The document defines standard enthalpy change, exothermic and endothermic reactions, and signs for exo/endothermic reactions. It provides examples of enthalpy calculations, experiments to determine enthalpy changes, and defines combustion enthalpy, formation enthalpy, and Hess's law. Hess's law states that the enthalpy change for a reaction is independent of the pathway and depends only on the initial and final states.
SOQL 201 for Admins & Developers: Slice & Dice Your Org’s Data With Aggregate...
Energetics1
1. Definitions
Standard 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.
4. Exo and Endo Energy Level
Diagrams
http://www.docbrown.info/page03/3_51energy.htm#3.
5. Energetics
Calculation of Enthalpy
http://www.youtube.com/watch?v=lxTmei2yrBg
6.
7. Experiments
Enthalpy of Neutralisation of HCl with NaOH.
Enthalpy of dissolution.
Enthalpy of precipitation.
8.
9. Define
Enthalpy of combustion
Enthalpy of formation
Draw energy level diagram for the process of
evaporation.
10. Hess’s Law
Students 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
11. 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.
12. Enthalpy types
Combustion
Energy released when one mole of a compound is burned
in excess oxygen.
Formation
Energy change when one mole of a compound is formed
under standard conditions from its constituent elements.
Bond Enthalpies – we will discuss later
13. Hess’ Law Defined
Hess’ Law: H for a process involving the
transformation of reactants into products is not
dependent on pathway. Therefore, we can pick
any pathway to calculate H for a reaction.
14. Shortcuts Learn Them
Enthalpies of formation :
ΔH = Σproducts – Σ reactants
Enthalpies of combustion :
ΔH = Σreactants– Σ products
Average Bond Enthalpies
Δ H = [bonds broken] – [bonds made]
16. Using Hess’ Law
When calculating H
N2 (g) + 2O2 (g) 2NO2 (g)
for a chemical
reaction as a single
step, we can use
2NO2 (g) combinations of
reactions as
q
“pathways” to
N2 (g) + 2O2 (g)
determine H for our
“single step” reaction.
17. Example (cont.)
Our reaction of interest is:
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
• This reaction can also be carried out in two
steps:
N2 (g) + O2 (g) 2NO(g) H = 180 kJ
2NO (g) + O2 (g) 2NO2(g) H = -112 kJ
18. 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) H = 180 kJ
2NO (g) + O2 (g) 2NO2(g) H = -112 kJ
N2 (g) + 2O2 (g) 2NO2(g) H= 68 kJ
19. Changes in Enthalpy
Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp>0. The reaction is endothermic
If H <0, then qp<0. The reaction is exothermic
20. Example (cont.)
Note the important things about this example,
the sum of H for the two reaction steps is
equal to the H for the reaction of interest.
We can combine reactions of known H to
determine the H for the “combined” reaction.
21. Hess’ Law: Details
Once can always reverse the direction of a reaction
when making a combined reaction. When you do
this, the sign of H changes.
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
2NO2(g) N2(g) + 2O2(g) H = -68 kJ
22. Details (cont.)
The magnitude of H 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 H is
also multiplied by the same integer.
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
N2(g) + 4O2(g) 4NO2(g) H = 136 kJ
23. Using Hess’ Law
When trying to combine reactions to form a reaction of
interest, one usually works backwards from the reaction
of interest.
Example:
What is H for the following reaction?
3C (gr) + 4H2 (g) C3H8 (g)
24. Example (cont.)
3C (gr) + 4H2 (g) C3H8 (g) H=?
• You’re given the following reactions:
C (gr) + O2 (g) CO2 (g) H = -394 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220
H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ
25. 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.
Initial:
C (gr) + O2 (g) CO2 (g) H = -394 kJ
Final:
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
26. Example (cont.)
Step 2. To get C3H8 on the product side of the
reaction, we need to reverse reaction 2.
Initial:
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220
Final:
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220
27. Example (cont.)
Step 3: Add two “new” reactions together to see what
is left:
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220
2
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 k
28. Example (cont.)
Step 4: Compare previous reaction to final reaction, and
determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038
kJ
H2 (g) + 1/2O2 (g) H2O (l) H = -286
kJ
3C (gr) + 4H2 (g) C3H8 (g)
Need to multiply second reaction by 4
29. Example (cont.)
Step 4: Compare previous reaction to final reaction, and
determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038
kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144
kJ
3C (gr) + 4H2 (g) C3H8 (g)
31. Changes in Enthalpy
Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp>0. The reaction is endothermic
If H <0, then qp<0. The reaction is exothermic
32. Another Example
Calculate H for the following reaction:
H2(g) + Cl2(g) 2HCl(g)
Given the following:
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
N2 (g) + 3H2 (g) 2NH3 (g) H = -92 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ
33. 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) H = -176 kJ
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ
34. 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) H = 352 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)
H = -277 kJ
35. 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) H = -277 kJ
( N2 (g) + 3H2(g) 2NH3(g) H = -92 kJ)
H2(g) + Cl2(g) 2HCl(g) H=?
Need to take middle reaction and reverse it
36. 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) H = -277 kJ
1
2NH3(g) 3H2 (g) + N2 (g) H = +92 kJ
H2(g) + Cl2(g) 2HCl(g) H = -185 kJ
37. Changes in Enthalpy
Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp>0. The reaction is endothermic
If H <0, then qp<0. The reaction is exothermic