1. 15.2 Heat
Calorimetry

2. A. Calorimetry qgained = - qlost
Calorimeter – an insulated device used to measure heat absorbed or released during a reaction
Measures the heat flow into or out of a system
Heat released by the system is equal to heat absorbed by the surroundings
qgained = - qlost

3. A. Calorimetry
We will use simple calorimeters of foam cups open to atmospheric pressure
Chemicals are placed in inner cup
Temperature will change until both chemicals reach same temperature
Coffee cup Calorimeter

4. Practice A. Calorimetry GIVEN: WORK: qwater = ? qwater = m c ΔT
A sample of an unknown metal is placed in a calorimeter that contains 125 g of water at 25.60oC. The 50.0 g metal is heated to 115.0oC and placed in the calorimeter. The metal transfers heat to the water until they both reach 29.30oC.
GIVEN:
qwater = ?
m = 125 g
c = J/goC
T = – = 3.70oC
WORK:
qwater = m c ΔT
qwater = (125)(4.184)(3.70)
qwater = 1940 J
CONTINUE

5. Practice A. Calorimetry GIVEN: WORK: qwater = 1940 J qwater = - qmetal
A sample of an unknown metal is placed in a calorimeter that contains 125 g of water at 25.60oC. The 50.0 g metal is heated to 115.0oC and placed in the calorimeter. The metal transfers heat to the water until they both reach 29.30oC.
GIVEN:
qwater = 1940 J
qmetal = J
m = 50.0 g
c = ?
T = – = oC
WORK:
qwater = - qmetal
- qmetal = m c ΔT
= (50.0) c (- 85.7)
cmetal = J/goC

6. Practice A. Calorimetry Are you ready for the lab tomorrow?
150.0 g sample of a metal at 75.0oC is added to g H2O at 15.0oC. The temperature of the water rises to 18.3oC. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

7. B. Enthalpy Negative enthalpy = exothermic
ENTHALPY: (H) the heat content of a system at constant pressure
The terms heat and enthalpy change are interchangeable
q = ΔH
qsys = ΔH = - m x C x ΔT
Negative enthalpy = exothermic
Positive enthalpy = endothermic

8. B. Enthalpy ΔHrxn = Hfinal – Hinitial ΔHrxn = Hproducts – Hreactants
Enthalpy of a reaction: (∆Hrxn) the change in enthalpy for a reaction
∆Hrxn is the heat absorbed or released in a chemical reaction
∆Hrxn = difference in the enthalpies that exist between substances at the end of a reaction and the substances at the beginning of a reaction
ΔHrxn = Hfinal – Hinitial
ΔHrxn = Hproducts – Hreactants

9. B. Enthalpy
Enthalpy changes for exothermic reactions are always negative
4Fe(s) + 3O2(g)  2Fe2O3(s) Hrxn = kJ
Enthalpy changes for endothermic reactions are always positive
NH4NO3(s)  NH4+(aq) + NO3-(aq) Hrxn = 27 kJ

10. B. Enthalpy

11. B. Enthalpy

12. 15.3 Thermochemical Equations
Pages 529 – 533

13. A. Thermochemical Equations
A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products, and energy change.
In a thermochemical equation, the enthalpy of change for the reaction can be written as either a reactant or a product

14. A. Thermochemical Equations
In a thermochemical equation, the enthalpy of change for the reaction can be written as either a reactant or a product
Endothermic (positive ΔH)
2NaHCO kJ Na2CO3 + H2O + CO2
Exothermic (negative ΔH)
CaO + H2O Ca(OH) kJ

15. B. Thermochemical Equation Practice
Write the thermochemical equation for the addition of oxygen to iron (III) if the ΔHrxn= kJ/mol Fe(s) + O2(g)→ Fe2O3(s) kJ Write the thermochemical equation for the decomposition of sulfur trioxide into sulfur dioxide and oxygen if the ΔHrxn= 198 kJ/mol SO3(g) kJ → SO2 (g) + O2 (g) 4 3 2 2 2

16. 15.4 Hess’s Law
Pages 534 – 540

17. A. Hess’s Law
Hess’s law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction
2S(s) + 3O2(g)  2SO3(g) H = ?

18. A. Hess’s Law

19. A. Hess’s Law N2(s) + 2O2(g)  2NO2(g) H = ? Build your equation:
Known:
N2 (g) + O2(g)  2NO(g) H = -180 kJ
2NO(g) + O2(g)  2NO2(g) H = -112 kJ
Build your equation:

20. A. Hess’s Law 2H2O2(l)  2H2O(l) + O2(g) H = ? Build your equation:
Known:
2H2(g) + O2(g)  2H2O(l) H = -572 kJ
H2(g) + O2(g)  H2O2(l) H = -188 kJ
Build your equation:

21. A. Hess’s Law 2S(s) + 3O2(g)  2SO3(g) H = ? Known:
S(s) + O2(g)  SO2(g) H = -297 kJ
2SO3(g)  2SO2(g) + O2(g) H = 198 kJ
Build your equation:

22. A. Hess’s Law Challenge 2B(s) + 3H2(g)  B2H6(g) H = ?
Known:
2B (s) + 3/2O2(g)  B2O3(s) H = kJ
B2H6(g) + 3O2(g)  B2O3 (s) + 3H2O (g) H = kJ
H2 (g) + ½ O2(g)  H2O (l) H = -286 kJ
H2O (l)  H2O (g) H = 44 kJ
Build your equation:

23. B. Standard Heats of Formation
Standard heat of formation, ΔHf, of a compound is the change in enthalpy that accompanies the formation of one mole of the compound from its elements
All substances at standard pressure and 25oC
ΔHf of free elements (and diatomic molecules) in standard state is set at 0

24. B. Standard Heats of Formation
Write the thermochemical equation for the formation of sodium oxide
Na(s) + O2(g)  Na2O(s)
-416 kJ 2 4 2
ΔHf =

25. B. Standard Heats of Formation
Provides an alternative to Hess’s Law of calculating heat of reaction indirectly
For reactions occurring at standard conditions, calculate heat of reaction, ΔHo , using standard heats of formation (given to you on appendix copies)
ΔHo = ΔHf (products) – ΔHf (reactants)

26. B. Standard Heats of Formation
What is the standard heat of reaction of CO(g) with O2(g) to form CO2 (g)?
2CO(g) + O2(g) → 2CO2(g) ΔH = ?
2(-393.5) – [2(-110.5) + 0]
ΔH = kJ/mol

27. B. Standard Heats of Formation
CaCO3(s) → CaO(s) + CO2 (g) ΔH = ?
(ΔHfCaO + ΔHfCO2) – (ΔHfCaCO3) =
( ) – (-1207)
ΔH =179 kJ/mol