Download Hess`s Law 5-1

Hess’s Law
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.
5-1
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.
Rules:
• 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
example
-then you can see what to do
C(s) + 1/2 O2(g)
CO(g)
enthalpy for this reaction is hard to measure
However
1. C(s) + O2(g)
2. 2CO(g) + O2(g)
So:
1.
-2/2
Add
C(s) + O2(g)
CO2 (g)
C(s) + 1/2 O2(g)
CO2(g) )HO = -393.5
2 CO2 (g) )HO = -566.0
CO2(g)
CO(g) + 1/2O2(g)
CO(g)
)HO = -393.5
)HO = +566.0/2
)HO = -110.5 kJ
5-2
Example when individual )HfO is unknown.
Calculate )HrxnO for: B2H6(g) + 6Cl2(g)
2BCl3(g) + 6HCl(g)
)Hrxn
Given:
1. BCl3(g) + 3H2O(l)
H3BO3(s) + 3HCl (g)
2. B2H6(g) + 6H2O(l)
2H3BO3(s) + 6H2
493.5
3. 1/2H2 (g) + 1/2 Cl2(g)
HCl(g)
Strategy: eqn 2 has B2H6 on LHS - leave it alone
eqn 3 has HCl on RHS - leave it
eqn 1 has BCl3 on LHS - reverse it
(2) + 2 (-1) + 12 (3)
-112.5
(g)
-
-92.3
)Hrxn = -1376.1 kJ
Work through this one carefully
5-3
)Hrxn from BOND ENERGIES
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)
example
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
5-4
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).
Since:
1/2 O2
O(g)
must use exactly half the energy of
O2(g)
2O(g)
5-5
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 1648 kJ
1 C-H Bond Enthalpy = 1648/4 = 412 kJ
more examples on use of BDE’s will follow soon
5-6
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
word?
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
5-7
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
5-8
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[Cl -(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.
5-9
And so…..
Cations
Ag+(aq)
Al3+(aq)
Ba2+(aq)
Ca2+(aq)
Cd2+(aq)
Cu2+(aq)
Fe 2+ (aq)
Fe3+(aq)
H+(aq)
K+(aq)
Li+(aq)
Mg2+(aq)
Mn2+(aq)
Na+(aq)
NH4+(aq)
Ni2+(aq)
Pb2+(aq)
Sn2+(aq)
Zn2+(aq)
)Hf (kJ/mol)
+105.9
-524.7
-538.4
-543.0
-72.4
+64.4
-87.9
-47.7
0.0
-251.2
-278.5
-462.0
-218.8
-239.7
-132.8
-64.0
+1.6
-10.0
-152.4
Anions
Br-(aq)
Cl-(aq)
ClO3-(aq)
ClO4-(aq)
CO32-(aq)
CrO42-(aq)
F-(aq)
HCO3-(aq)
H2PO4-(aq)
HPO42-(aq)
I-(aq)
MnO4-(aq)
NO3-(aq)
OH-(aq)
PO43-(aq)
S2-(aq)
SO42-(aq)
)Hf (kJ/mol)
-120.9
-167.4
-98.3
-131.4
-676.3
-863.2
-329.1
-691.1
-1302.5
-1298.7
-55.9
-518.4
-206.6
-229.9
-1284.1
+41.8
-907.5
5-10
example
Calculate ) H for the ionization of hydrogen bromide:
HBr(g) --> H+(aq) + Br-(aq)
Solution
) 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
5-11
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
5-12
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
Heat of vaporization liquid hydrazine is +48 kJ
Bond Energies:
N/N is +946 kJ
H-H is +436 kJ
N-H is +389 kJ
)H (BE)
1. N2(g) + 2H2(g)
2. N2H4(l)
3. N2(g)
4. H2(g)
5. N2H4(g)
N2H4(l)
N2H4(g)
2N(g)
2H (g)
2N(g) + 4H(g)
51
48
946
436
4(389) + (N-N)
5 is the target equation
5-13
We can use Hess’s Law to solve this:we need to rearrange so N2H4(g) is on LHS
-1
-2
3
4x2
add
N2H4(l)
N2H4(g)
N2(g)
2H2(g)
5. N2H4(g)
N2(g) + 2H2(g)
N2H4(l)
2N(g)
4H (g)
2N(g) + 4H(g)
but. N2H4(g)
2N(g) + 4H(g)
-51
-48 Cancels out N2H4(l)
946 }N and H on RHS
872
1719
4(389) + (N-N)
4(389) + (N-N) = 1719
N-N BE = 163 kJmol-1
5-14
homework
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
5-15