1. Thermochemistry Heat and Chemical Change

2. Thermochemistry
The study of the changes in heat energy that accompany chemical reactions and physical changes

3. Exothermic and Endothermic Processes
In studying heat changes of chemical reactions, think of these two parts:
the reactants
the products

4. Exothermic and Endothermic
Every chemical reaction has an energy change associated with it
Exothermic reactions release energy, usually in the form of heat.
Endothermic reactions absorb energy in the form of heat.

5. Thermochemical Equation
An equation that includes the quantity of energy released or absorbed as heat during a reaction.
CH4 (g) + 2O2 (g) ® CO2 (g) + 2H2O (g) kJ
This combustion reaction gives off kJ of energy (energy on the product side) – Exothermic Reaction

6. 2H2O (g) kJ ® 2H2 (g) + O2 (g)
Energy that appears on the reactant side of the equation is absorbed during the reaction.
Endothermic Reaction

7. Exothermic The products are lower in energy than the reactants
Reaction releases energy

8. C + O2 ® CO2
+ 395 kJ
Energy
Reactants
Products
C + O2
395kJ
C O2

9. Endothermic The products are higher in energy than the reactants
Reaction absorbs energy

10. CaCO3 + 176 kJ ® CaO + CO2 CaCO3 ® CaO + CO2 Energy Reactants Products

11. Thermochemical Reaction
q = H The energy released or absorbed as heat during a chemical reaction is represented by H
Enthalpy = H
H is negative for an exothermic reaction (loses energy)
H is positive for an endothermic reaction (gains energy)

12. Summary, so far...

13. Enthalpy Symbol is H Change in enthalpy is DH (delta H)
If heat is released, the heat content of the products is lower
DH is negative (exothermic)
If heat is absorbed, the heat content of the products is higher
DH is positive (endothermic)

14. Exothermic Reaction Energy Change is down DH is <0 Reactants ®
Products
Reactants
Products

15. Endothermic Reaction Energy Change is up DH is > 0 Reactants ®
Products
Energy
Change is up
DH is > 0
Reactants
Reactants
Products

16. Heat of Reaction
The heat that is released or absorbed in a chemical reaction
Equivalent to DH
C + O2(g) ® CO2(g) kJ
C + O2(g) ® CO2(g) DH = kJ
In thermochemical equation, it is important to indicate the physical state
H2(g) + 1/2O2 (g)® H2O(g) DH = kJ
H2(g) + 1/2O2 (g)® H2O(l) DH = kJ

17. Calculating Heat of Reaction
Methods:
1) Doing it yourself (calorimetry)
2) Refer to published tables
3) Using standard heats of formation; DHf
4) Hess’s Law

18. Calorimetry
Calorimetry – The energy absorbed or released as heat (q) in a chemical reaction is measured in a calorimeter.
q(reaction) + q(solution) = 0
(Cp)(mass) (DT) + (Cp)(mass) (DT) = 0

20. Suppose you place grams of magnesium chips in a calorimeter and then add mL of 1.00 M HCl.
The temperature of the solution increases from K to K. What is the enthalpy change for the reaction.
The specific heat capacity of the solution is 4.20 J/g•K and the density of HCl is 1.00 g/mL

21. q(reaction) + q(solution) = 0
Mg HCl ® H2 + MgCl2
q(solution) = (Cp)(mass) (DT)
q(solution) = (100g g) (4.20) (297.6 – 295.4)
= 945 J
q(reaction) + q(solution) = 0
q(reaction) J = 0
q(reaction) = J

22. Using Standard Heats of Formation
You can calculate the heats of reaction (DH) using heats of formation data.
Heats of formation are heat changes for the formation of a substance from its elements.
Symbol is DHf
The standard heat of formation of an element, DHf = 0
This includes the diatomic elements
Must use the compound in correct state (g, l or s)

23. Thermochemical Equations
A enthalpy change, H, is the heat change for the reaction from reactants to products.
H = Hproducts – H reactants

24. What good are they? DHo DHf (Products) - DHf (Reactants) =
Standard heats of formation for some compounds are listed in tables.
The heat of a reaction can be calculated by:
subtracting the heats of formation of the reactants from the products
DHo
DHf (Products) -
DHf (Reactants) =

25. Example 1
CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)
CH4 (g) DHf = kJ/mol
O2(g) DHf = 0 kJ/mol
CO2(g) DHf = kJ/mol
H2O(g) DHf = kJ/mol
DH= [ (-241.8)] - [ (0)]
DH= kJ

26. Example 2 C3H5(NO3)3 DHf = - 364 kJ/mol O2(g) DHf = 0 kJ/mol
2C3H5(NO3)3 ® 3N2 + ½ O2(g) + 6CO2(g) + 5 H2O(g)
C3H5(NO3) DHf = kJ/mol
O2(g) DHf = 0 kJ/mol
N2(g) DHf = 0 kJ/mol
CO2(g) DHf = kJ/mol
H2O(g) DHf = kJ/mol
DH= [6(-393.5) + 5(-241.8)] - [2(-364)]
DH= kJ

27. Example 3 C6H5(l) DHf = + 49.0 kJ/mol O2(g) DHf = 0 kJ/mol
C6H5 (l) /2 O2(g) → 6 CO2(g) + 3 H2O(l)
C6H5(l) DHf = kJ/mol
O2(g) DHf = 0 kJ/mol
CO2(g) DHf = kJ/mol
H2O(l) DHf = kJ/mol
DH= [6(-393.5) + 3(-285.8)] - [+ 49.0]
DH= kJ

28. Hess’s Law
The overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the reaction.

29. Why Does It Work? If you turn an equation around, you change the sign:
If H2(g) + ½ O2(g) ® H2O(l) DH= kJ
then, H2O(l) ® H2(g) + ½ O2(g) DH = kJ
also,
If you multiply the equation by a number, you multiply the heat by that number:
2 H2O(l) ® 2 H2(g) + O2(g) DH = kJ

30. Example C(s) + 2 H2(g) ® CH4(g) DH= ? C(s) + O2(g) ® CO2(g)
DH = kJ/mol
H2 (g)+ ½ O2(g) ® H2O(l)
DH = kJ/mol
CH4 (g)+ 2O2(g) ® CO2 + 2H2O(l)
DH = kJ/mol
Because CH4 appears on the product side of our original
reaction we must reverse the reaction and the value of DH
CO2 + 2H2O(l) ® CH4 (g)+ 2O2(g)
DH = kJ/mol

31. Example C(s) + 2 H2(g) ® CH4(g) DH= ? C(s) + O2(g) ® CO2(g)
DH = kJ/mol
H2 (g)+ ½ O2(g) ® H2O(l)
DH = kJ/mol
CO2 + 2H2O(l) ® CH4 (g)+ 2O2(g)
DH = kJ/mol
Since there are 2 moles of hydrogen used as a reactant,
and there is only 1 mole in equation 2,
we must multiply the hydrogen reaction and DH value by 2.

32. Example C(s) + 2 H2(g) ® CH4(g) DH= ? C(s) + O2(g) ® CO2(g)
DH = kJ/mol
2H2 (g)+ O2(g) ® 2H2O(l)
DH = kJ/mol
CO2 + 2H2O(l) ® CH4 (g)+ 2O2(g)
DH = kJ/mol
Now add the three equations together cancelling out
compounds that appear on both sides of the equation.

33. C(s) + 2 H2(g) ® CH4(g) DH◦ = -74.3 KJ/mole
Example
C(s) + O2(g) ® CO2(g)
DH = kJ/mol
2H2 (g)+ O2(g) ® 2H2O(l)
DH = kJ/mol
CO2 + 2H2O(l) ® CH4 (g)+ 2O2(g)
DH = kJ/mol
C(s) + 2 H2(g) ® CH4(g) DH◦ = KJ/mole

34. Problem
Use Hess’s law to calculate the enthalpy change for the formation of CS2 from C and S: (C + 2S → CS2)
C + O2 → CO = KJ/mole
S + O2 → SO2 DH = KJ/mole
CS2 + 3O2 → CO SO DH = KJ/mole
DH

35. (C + 2S → CS2) C + O2 → CO2 DH = - 393.5 KJ/mole
S + O2 → SO2 DH = KJ/mole
CS2 + 3O2 → CO SO DH = KJ/mole
Strategy:
Equation 2: Double the moles and enthalpy value.
Equation 3: Reverse the reaction and the sign of the enthalpy.

36. (C + 2S → CS2) C + O2 → CO2 DH = - 393.5 KJ/mole
2S + 2O2 → 2SO2 DH = KJ/mole
CO SO2 → CS2 + 3O DH = KJ/mole
C + 2S → CS DH = KJ/mole