CHEMICAL EQILIBRIUM

 

1.         MANY CHEMICAL REACTIONS ARE REVERSIBLE.  THIS MEANS THE PRODUCTS CAN REGENERATE THE REACTANTS.

 

2.         THIS IS SHOWN BY ARROWS IN BOTH DIRECTIONS IN A CHEMICAL EQUATION.

 


                                    2  N02   (g)                            N204  (g)

 

 

3.         N02 GOING TO N204 IS CALLED THE FORWARD REACTION.  N204 GOING TO N02 IS CALLED THE REVERSE REACTION.

 

4.         AT THE BEGINNING OF THE FORWARD REACTION THERE IS NO REVERSE REACTION.  AS THE FORWARD REACTION CONTINUES THE REVERSE REACTION BEGINS, SPEEDING UP WHILE THE FORWARD REACTION SLOWS DOWN, UNTIL THE FORWARD AND REVERSE REACTION RATES EQUAL EACH OTHER.  THIS IS CALLED DYNAMIC CHEMICAL EQUILIBRIUM.

 

                                    THE LAW OF CHEMICAL EQUILIBRIUM

 

1.         THIS WAS DISCOVERED BY TWO NORWEGIAN CHEMISTS WHO FOUND THE FOLLOWING RELATIONSHIP EXISTED IN EQUILIBRIUM REACTIONS.

 

GIVEN THE FOLLOWING GENERAL REACTION:

 

                        aA    +   bB                      cC     +    dD

 

Where a, b, c, d, ARE THE COEFFICIENTS AND A, B, C, D ARE THE REACTANTS AND PRODUCTS.

 

THISRELATIONSHIP WAS FOUND

           

                                    Keq    =     [C]c________[D]d

                                                       [A]a               [B]b

 

            Keq  =  THE EQUILIBRIUM CONSTANT FOR THE EQUILIBRIUM EQUATION BASED ON THE RATIO OF THE PRODUCT CONCENTRATIONS TO REACTANT CONCENTRATIONS.   (A,    B,   C,   D)   =  CONCENTRATION IN TERMS OF MOLARITY.

 

2.         EVERY REVERSIBLE REACTION OBEYS THIS RELATIONSHIP AND HAS A SPECIFIC CONSTANT Keq WHEN THE CONCENTRATIONS HAVE REACHED EQUILIBRIUM AT A GIVEN TEMPERATURE.

3.         THE CHEMICAL BQUILIBRIUM CONSTANT IS A MEASURE OF THE EXTENT TO WHICH A REACTION PROCEEDS TO COMPLETION.

 

REACTANTS             A     K    >    1   MEANS THERE ARE MORE PRODUCTS AND THAN           

 

PRODUCTS                B      K    <    1  MEANS THERE ARE MORE REACTANTS THAN

 

                                    C      K    =    1  MEANS PRODUCTS AND REACTANTS ARE APPROX. EQUAL       

 

 

HETEROGENBOUS EQUILIBRIA

 

1.         SOLIDS AND LIQUIDS ARE SO CONCENTRATED THAT THEIR CONCENTRATION IS CONSTANT IN COMPARISON TO GASES.  THEREFORE THEIR CONCENTRATIONS ARE NOT INCLUDED IN THE EQUATION.

 

            NH4Cl (s)                     NH3 (g)        +      HCl (g)

 

                        Keq         =      [NH3]    [HCl]

 

 

REACTION QUOTIENT

 

1.         A REACTIONS CONCENTRATION CAN BE FOUND AND CALCULATED IN THE EQUATION AT ANY POINT.  THIS POINT DOES NOT HAVE TO BE AT EQUILIBRIUM.  THE ANSWER TO THIS EQUATION IS “Q,” THE REACTION QUOTIENT.

 

                                    Q      =       [NH3]

                                                  

 

2.         IF “Q” IS THE SAME AS Keq THE REACTION IS AT EQUILIBRIUM.  IF IT IS GREATER THAN Keq THERE IS MORE NH3.  IF IT IS LESS THAN Keq THERE IS MORE N2 AND H2 THAN THE EQUILIBRIUM.

 

 

 

 

 

 

 

 

 

 

 

 

 

LE CHATELIER’S PRINCIPLE

 

1.         THIS PRINCIPLE STATES THAT IF A CHANGE IN CONDITIONS ON A CHEMICAL REACTION EQUILIBRIUM IS MADE, THE REACTION WILL MOVE IN A DIRECTIOIN THAT WILL REDUCE THE CHANGE.

 

2.         THIS CAN BE THOUGHT OF AS SPEEDING UP OR SLOWING DOWN THE FORWARD OR REVERSE REACTION.

 

USING THE FOLLOWING REACTION

 

            HEAT   +   A(g)  +  B(g)                  AB(g)

 

CHANGE OF              INCREASE OF                                      REACTION INCREASES

CONDITIONS            FORWARD/REVERSE REACTION  PRODUCTS/REACTANTS

 

 

ADD REACTANTS                X                                             X

ADD PRODUCTS                                         X                                               X

REMOVE REACTANTS                               X                                               X

REMOVE PRODUCTS          X                                             X

INCREASE IN HEAT             X                                             X

DECREASE IN HEAT                                   X                                               X

 

AN INCREASE OF PRESSURE MAKES THE REACTION REDUCE THE PRESSURE OR MOVE TO THE SIDE WITH LESS PARTICLES.  REDUCING THE PRESSURE CAUSES THE REVERSE.

 

3.         THE HABER PROCESS IS A PRIME EXAMPLE OF THIS PRINCIPLE.

 

                                    N2(g)       +      3H2(g)             2 NH3(g)  + HEAT

 

THIS PROCESS WAS CREATED DURING WWII TO GET NH3 FOR EXPLOSIVES FOR GERMANY WHEN THE US BLOCKAIDED THE CHILIAN SOURCE OF SALT PETER (FOR NITOGEN).

 

IT WAS NOT DONE BEFORE BECAUSE THE REVERSE REACTION WAS SO GREAT THAT VERY LITTLE NH3 WAS FORMED.

 

 

 

 

 

 

 

 

 

ACIDS AND BASES

 

1.         THE MOST COMMON DEFINITION OF ACIDS IS THE ARRENIUS DEFINITION.

 

AN ACID IS ANYTHING THAT GIVES OFF H                  HC1

A BASE IS ANYTHING THAT GIVES OFF OH                  NaOH

 

2.         SINCE THIS DEFINITION DIDN’T COBER EVERYTHING THAT ACTED LIKE AN ACID OR BASE, BRONSTED AND LOWRY CREATED A NEW DEFINITION OF ALL THINGS CALLED THE BRONSTED-LOWERY THEORY.

ACIDS ARE ANYTHING THAT GIVES OFF H.  WHILE BASES ARE ANYTHING THAT TAKES ON H.

 

3.         THERE HAS TO BE ANOTHER ADJUCTMENT IN OUR THINKING.  WHEN H+ IS GIVEN OFF IT COMBINES WITH A WATER MOLECULE TO FORM H3O+ THIS IS CALLED A HYDRONIUM MOLECULE.

 

4.         ACIDS AND BASES ALSO FORM EQUILIBRIUM REACTIOINS.  THEREFORE THEIR Ka OR Kb CAN BE USED TO DETERMINE THEIR STRENGTHS.

 

                                    HA     +     H2O (l)           H3O+        =      A-

 

                                                      Ka         =        

                                                                                                    

 

IF THE Ka IS LESS THAN 1 THERE IS MORE UNIONIZED ACID MEANING IT IS WEAK NUMBERS GREATER THAN 1 MEANS THERE IS MORE H3O MAKING IT A STRONG ACID.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PH AND POH SCALE

 

1.         WATER ALSO HAS ITS OWN EQUILIBRIUM

 

                                    2H2O(l)                  H3O+ (aq)    +   OH-(aq)

 

ALTHOUGH THIS IS VERY VERY SMALL IT DOES EXIST AND AN EQUILIBRIUM EQUATION CAN BE WRITTEN.

 

                                    Kw     =     [H3O+]    [OH-]

 

WATER DROPS OUT BECAUSE IT IS A LIQUID.

 

            Kw EQUALS    1 X 10-14        H3O+  =  1  X 10 -7    OH - = 1  x   10 -7

 

2.         WHILE THIS IS A VERY SMALL NUMBER, IT STILL EXISTS EVEN IN SOLUTIONS WHERE THERE IS STRONGER ADICS OR BASES IN SOLUTION WITH WATER AND CAN BE USED TO CALCULATE CONCENTRATIONS.

 

            EXAMPLE:   WHAT IS THE CONCENTRATION OF OH- IN AN ACID SOLUTION OF 1 x 10-2M CONCENTRATION OF ACID.

 

                        Kw    =   H3O+    OH

 

                        1  x  10-14   =   1  x  10-2 OH

 

                        OH     =    1_x_10-14

                                                        1  x  10-2       =     1  x  10-12

 

3.         THIS RELATIONSHIP WAS CONDENSED TO SIMPLER NUMBERS BY USING LOGARITHMS CALLED THE PH SCALE.

 

                                    PH = -LOG[H3O+]      POH = -LOG[OH-]

 

4.         LOGS ARE BASED ON THE BASE 10, THEREFORE THE LOG OF 104 = 4.

THE –LOG OF 104 IS -4.  THE –LOG OF 104 IS 4.  FOR WHOLE NUMBERS YOU HAVE TO USE YOUR CALCULATOR OR LOG TABLES IN THE OLD DAYS.

 

 

 

 

 

 

 

 

 

 

5.         SINCE WATER HAS A CONCENTRATION OF H3O+ OF 1 x 10-7 IN H2O

 

ITS PH IS 7-0 = 7    THIS MEANS THE PH OF 7 IS NEUTRAL.  THE SAME IS TRUE FOR POH OF 7.  THEREFORE ANY CONCENTRATION OF ACID LARGER THAN 10-7 WILL GIVE A PH OF LESS THAN 7 MAKING IT AN ACID.  ANY  CONCENTRATION OF ACID LESS THAN 10-14 WILL GIVE IT A PH OF MORE THAN 7 MAKING IT A BASE.

 

6.         THE –LOG  Kw   =    -LOG [H3O]   +   -LOG [OH]

                                                         PH          +         POH

                        -LOG 1 x 10   =  -LOG 1 x 10    +  -LOG  1 x 10

 

                               14            =           7            +          7

                        FOR WATER             PH         +         POH

THIS RELATIONSHIP CAN BE USED IN CALCULATING PH AND POH FROM ANY ACID OR BASE IN SOLUTION WITH WATER.

 

EXAMPLE:  A VEGETABLE JUICE HAS A H3O CONCENTRATION OF 7.3 x 10.  WHAT IS ITS PH AND POH.  IS THE JUICE ACIDIC OR BASIC

 

                                    PH  =  -LOG [H3O+]

 

                                           =  -LOG 7.3 x 10-5

 

                                           =  5.833229 = 4.13366771

 

USING SIGNIFICANT NUMBERS THE PH IS 4.14 (ROUND TO USING FIRST EXTIMATED PLACE)

 

                                    SINCE         PH + POH  =  14

                                                                  POH  =  14 – PH

                                                                                 14 – 4.14  =  9.86

 

SINCE THE PH IS LESS THAN 7 IT IS AN ACID.