Hess's Law (Ebbing 242) by lzy18804

VIEWS: 281 PAGES: 15

									Hess’s Law (Ebbing 242)
For a chemical equation that can be written as the sum of two or more steps
the enthalpy change for the overall equation equals the sum of the enthalpy
changes of the individual steps

     The enthalpy change for the reaction of NO2 to produce N2O4 can be
     determined by using a two-step path. In the first step, NO2 decomposes
     to N2 and O2. In the second step, the elements react to form N2O4.

Using Hess’s Law: there are various ways to do this examples are now given
to illustrate its use. In the example below the strategy is to combine equations
so that the reactants are on the LHS only.
• if you reverse direction of eqn. then sign of energy is reversed
• multiples : )H is reversed
                                                      Write down the target equation
• same stuff on both sides cancels                    first
C(s) + 1/2 O2(g)                       CO(g)          -then you can see what to do
enthalpy for this reaction is hard to measure

1. C(s) + O2(g)                      CO2(g) )HO = -393.5
2. 2CO(g) + O2(g)                    2 CO2 (g) )HO = -566.0
      1.      C(s) + O2(g)              CO2(g)                      )HO = -393.5
      -2/2    CO2 (g)           CO(g) + 1/2O2(g)             )HO = +566.0/2

Add          C(s) + 1/2 O2(g)           CO(g)     )HO = -110.5 kJ
Example when individual )HfO is unknown.
Calculate )HrxnO for: B2H6(g) + 6Cl2(g)               2BCl3(g) + 6HCl(g)
Given:                                                         )Hrxn

1. BCl3(g) + 3H2O(l)                H3BO3(s) + 3HCl (g)       -112.5
2. B2H6(g) + 6H2O(l)                2H3BO3(s) + 6H2(g)        -493.5
3. 1/2H2 (g) + 1/2 Cl2(g)           HCl(g)                    -92.3

  Strategy: eqn 2 has B2H6 on LHS - leave it alone
            eqn 3 has HCl on LHS - leave it
            eqn 1 has BCl3 on LHS - reverse it

   (2) + 2 (-1) + 2 (3)          )Hrxn = -1376.1 kJ

         Work through this one carefully

 Recall: In order to break a chemical bond we must supply
 energy ( think of the bond as a spring - you have to work on it
 to break it)
                  C-----H : needs energy

 Bond enthalpies : (Bond dissociation enthalpies)

 H2(g)            2H(g) )H = +436 kJ

 energy required to pull apart 1 mole of H-H bonds is 436kJ

 the bond enthalpy of H2 is 436 kJ

  Note: bond enthalpies are for the gas phase only
What about F2(g), C 2(g), [Br2(g)], N2(g), O2(g)?
       No Problem
O2(g) 2 O(g) )Ho = 496 kJ mol-1
Hence the bond dissociation enthalpy of the O-O bond
in molecular oxygen is 496 kJ mol-1. Notice that this
equation also defines the standard enthalpy of formation
of two moles of atomic oxygen; the bond dissociation enthalpy
of O2(g) is exactly twice the heat of formation of O(g).


1/2 O2            O(g)

must use exactly half the energy of

O2(g)             2O(g)
What about an O-H bond?
        H2O(g)      2H(g) + O(g), )H = 926 kJ
so that one mole of O-H bonds would require
                 926/2 = 463 kJ

What about C-H?
No C-H molecules
but CH4(g)         C(g) + 4H(g), )H = +1648 kJ
i.e. to break FOUR moles of C-H bonds cost 1656 kJ
   1 C-H Bond Enthalpy = 1648/4 = 412 kJ

more examples on use of BDE’s will follow soon

Review:Descriptions of reactions you should know

Heat of Fusion
Enthalpy change when we fuse (i.e., melt) 1 mole solid liquid
1.01×105 Pa: e.g., H2O(s)            H2O(l )

Heat of Sublimation:Enthalpy change when we sublime 1 mole solid gas at
1.01×105 Pa: e.g. I2(s)         I2(g)

Heat of Combustion:Enthalpy change when we burn 1 mole completely in O2,
1.01×105 Pa.:e.g., CH4 (g) + 2 O2 (g)       CO2 (g) +2 H2O(l)

Heat of Reaction: General description when you can't think of the correct

Heat of vaporization Enthalpy change when we vaporize 1 mole liquid at
1.01×105 Pa: e.g., H2O(l)           H2O(g )

Heat of Atomization Enthalpy change when 1 mole of element in standard
state is converted to 1 mole of gas phase atoms
Standard Molar Enthalpies of Formation of IONS in Aqueous Solution.
Consider the reaction:

NaCl (s)         Na+(aq) + Cl -(aq) (infinite dilution)

)H for this reaction is known to be +3.9kJmol -1

)H f(NaCl (s)) is known as -411.2 kJ mol-1
We can then write:
 )H rxn = {)H f[Na+(aq)] + )H f[Cl -(aq)]} - {)H f[NaCl (s)]}

from which we can deduce by simple arithmetic that
{)H f[Na+(aq)] + )H f[Cl -(aq)]}
         = -407.3 kJ mol-1
Although we can obtain the sum this way, the separate contributions of the
two ions cannot be obtained by any experiment yet devised.

So - what do we do? ---- next page
What to do?

Define )H f(H+(aq)) = 0
and consider the reaction:-
        HCl (g)        H+(aq) + Cl -(aq) (infinite dilution)
 )H rxn = {)H f ) [H+(aq)] + )H f[C -(aq)]} -            {) H f[HCl (g)]}

)H rxn can be measured
 )H f[H+(aq)] = 0 (by definition)
 )H f[HCl(g)] is known (tables or measure)
which only leaves )H f[Cl -(aq)] (get by difference).

Now that we have )H f[Cl -(aq)] we can get )H f[Na+(aq)] from the first
experiment and from similar experiments with NaBr, KF, etc, we can deduce
the )H f values for all other ions in aqueous solution.

    And so…..

Cations         )Hf (kJ/mol)   Anions       )Hf (kJ/mol)
Ag+(aq)         +105.9         Br-(aq)      -120.9
Al3+(aq)        -524.7         Cl-(aq)      -167.4
Ba2+(aq)        -538.4         ClO3-(aq)    -98.3
Ca2+(aq)        -543.0         ClO4-(aq)    -131.4
Cd2+(aq)        -72.4          CO32-(aq)    -676.3
Cu2+(aq)        +64.4          CrO42-(aq)   -863.2
Fe 2+ (aq)      -87.9          F-(aq)       -329.1
Fe3+(aq)        -47.7          HCO3-(aq)    -691.1
H+(aq)          0.0            H2PO4-(aq)   -1302.5
K+(aq)          -251.2         HPO42-(aq)   -1298.7
Li+(aq)         -278.5         I-(aq)       -55.9
Mg2+(aq)        -462.0         MnO4-(aq)    -518.4
Mn2+(aq)        -218.8         NO3-(aq)     -206.6
Na+(aq)         -239.7         OH-(aq)      -229.9
NH4+(aq)        -132.8         PO43-(aq)    -1284.1
Ni2+(aq)        -64.0          S2-(aq)      +41.8
Pb2+(aq)        +1.6           SO42-(aq)    -907.5
Sn2+(aq)        -10.0
Zn2+(aq)        -152.4                                     5-10
   Calculate ) H for the ionization of hydrogen bromide:

   HBr(g) --> H+(aq) + Br-(aq)


   ) H for a reaction is equal to the sum of the heats of formation of the
   product compounds minus the sum of the heats of formation of the
   reactant compounds:

   ) H = Sum( ) Hf products) - Sum ( ) Hf reactants)

   Remember, the heat of formation of H+ is zero. The equation becomes:

   ) H = ) Hf Br-(aq) - ) Hf HBr(g)

   The values for ) Hf may be found in the Heats of Formation of
   Compounds of Ions table. Plugging in these numbers:
   ) H = -120.9 kJ - (-36.2 kJ)                 ) H = -84.7 kJ
    More examples
The enthalpy of formation of KCl(s) is -436.7 kJ/mol. Using the table on page 5-10.
Determine the enthalpy of solution for KCl dissolving.

           KCl(s)                  K+ (aq) + Cl- (aq)

              )Horeaction = E )HfO (products) -E )HfO (reactants)

            So: -167.4 + (-251.2) - (-436.7) = 18.1 kJ/mol

      Bond energies: another example

Given the following, determine a value for the N-N bond enthalpy in Hydrazine
N2H4 (a rocket fuel)                                  Bond energies

Heat of formation liquid hydrazine is +51 kJ             N/N is +946 kJ
Heat of vaporization liquid hydrazine is +48 kJ          H-H is +436 kJ
Bond Energies:                                           N-H is +389 kJ

                                                      )H (BE)
   1. N2(g) + 2H2(g)             N2H4(l)               51
   2. N2H4(l)                    N2H4(g)               48
   3. N2(g)                      2N(g)                  946
   4. H2(g)                      2H (g)                436
   5. N2H4(g)                   2N(g) + 4H(g)           4(389) + (N-N)

          5 is the target equation
We can use Hess’s Law to solve this:we need to rearrange so N2H4(g) is on LHS

     -1    N2H4(l)             N2(g) + 2H2(g)            -51
     -2    N2H4(g)             N2H4(l)                   -48 Cancels out N2H4(l)
     3     N2(g)                2N(g)                    946 }N and H on RHS
     4x2   2H2(g)                4H (g)                  872

 5. N2H4(g)                 2N(g) + 4H(g)            1719

    but. N2H4(g)             2N(g) + 4H(g)     4(389) + (N-N)

              4(389) + (N-N) = 1719

              N-N BE = 163 kJmol-1                                  5-14
  Given the following (kJ) data, estimate a value for the CS bond enthalpy
  in CS2(g).
  [CS2 is isostructural with CO2]
  )H(combustion) CS2(R) -1108, C(s) -393, S(s) -297
  )H(atomization) C(s) +711, S(s) +280 kJ
  )H(vaporisation) CS2(R) +27 kJ

                                             Answer : 561 kJmol-1

       Full solution next lecture


To top