1. Unit 2: Thermochemistry

2. Energy The capacity to do work or transfer heat
Law of conservation of energy states that energy cannot be create nor destroyed but can change forms
Energy lost by a system must be gained by the surroundings and vice versa

3. Types of Energy Radiant energy Thermal energy Chemical energy
From electromagnetic radiation
Thermal energy
From heat (movement of particles)
Chemical energy
Stored in the bonds of chemical compounds, released in reactions

4. Heat and Thermochemistry
Heat is the energy used to cause the temperature of an object to increase
Thermodynamics is the study of energy and its transformations
Thermochemistry is a portion of thermodynamics that studies chemical reactions and energy changes that involve heat

5. System Vs. Surroundings
System is the portion of the universe we single out for study
In chemistry the reactants and products are the system
Surroundings are everything else besides the system

6. Open Vs. Closed System
Open systems can exchange matter and energy with the surroundings
Uncovered pot of boiling water on the stove
Closed systems can exchange energy but not matter with the surroundings
Piston with H2 and O2 reacting to produce H2O and energy in the form of heat and work

7. Exothermic Vs. Endothermic
Exothermic process – the system loses heat to the surroundings (spontaneous)
Combustion of gasoline
Endothermic process – the system gains heat from the surroundings (not spontaneous)
Melting ice cubes

8. Enthalpy (“H”)
Internal energy plus the product of pressure and volume of the system
𝐻=𝐸+𝑃𝑉
Heat evolved (absorbed or released) by a reaction is the change in enthalpy (ΔH)
∆𝐻= 𝑞 𝑃 =∆𝐸+𝑃∆𝑉
qP = heat or energy evolved from a system at a constant pressure

9. Enthalpy of Reaction (ΔHrxn)
“Heat of reaction”
Unit are in Joules (J)/kilojoules (kJ)
Enthalpy change associated with a chemical reaction
-ΔH = exothermic = gives off heat
+ΔH = endothermic = absorbs heat
∆𝐻= 𝑞 𝑛
𝑞 𝑟𝑥𝑛 = −𝑞 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

10. Thermochemical Equations
In your data sheets booklet these equations are given
We must incorporate mol calculations into these questions

11. Thermochemical Equations
Consider the following reaction:
2Mg(s) + O2(g)  2MgO(s)
ΔH=-1204kJ
Is the reaction exothermic or endothermic?
Calculate the amount of heat transferred when 3.55g of Mg(s) reacts at a constant pressure.
How many grams of MgO(s) are produced during an enthalpy change of -234kJ?
How many kJ of heat are absorbed when 40.3g of MgO(s) is decomposed into Mg(s) and O2(g) at constant pressure?

12. Thermochemical Equations
Exothermic (Δ𝐻=−)
3.55𝑔 𝑀𝑔 1𝑚𝑜𝑙 𝑀𝑔 𝑔 𝑀𝑔 −1204𝑘𝐽 2𝑚𝑜𝑙 𝑀𝑔 =−87.9𝑘𝐽
−234kJ 2𝑚𝑜𝑙 𝑀𝑔𝑂 −1204𝑘𝐽 𝑔 𝑀𝑔𝑂 1𝑚𝑜𝑙 𝑀𝑔𝑂 =15.7𝑔 𝑀𝑔𝑂
40.3𝑔 𝑀𝑔𝑂 1𝑚𝑜𝑙 𝑀𝑔𝑂 𝑔 𝑀𝑔𝑂 𝑘𝐽 2𝑚𝑜𝑙 𝑀𝑔𝑂 =+602𝑘𝐽

13. Calorimetry Measurement of heat flow
Carried out in/with a calorimeter by measuring the magnitude of temperature change the heat flow produces
In this course we do calculations using a specific type of calorimeter (coffee cup calorimeter)
𝑞=𝑚𝐶∆𝑇

14. Heat Capacity (“C”)
Amount of heat required to raise its temperature by 1K (or 1℃)
Greater C means more heat required top produce a temperature increase
Sand vs. water

15. Specific Heat (“CS”)
Heat capacity for pure substances is usually for a given amount
Specific heat is the amount of energy needed to raise the temperature of 1g of a substance by 1K.

16. Calorimetry
Calorimeters are vessels used to carry out experiments that allow us to analyze temperature changes
This allows us to see the energy being created by a reaction

17. Bomb Calorimeter Vs. Coffee Cup Calorimeter
Used to study combustion
Constant volume
Used to study principles of calorimetry
Mostly solutions
Constant pressure

18. Bomb Calorimeter Vs. Coffee Cup Calorimeter

19. Calorimetry Calculations
For calculations we use:
q = heat
m = mass
C = specific heat (or heat capacity)
ΔT = change in temperature

20. Coffee Cup Calorimetry
The specific heat of octane (C8H18) is 2.22J/g•K
How many joules of heat are needed to raise the temperature of 80.0g of octane from 10.0 to 25.0℃?
Which will require more heat, increasing the temperature of 1mol of C8H18 by a certain amount or increasing the temperature of 1mol of H2O by the same amount?

21. Coffee Cup Calorimetry
𝑞=𝑚𝐶∆𝑇= 80.0𝑔 𝐽 𝑔∙𝐾 15𝐾 =2.66× 10 3 𝐽
1𝑚𝑜𝑙 𝐶8𝐻 𝑔 𝐶8𝐻18 1𝑚𝑜𝑙 𝐶8𝐻 𝐽 𝑔∙𝐾 = 𝐽 𝐾
1𝑚𝑜𝑙 𝐻2𝑂 𝑔 𝐻2𝑂 1𝑚𝑜𝑙 𝐻2𝑂 𝐽 𝑔∙𝐾 = 𝐽 𝐾
It will take more energy to increase one mol of C8H18 that one mol of H2O

22. Coffee Cup Calorimetry
When 50mL of 1.0M HCl and 50mL of 1.0M NaOH is combine in a coffee cup calorimeter the temperature of the resultant solution increases from 21.0℃ to 27.5℃. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses only negligible quantity of heat, that the total volume of the solution is 100mL, that its density is 1.0g/mL, and that its specific heat is 4.18J/g•K.

23. Coffee Cup Calorimetry
𝑚= 100𝑚𝐿 1.0𝑔 1𝑚𝐿 =100𝑔
∆𝑇=27.5℃−21.0℃=6.5℃=6.5𝐾
𝑞 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟 =𝑚𝐶∆𝑇= 100𝑔 𝐽 𝑔∙𝐾 6.5𝐾
=2717𝐽=2.7𝑘𝐽=-ΔH (exothermic)
0.050𝐿 𝐻𝐶𝑙 1.0𝑚𝑜𝑙 𝐻𝐶𝑙 1𝐿 =0.050𝑚𝑜𝑙𝐻𝐶𝑙
∆𝐻 𝑚𝑜𝑙 = −2.7𝑘𝐽 0.050𝑚𝑜𝑙 =−54𝑘𝐽/𝑚𝑜𝑙

24. Enthalpy Diagrams
Used to give a visual representation of energy changes throughout a reaction

25. Enthalpy Diagrams Things to remember Enthalpy is an extensive property
Changes with the number of mols being reacted
Enthalpy stays the same magnitude for the reverse reaction but changes the sign
Enthalpy change depends on the states of the reactants and products
Enthalpy of gases>liquids>solids
+ΔH = endothermic (products higher than reactants)
-ΔH = exothermic (products lower than reactants)

26. Enthalpy Diagrams

27. 2H2(g) + O2(g) 2H2O(g) ΔH = -483.6kJ
Enthalpy Diagrams
Draw the enthalpy diagram for the following equation:
2H2(g) + O2(g) 2H2O(g) ΔH = kJ
Remember to include:
Direction of increasing enthalpy
Products (with proper coefficients and states)
Reactants (with proper coefficients and states)
Direction of reaction
ΔH (heat of reaction) with proper sign
“Exothermic” or “endothermic”

28. Hess’s Law
No matter how many steps a reaction has the total enthalpy change is equal to the sum of all enthalpy changes
ΔHtotal rxn = ΔHall steps

29. Hess’s Law Calculate ΔH for the reaction: 2C(s) + H2(g)  C2H2(g)
given the following chemical equations and their respective enthalpy changes:
C2H2(g) O2(g)  2CO2(g) + H2O(l) ΔH= kJ
C(s) + O2(g)  CO2(g) ΔH=-393.5kJ
H2(g) O2(g)  H2O(l) ΔH=-285.8kJ

30. Hess’s Law
2CO2(g) + H2O(l)  C2H2(g) O2(g) ΔH=1299.6kJ 2C(s) + 2O2(g)  2CO2(g) ΔH=-787.0kJ H2(g) O2(g)  H2O(l) ΔH=-285.8kJ C(s) + H2(g)  C2H2(g) ΔH=226.8kJ

31. Hess’s Law Things to remember
Products and reactants should be on same side in component equations as in net equation
Mol ratio should add up to be the same in component and net equations
Reversing an equation changes the sign of ΔH for that equation
Add ΔH’s with appropriate signs
When multiplying an equation by any number multiply the ΔH as well