Energetics1
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Transcript of Energetics1
DefinitionsStandard Enthalpy Change
the change of enthalpy that accompanies the formation of 1 mole of a substance in its standard state from its constituent elements in their standard states.
Exothermic Rxn Typical examples; combustion, neutralisation,
respiration
Endothermic Rxn Typical examples; melting, evaporating, photosynthesis.
Signs Exothermic
Endothermic
Energy Level Diagrams
Exo and Endo Energy Level Diagrams
http://www.docbrown.info/page03/3_51energy.htm#3.
EnergeticsCalculation of Enthalpy
http://www.youtube.com/watch?v=lxTmei2yrBg
ExperimentsEnthalpy of Neutralisation of HCl with NaOH.
Enthalpy of dissolution.
Enthalpy of precipitation.
DefineEnthalpy of combustion
Enthalpy of formation
Draw energy level diagram for the process of evaporation.
Hess’s LawStudents should be able to use simple enthalpy
cycles and enthalpy level diagrams and to manipulate equations. Students will not be required to state Hess’s law…..but I’ll tell you anyway
Energy can not be created or destroyed. It can only be converted from one form to another – first law of thermodynamics
Conventions -ve for exothermic
+ve for endothermic
If forward reaction is exo the reverse reaction is endo and of identical magnitude.
Hess’s Law states that the enthalpy change for a reaction is independent of the route the reaction takes.
The overall enthalpy change depends only on the initial and final stages.
Direct measurement of enthalpy is impossible.
Enthalpy typesCombustion
Energy released when one mole of a compound is burned in excess oxygen.
FormationEnergy change when one mole of a compound is
formed under standard conditions from its constituent elements.
Bond Enthalpies – we will discuss later
Hess’ Law Defined
Hess’ Law: DH for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate DH for a reaction.
Shortcuts Learn Them Enthalpies of formation :
ΔH = Σproducts – Σ reactants
Enthalpies of combustion :ΔH = Σreactants – Σ products
Average Bond EnthalpiesΔ H = [bonds broken] – [bonds made]
Hess’ Law: An Example
Using Hess’ LawWhen calculating
DH for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine DH for our “single step” reaction.
2NO2 (g)
N2 (g) + 2O2 (g)
q
2NO2 (g)N2 (g) + 2O2 (g)
Example (cont.)Our reaction of interest is:
N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ
• This reaction can also be carried out in two steps:
N2 (g) + O2 (g) 2NO(g) DH = 180 kJ2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ
Example (cont.)
If we take the previous two reactions and add them, we get the original reaction of interest:
N2 (g) + O2 (g) 2NO(g) DH = 180 kJ 2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ
N2 (g) + 2O2 (g) 2NO2(g) DH = 68 kJ
Changes in EnthalpyConsider the following expression for a chemical process:
DH = Hproducts - Hreactants
If DH >0, then qp >0. The reaction is endothermic
If DH <0, then qp <0. The reaction is exothermic
Example (cont.)Note the important things about this
example, the sum of DH for the two reaction steps is equal to the DH for the reaction of interest.
We can combine reactions of known DH to determine the DH for the “combined” reaction.
Hess’ Law: DetailsOnce can always reverse the direction of a
reaction when making a combined reaction. When you do this, the sign of DH changes.
N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ
2NO2(g) N2(g) + 2O2(g) DH = -68 kJ
Details (cont.)The magnitude of DH is directly
proportional to the quantities involved (it is an “extensive” quantity).
As such, if the coefficients of a reaction are multiplied by a constant, the value of DH is also multiplied by the same integer.
N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ
2N2(g) + 4O2(g) 4NO2(g) DH = 136 kJ
Using Hess’ LawWhen trying to combine reactions to form a
reaction of interest, one usually works backwards from the reaction of interest.
Example:
What is DH for the following reaction?
3C (gr) + 4H2 (g) C3H8 (g)
Example (cont.) 3C (gr) + 4H2 (g) C3H8 (g) DH = ?
• You’re given the following reactions:
C (gr) + O2 (g) CO2 (g) DH = -394 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ
H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ
Example (cont.)Step 1. Only reaction 1 has C (gr). Therefore,
we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation.
C (gr) + O2 (g) CO2 (g) DH = -394 kJ
3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ
Initial:
Final:
Example (cont.)Step 2. To get C3H8 on the product side of the
reaction, we need to reverse reaction 2.
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJInitial:
Final:
Example (cont.)Step 3: Add two “new” reactions together to see
what is left:
3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ2
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ
Example (cont.)
Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ
H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ
3C (gr) + 4H2 (g) C3H8 (g)
Need to multiply second reaction by 4
Example (cont.)
Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ
3C (gr) + 4H2 (g) C3H8 (g)
Example (cont.)• Step 4 (cont.): 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH =
+1038 kJ 4H2 (g) + 2O2 (g) 4H2O (l) DH =
-1144 kJ3C (gr) + 4H2 (g) C3H8 (g) DH = -106 kJ
Changes in EnthalpyConsider the following expression for a chemical process:
DH = Hproducts - Hreactants
If DH >0, then qp >0. The reaction is endothermic
If DH <0, then qp <0. The reaction is exothermic
Another Example
Calculate DH for the following reaction:
H2(g) + Cl2(g) 2HCl(g)
Given the following:
NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ
N2 (g) + 3H2 (g) 2NH3 (g) DH = -92 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ
Another Example (cont.)Step 1: Only the first reaction contains the
product of interest (HCl). Therefore, reverse the reaction and multiply by 2 to get stoichiometry correct.
NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ
2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ
Another Example (cont.)Step 2. Need Cl2 as a reactant, therefore, add
reaction 3 to result from step 1 and see what is left.
2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)
DH = -277 kJ
Another Example (cont.)Step 3. Use remaining known reaction in
combination with the result from Step 2 to get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ
( N2 (g) + 3H2(g) 2NH3(g) DH = -92 kJ)
H2(g) + Cl2(g) 2HCl(g) DH = ?
Need to take middle reaction and reverse it
Another Example (cont.)Step 3. Use remaining known reaction in
combination with the result from Step 2 to get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ
2NH3(g) 3H2 (g) + N2 (g) DH = +92 kJ
H2(g) + Cl2(g) 2HCl(g) DH = -185 kJ
1
Changes in EnthalpyConsider the following expression for a chemical process:
DH = Hproducts - Hreactants
If DH >0, then qp >0. The reaction is endothermic
If DH <0, then qp <0. The reaction is exothermic