the walls of the container.

            Gas particles collide with the walls of the container and each collision exerts a

                        force on the wall.      Force/Area = Pressure

                                    units for measuring pressure:

                                    1 atm= 760mmHg = 760 torr

            Temperature is a measure of the average velocity of the particles.

            A Gas is mostly empty space in which particles are moving in random, straight-

                        line motion until they collide.

            An Ideal Gas is one in which the particles collide with perfectly elastic collisions.

 

How do gases behave?

            As pressure is increases                         ?volume

            As temperature increases,                       ?volume

            As the # of gas particles increases       ?volume

 

The Ideal Gas Law:  [Check problems on pp 172-192.]

            PV = nRT

 

Using the Ideal Gas Law:

            P1V 1   = R   and   P2V2= R

            n1 T1                      n2T2

 

            Therefore,     P1V 1   = P2V2

                                    n1 T1      n2T2

 

                                           

Boyles’ Law:  If the temperature of a sample of gas is constant, T1=T2 and n1=n2.

           

                        P1V 1   = P2V2

                        n1 T1      n2T2

 

                                P1V 1= P2V2

 

 

 

 

Charles’ Law:  If the pressure of a sample of gas is constant, P1 = P2 and n1=n2.

                        P1V 1   = P2V2

                        n1 T1      n2T2

 

                                V 1   = V2

                        T1        T2

 

            Absolute Zero from Charles’ Law data;

 

 

 

 

Avogadro’s Law:  If the pressure and temperature of a sample of gas are constant:

                        P1V 1   = P2V2

                        n1 T1      n2T2

                       

                        V 1    =   V2

                        n1           n2

 

 

 

 

 

Things you can determine about gases using the Ideal Gas Law:

 

            1.  Density of a gas:          

                                                PV = nRT       We know that n= # moles = #grams/MM

                       

                                                PV = (#g/MM)RT

 

                        Rearrange:  PV(MM) = #g RT

 

                                                P(MM) = (#gRT)/V

 

                                                (P(MM))/(RT) = #g/V = mass/volume= density

 

 

  1. Molar Mass of a Gas:

PV = nRT       We know that n= # moles = #grams/MM

                       

                                                PV = (#g/MM)RT

 

                        Rearrange:  PV(MM) = #g RT

 

                                                MM = (#gRT)/(PV)  or   (#g/V) (RT/P)

 

                        MM = (density RT)/P

 

 

  1. Stoichiometry:

PV = nRT

 

n= (PV)/(RT)= # moles of gas

           

Another way to fine the # moles of material in a reaction vessel and from which you can do stoichiometry:

 

                        Example:  Given the reaction for the burning of acetylene:

  2 C2H2(g)  +   5 O2(g) à  4 CO2(g)   +  2 H2O(g)

 

What volume of oxygen gas at STP(0oC and 1.00 atm) will be needed to react completely with 7.64 Liters of acetylene at STP? (p. 184)