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Topic 5.3 and 5.4Hess’s Law and Bond Enthalpies
Hess’s LawTopic 5.3
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890 KJ
• shows three different pathways:A BA C BA 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
6
energy in reactants
energy in products
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 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) H1 = +181 kJ2 NO (g) + O2 (g) 2 NO2 (g) H2 = -113 kJ
• Find the enthalpy change for: N2 (g) + 2 O2 (g) 2 NO2 (g)
• H = H1 + H2 = +181 kJ +(-113) = + 68 kJ
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Example 22 2 3
2 2 2
2 N ( ) 6 H ( ) 4 NH ( ) H = 184 kJ
6 H O( ) 6 H ( ) 3 O ( ) H = +1452 kJ
-
g g g
g g g
2 2 2 32 N ( ) 6 H O( ) 3 O ( ) 4 NH ( ) g g g g
ΔH = (- 184 kJ) + (+ 1452 kJ) + 1268 kJ
Example 3• Given:
C (s) + O2 (g) CO2 (g) H1 = - 393 kJ mol-1
H2 (g) + ½O2 (g) H2O (g) H2 = - 286 kJ mol-1
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) H2 = - 890 kJ mol-1
• Find the enthalpy change for: C (s) + 2H2 (g) CH4 (g)
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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) H1 = - 393 kJ mol-1
H2 (g) + ½O2 (g) H2O (g) H2 = - 286 kJ mol-1
CO2 (g) + 2H2O (g) CH4 (g) + 2O2 (g) H2 = + 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
12 2 - 572
Calculate H° for the reaction
2 Al (s) + 3 Cl2 (g) 2 AlCl3 (s)
Use the following data:2 Al (s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g) H° = -1049. kJ mol-1
HCl (g) HCl (aq) H° = -74.8 kJ mol-1
H2 (g) + Cl2 (g) 2 HCl (g) H° = -1845. kJAlCl3 (s) AlCl3 (aq) H° = -323. kJ mol-1
= - 6,981.8 kJ
3 3 6 - 5,535 kJ mol-1
Example 4
• 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 so energy is released
Bond Enthalpies. Topic 5.4
• 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 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 just an average value, may vary, and is not very accurate
The average bond enthalpies for several types of chemical bonds are shown in the table below:
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• Page 326 in text
• Also see Chemistry Data Booklet p. 11
• Bonds broken (requires energy)• 1 N N = 945 kJ• 3 H-H 3(435) = 1305 kJ• Total = 2250 kJ
• Bonds formed (releases energy)• 2x3 = 6 N-H 6 (390) = - 2340 kJ
• Net enthalpy change• = (+ 2250) + (- 2340) = - 90 kJ (exothermic)
Calculate the enthalpy change for the reaction. Is it endo or exothermic?
N2 + 3 H2 2 NH3
Bond Enthalpy CalculationsExample 1:
H-H
Example 2
energy
course of reaction
2H2 + O2
2H2O
Working out ∆HShow all the bonds in the reactants
energy
course of reaction
2H2O
H―HH―H O=O+
Working out ∆HShow all the bonds in the products
energy
course of reaction
H―HH―H
O=O+
H HO
H HO
Working out ∆HShow the bond energies for all the bonds
energy
course of reaction
436436
O=O+
H HO
H HO
Working out ∆HShow the bond energies for all the bonds
energy
course of reaction
436436
498+
H HO
H HO
Working out ∆HShow the bond energies for all the bonds
energy
course of reaction
436436
498+
H HO
464 464+
Working out ∆HShow the bond energies for all the bonds
energy
course of reaction
436436
498+
464 464+
464 464+
Working out ∆HAdd the reactants’ bond energies together
energy
course of reaction
464 464+
464 464+
1370
Working out ∆HAdd the products’ bond energies together
energy
course of reaction
1370
1856
Working out ∆H∆H = energy in ― energy out
energy
course of reaction
1370
1856
13701856-
+
Working out ∆H∆H = energy in ― energy out
ener
gy
course of reaction
1370
1856
13701856- 486
-+
Working out ∆H∆H = energy in ― energy out
energy
course of reaction
1370
1856
∆H = -486exothermic
The average bond enthalpies for several types of chemical bonds are shown in the table below:
32
• Page 326 in text
• Also see Chemistry Data Booklet p. 11
Example 3• Calculate the enthalpy change for the following reaction:
• CH4 + Cl2 CH3Cl + HCl
• 4(414) + 243 -3(414) & (-397) + (-431)• 1,899 - 2,070• - 171 kJ mol-1 (exothermic)
Cl-Cl H-Cl