Topic 5.1 and 5.2
Hess’s Law and Bond Enthalpies
Hess’s Law
Topic 5.2
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
H = -890 KJ
• shows three different
pathways:
AB
ACB
ADEB
– enthalpy change
from reactants to
products for all of
these is the same
• if a series of reactions are added together, the
enthalpy change for the net reaction (Hfinal) will be
the sum of the enthalpy change for the individual
reactions (Hind + Hind + Hind ….)
– the change in enthalpy is the same whether the
reaction takes place in one step, or in a series of
steps
– H is independent of the reaction pathway
– depends only on the difference between the
enthalpy of the products and the reactants
• H = Hproducts − Hreactants
• provides a way to calculate enthalpy changes even
when the reaction cannot be performed directly
8
energy in products
energy in reactants
Problem-Solving Strategy
–
–
–
work backwards from the final reaction, using the
reactants and products to decide how to manipulate
the other given reactions at your disposal
if a reaction is reversed the sign on ΔH is reversed
– N2 (g) + 2O2 (g) → 2NO2 (g) ΔH = 68kJ
– 2NO2 (g) → N2 (g) + 2O2 (g) ΔH = - 68kJ
multiply reactions by a coefficient to give the correct
numbers of reactants and products in order to get
the final reaction
–
–
the value of Δ H is also multiplied by the same
integer
identical substances found on both sides of the
summed equation cancel each other out
Example 1
• Given:
N2 (g) + O2 (g)  2 NO (g)
2 NO (g) + O2 (g)  2 NO2 (g)
• Find the enthalpy change for:
N2 (g) + 2 O2 (g)  2 NO2 (g)
H1 = +181 kJ
H2 = -113 kJ
• H = H1 + H2 = +181 kJ + (-113 kJ) = + 68 kJ
Example 2
2 N2 (g )  6 H2 ( g ) 
 4 NH3 ( g )
H = - 184 kJ
6 H2O(g ) 
 6 H2 ( g )  3 O2 ( g )
H = +1452 kJ
2 N2 (g )  6 H2O(g ) 
 3 O2 ( g )  4 NH3 ( g )
ΔH = (- 184 kJ) + (+ 1452 kJ) = + 1268 kJ
Example 3
• Given:
C (s) + O2 (g)  CO2 (g)
H2 (g) + ½O2 (g)  H2O (g)
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
H1 = - 393 kJ mol-1
H2 = - 286 kJ mol-1
H3 = - 890 kJ mol-1
• Find the enthalpy change for:
C (s) + 2H2 (g)  CH4 (g)
This equation needs
to be “flipped”. The
CH4 is on the wrong
side of the equation
Example 3
• Given:
C (s) + O2 (g)  CO2 (g)
2 H2 (g) + 1½O2 (g)  2 H2O (g)
CO2 (g) + 2H2O (g)  CH4 (g) + 2O2 (g)
H1 = - 393 kJ mol-1
H2 = -- 572
286 kJ mol-1
H3 = + 890 kJ mol-1
• Find the enthalpy change for:
C (s) + 2H2 (g)  CH4 (g)
• (- 393 kJ mol-1) + (- 572 kJ mol-1 ) + (+ 890 kJ mol-1 )
= - 75 kJ mol-1
This reaction need
to “run
twice”/doubled
because of the 2H2
in the final equation
Using enthalpy cycles instead…
?
• counter-clockwise reaction energy (-3,222) has to
equal clockwise reaction energy (-3,267 + ΔH)
• ΔH = +45 KJ/mol
?
• clockwise needs to equal counter-clockwise
• -109 = 52.2 + (-92.3) + ΔH
• ΔH = - 68.9 kJ mol-1
Bond Enthalpies
• can be used to calculate the enthalpy change for a
chemical reaction if we know the energy necessary to
break or form bonds in the gaseous state
– breaking bonds
• energy is required so enthalpy is positive (endothermic)
• the molecule was stable so energy was necessary to
break apart the molecule
– forming bonds
• energy is released so enthalpy is negative (exothermic)
• the new molecule is more stable than the individual
atoms (unstable) that make it up, so energy is released
• a molecule with strong chemical bonds
generally has less tendency to undergo
chemical change than does one with weak
bonds
– SiO bonds are among the strongest ones that silicon
forms
• it is not surprising that SiO2 and other substances containing SiO bonds
(silicates) are so common
• it is estimated that over 90 percent of Earth's crust is composed of SiO2
and silicates
• we use average bond enthalpies
– again, in the gaseous state
– different amount of energy can be required to
break the same bond
• example- methane, CH4
– if you took methane to pieces, one hydrogen at a
time, it needs a different amount of energy to break
each of the four C-H bonds
– every time you break a hydrogen off the carbon, the
“environment” (think bond angles) of those left
behind changes, and the strength of the remaining
bonds is affected
– therefore, the 412 kJ mol-1 needed to break C-H is
an average value and therefore the one used in
calcuations
The average bond enthalpies for several types of chemical
bonds are shown in the table below:
20
Bond Enthalpy Calculations
Example 1:
Calculate the enthalpy change for the
reaction. Is it endo or exothermic?
N2 + 3 H2  2 NH3
• Bonds broken
• 1N N
• 3 H-H 3(435)
• Total
H-H
= 945 kJ
= 1305 kJ
= 2250 kJ
• Bonds formed
• 2x3 = 6 N-H:
6 (390) = - 2340 kJ
• Net enthalpy change
• = (+ 2250) + (- 2340) = - 90 kJ (exothermic)
Example 2
energy
2H2 + O2
2H2O
course of reaction
Working out ∆H
Show all the bonds in the reactants
energy
H―H
H―H
+ O=O
2H2O
course of reaction
Working out ∆H
Show all the bonds in the products
energy
H―H
H―H
+ O=O
O
H
H
course of reaction
O
H
H
Working out ∆H
Show the bond energies for all the bonds
energy
436
436
+ O=O
O
H
H
course of reaction
O
H
H
Working out ∆H
Show the bond energies for all the bonds
energy
436
436
+ 498
O
H
H
course of reaction
O
H
H
Working out ∆H
Show the bond energies for all the bonds
energy
436
436
+ 498
464 + 464
O
H
course of reaction
H
Working out ∆H
Show the bond energies for all the bonds
energy
436
436
+ 498
464 + 464
464 + 464
course of reaction
Working out ∆H
Add the reactants’ bond energies together
energy
1370
464 + 464
464 + 464
course of reaction
Working out ∆H
Add the products’ bond energies together
+ 1370
energy
- 1856
course of reaction
Notice the
negative due to
bonds being
formed
Working out ∆H
∆H = energy in ― energy out
energy
+ 1370
+ 1370
- 1856
- 1856
course of reaction
Working out ∆H
∆H = energy in ― energy out
+ 1370
+ 1370
- 1856
- 486
energy
- 1856
course of reaction
Working out ∆H
∆H = energy in ― energy out
+ 1370
∆H = -486
exothermic
energy
- 1856
course of reaction