1. Ch. 17: Reaction Energy and Reaction Kinetics
Heat of Reaction
Heat of Formation
Hess’ Law

2. Heat of Reaction
amount of energy released or absorbed during a chemical reaction
the difference between the stored energy of reactants and products
energy can be a reactant or product
same as the change in enthalpy (H) at constant pressure
∆H = Hproducts - Hreactants
endo: + ∆H exo: - ∆H

3. Heat of Reaction / Enthalpy Change

4. Standard Enthalpy of Formation
∆Hf°
change in enthalpy that accompanies the formation of one mole of a compound from its elements in standard states
° means that the process happened under standard conditions so we can compare values more easily
room temperature (298 K) and 1 atm
values given p. 902

5. Writing Formation Equations
always write equation where 1 mole of compound is formed (even if you must use non-integer coefficients)
NO2(g):
½N2(g) + O2(g)  NO2(g)
∆Hf°= 34 kJ/mol
CH3OH(l):
C(s) + 2H2(g) + ½ O2(g) CH3OH(l)
∆Hf°= -239 kJ/mol

6. Stability
Would a large negative value, like ∆Hf°= -239 kJ/mol, indicate a very stable or unstable compound ?
If the compound is very stable, lots of energy will be released when it is formed so it will have a large negative value.
If the compound is unstable, it will have a positive value and will most likely decompose into its elements

7. Hess’ Law
thermochemical equations can be rearranged and added together to give enthalpy changes for reactions not given in the Appendix
Hess’ Law- the overall enthalpy change is equal to the sum of the enthalpy changes for individual steps in a reaction

8. Rules
If a reaction is reversed, the sign of ∆H must be reversed as well.
because the sign tells us the direction of heat flow
The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction.
If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way.

9. Example 1
Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to diamond.
Cgraphite (s)  Cdiamond (s) ∆H=?

10. Example 1 (1) Cgraphite(s) + O2(g)  CO2(g) ∆H=-394kJ/mol
(2) Cdiamond(s) + O2(g)  CO2(g) ∆H=-396kJ/mol
to get the desired equation, we must reverse 2nd equation:
-(2) CO2(g)  Cdiamond(s) + O2(g) ∆H=396kJ/mol
Cgraphite (s)  Cdiamond (s) ∆H=
∆H=2 kJ/mol

11. 2N2(g) + 6H2O(g)  3O2(g) + 4NH3(g)
Example 1b
Find the enthalpy change for the following reaction.
2N2(g) + 6H2O(g)  3O2(g) + 4NH3(g)
Given:
NH3(g)  ½N2(g) + 3/2H2(g) ∆H=46 kJ
2H2(g) + O2(g)  2H2O(g) ∆H=-484 kJ

12. Example 1b NH3: (1) must be reversed and multiplied by 4
H2O: (2) must be reversed and multiplied by 3
2N2(g) + 6H2(g)  4NH3(g) ∆H=-4(46 kJ)
6H2O(g)  6H2(g) + 3O2(g) ∆H=-3(-484 kJ)
∆H= kJ

13. Example 2 (1) H2(g) + ½O2(g)  H2O(l) ∆H=-285.8kJ/mol
Calculate the change in enthalpy for the following reaction:
2N2(g) + 5O2(g)  2N2O5(g)
Given
(1) H2(g) + ½O2(g)  H2O(l) ∆H=-285.8kJ/mol
(2) N2O5(g) + H2O(l)  2HNO3(l) ∆H=-76.6kJ/mol
(3) ½H2(g) + ½N2(g) +3/2O2(g) HNO3(l)
∆H=-174.1kJ/mol

14. Example 2 -2(∆H1)= (-2)(-285.8kJ/mol)= 571.6
for N2O5: multiply (2) by 2 and reverse it
for H2O: multiply (1) by 2 and reverse it
for HNO3: multiply (3) by 4
-2(∆H1)= (-2)(-285.8kJ/mol)= 571.6
-2(∆H2)= (-2)(-76.6kJ/mol)= 153.2
4(∆H3)= (4)(-174.1kJ/mol)=
28.4 kJ/mol

15. Example 3 Find ∆H for the synthesis of B2H6, diborane:
2B(s) + 3H2(g)  B2H6(g)
Given:
2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ
B2H6(g) + 3O2(g)  B2O3(s) + 3H2O(g) ∆H2=-2035kJ
H2(g) + ½O2(g)  H2O(l) ∆H3=-286kJ
H2O(l)  H2O(g) ∆H4=44 kJ

16. Example 3
Start by paying attention to what needs to be on reactants and products side
(1) 2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ
-(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g) ∆H2=2035kJ
(3) H2(g) + ½O2(g)  H2O(l) ∆H3=-286kJ
(4) H2O(l)  3H2O(g) ∆H4=44 kJ

17. Example 3
Underline what you want to keep- that will help you figure out how to cancel everything else:
(1) 2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ
-(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g) -∆H2=2035kJ
(3) H2(g) + ½O2(g)  H2O(l) ∆H3=-286kJ
(4) H2O(l)  3H2O(g) ∆H4=44 kJ

18. Example 3 Need 3 H2 (g) so 3 x (3) Need 3 H2O to cancel so 3 x (4)
(1) 2B(s) + 3/2O2(g)  B2O3(s) ∆H1=-1273kJ
-(2) B2O3(s) + 3H2O(g)  B2H6(g) + 3O2(g) ∆H2=-(-2035kJ)
3x(3) 3H2(g) + 3/2O2(g)  3H2O(l) 3∆H3=3(-286kJ)
3x(4) 3H2O(l)  3H2O(g) 3∆H4=3(44 kJ)
2B(s) + 3H2(g)  B2H6(g)
∆H = (-2035) + 3(-286) + 3(44) = 36kJ

19. Entropy - PAP Only
Defined as the measure of randomness of particles in a system.
Higher energy results in higher entropy of the same particle.
i.e. a particle in the gas phase will have a higher entropy than a particle in the liquid phase.
At absolute zero, a particle would have zero entropy

20. Gibbs Free Energy - PAP Only
Used to determine the spontaneity of a reaction
A positive DG value indicates the reaction is not spontaneous at this condition.
A negative DG value indicates the reaction is spontaneous at this condition.

21. Relationships - PAP Only
Using Gibbs Free Energy to find out if a reaction is spontaneous or
non-spontaneous. DG DH DS
Reaction driven by
Favorable temperature
Sign
Meaning
+ or -
???????? +
Endothermic
Increase disorder
Entropy
High temperatures
Non-spontaneous -
Decrease disorder
Nothing
None
spontaneous
Exothermic
Enthalpy and entropy
All temperatures
Enthalpy
Low temperatures