1. Heat, what does it really mean?
Thermodynamics
Heat, what does it really mean?

2. What is thermochemistry?
-heat changes that occur during chemical reactions
REMEMBER: Energy is the capacity to do work
I. Heat (q) – energy that transfers from one object to another due to a temperature difference (can only see changes)
-we are back to Celsius when we deal with thermodynamics

3. A. Endothermic – absorbs heat from surroundings
II. Change in Heat
A. Endothermic – absorbs heat from surroundings
DEMO: Predict what will happen to the temperature of the system during an endothermic reaction
B. Exothermic – heat leaves system (temperature rises)
DEMO: Predict what will happen to the temperature of the system during an exothermic reaction
Endothermic: temperature decreased due to absorbing energy, but not enough
Exothermic: temperature increased due to release of heat

4. (Calorie, what you see listed on food labels, is 1000 calories)
III. Heat Capacity and Specific Heat
-calorie – amount of heat needed to raise the temperature of 1g of pure water by 1°C
(Calorie, what you see listed on food labels, is 1000 calories)
We can relate the calorie to the Joule
1 J = cal
4.184 J = 1 cal

5. A. Heat Capacity – amount of heat required to increase the
A. Heat Capacity – amount of heat required to increase the temperature of an object exactly 1°C
-the greater the mass, the greater its heat capacity
DEMO: Predict what will happen when the gas only balloon touches the candle flame and when the water balloon touches the flame
Gas balloon = popped Water balloon = not popped
????? – because water is a good insulator, the heat is absorbed by the water, protecting the latex of the balloon
DEMO: Predict what happens when water balloon is tilted

6. Practical example: On a hot, summer day you walk barefoot across the grass. What do your feet feel?
Cool, right? That’s because grass has a high heat capacity, not allowing the heat absorbed from the sun to change the temperature of the grass much
Now, same sunny day. You run across the street to get the mail, and step on a drain cover. How do your feet feel now?
HOT!!! That’s because iron has a low heat capacity, allowing the heat absorbed to change the temperature dramatically

7. B. Specific Heat Capacity – the amount of heat it takes to
B. Specific Heat Capacity – the amount of heat it takes to raise the temperature of 1g of a substance by 1C
-now size of object doesn’t matter. Only composition impacts temperature change
1. Heat affects substances with a high specific heat much less than those with a low specific heat
Ex. 1 gram of H2O is raised by 1°C with 1 calorie of heat added, while 1 gram of iron has a temperature change of 9°C with the same 1 calorie of heat added.
-pan gets hot before water inside does

8. q = mCΔT
q = heat (J or cal)
m = mass (g)
C = Specific Heat (J/g °C or cal/g °C)
ΔT = change in temperature (°C)
Ex. The temperature of a piece of copper with a mass of 95.4g increases from 25.0°C to 48.0°C when the metal absorbs 849J of heat. What is the specific heat of copper?
1st – Rearrange the equation to solve for C
C = q/mΔT

9. C = 849J/[95.4g x (48.0°C – 25.0°C)] = 849J/(95.4g x 23.0°C)
= J/g°C
CHECK: waters’ specific heat is 4.18J/g°C. Since metal has a lower specific heat, J/g°C is reasonable and expected
C. Calorimetry – accurate and precise measurement of heat change for chemical and physical processes
**heat released by the system is equal to the heat absorbed by its surroundings – The Law of Conservation of Matter

10. q = ΔH heat and q can be used interchangeably
1. Calorimeter – insulated device used to measure the absorption or release of heat during a reaction
2. Enthalpy (H) – heat property when pressure is constant
q = ΔH heat and q can be used interchangeably
ΔH = m°CΔT
-where each variable refers to water
ΔH = +, endothermic
ΔH = -, exothermic
q = +, endothermic
q = -, exothermic

11. (100mL)(1.0g/mL*) = 100g *density of water
-When we have a solution, we can assume that no heat escapes, but it transferred from the reaction to the solution
q soln = (C of soln) x (g of soln) x ΔT = -qrxn
Ex. When student mixes 50 mL HCl & 50 mL NaOH in calorimeter, T changes from 21.0 to 27.5°C, What is heat when C=4.18J/g°C?
HCl + NaOH H2O + NaCl
*total amount of solution = 50mL + 50mL
(100mL)(1.0g/mL*) = 100g *density of water
ΔT = 27.5 – 21.0˚C = 6.5˚C = 6.5K

12. qrxn = C x m x ΔT (4.18J/g°C)(100g)(6.5°C)
= 2.7 x 103 J = -2.7 x 103 J or -2.7kJ
-our calculation was really using the information for water, but since the heat lost is equal to the heat gain, by changing the sign we show the heat for our substance
-when you are putting a solid object (usually metal) in a calorimeter, you will first calculate the heat of water (q) to obtain amount of heat gained. Change the sign to reflect heat lost by metal, and use to calculate unknown of metal.

13. 3. Thermochemical Equations
a. Heat released or absorbed during a reaction can be included in the reaction itself
Ex. CaO(s) + H2O(l)  Ca(OH)2(s) kJ OR
CaO(s) + H2O(l)  Ca(OH)2(s) ΔH=-65.2kJ
b. Heat shown in a reaction us called the heat of reaction, and is shown in a thermochemical equation
- This reactions shows that 1 mole of CaO reacts with 1 mole of H2O and releases 65.2kJ of heat

14. -works the same for endothermic reactions
Ex. 2NaHCO3(s) + 129kJ  Na2CO3(s) + H2O(g) + CO2(g) OR
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g) ΔH = +129kJ
Ex. Using the equation above, calculate the kJ of heat required to decompose 2.24moles 2NaHCO3(s)
2.24mol NaHCO3(s) 129kJ
2 moles NaHCO3(s)
=144kJ
**remember to use the coefficient in the balanced equation when using a thermochemical equation to get energy value

15. -states of matter become important, as it takes a different amount of energy to convert a solid to a liquid than a solid to a gas
c. Heat of combustion – energy required to completely burn 1 mole of a substance
-will use a table of values when energy of reaction is not given in problem
-can use table to convert to grams or moles, depending on question asked

17. d. Heat of fusion (ΔHfus) – molar heat would be the heat
d. Heat of fusion (ΔHfus) – molar heat would be the heat required to convert 1 mole of a substance from a solid to a liquid (melting)
e. Heat of solidification (ΔHsolid) – molar heat would be the heat required to convert 1 mole of a substance from a liquid to a solid (freezing, or solidification)
** heat absorbed by a melting solid is exactly the same as the quantity of heat lost when the liquid solidifies
ΔHfus = -ΔHsolid

18. Ex. How many grams of ice at 0C and 101
Ex. How many grams of ice at 0C and 101.3kPa could be melted by the addition of 2.25kJ of heat?
Since ice is formed by a liquid becoming a solid, we call that fusion. The ΔHfus = 6.01 kJ/mol (again, this would come from the problem or from a table)
2.25kJ 1 mol ice g
6.01kJ I mol ice
= 6.74g ice
4. Heats of Vaporization and Condensation
a. Vaporization is a cooling process, as heat from the surroundings flow to the liquid

19. b. Molar heat of vaporization (ΔHvap) – amount of heat
b. Molar heat of vaporization (ΔHvap) – amount of heat necessary to vaporize one mole of a given liquid. Tables are at standard conditions
c. Molar heat of condensation (ΔHcond) – the amount of heat necessary to convert 1 mole of a given gas to a liquid
Ex. How much heat (in kJ) is absorbed when 24.8g H2O(l) at 100°C is converted to steam at 100°C?
24.8g H2O 1mol H2O kJ
18.013g H2O 1mol H2O
= 56.1kJ

20. 5. Heat of Solution
a. Molar heat of solution (ΔHsoln) – the amount of heat needed to dissolve one mole of a substance
6. Can combine calorimetry and phase change enthapies to determine total energy change
q = mΔCT is used when temperature is changing (A-B)
kJ/mol conversion (fence panel) is used when phase change is occuring (B-C)

21. q = (50g)(4.184J/g°C)(0°C-(-34°C)
Ex. Calculate the energy required to melt 50g of ice from -34°C to 67°C water? ΔH = 6.01kJ/mol
1st – use mCΔT to calculate heat required to change temperature of ice from -34°C to 0°C (we always use standard freezing and boiling point of water unless told otherwise)
q = (50g)(4.184J/g°C)(0°C-(-34°C)
= J
2nd – use conversion to calculate energy for melting ice
50g H2O 1mol H2O 601kJ
18.013g H2O 1mol H2O
=5.413kJ
=5.413x108J

22. 3rd – Use mCΔT to convert from 0C water to 67C water
q = 50g)(4.184J/gC)(67°C-0°C)
= J
4th – total all answers: x104J x 108J x104J
= 5.4x108J
IV. Calculating Heat Changes
A. Hess’s Law – if you add 2 or more thermochemical equations to give a final equation, you can also add the heats of reaction to give the final heat of reaction
-used when heats of reaction cannot be directly determined (for in between (intermediate) steps

23. Ex. C(diamond)  C(graphite)
This process cannot be measure directly because it takes millions of years, so we use Hess’s Law
C(s, graphite) + O2(g)  CO2(g) ΔH = kJ
C(s, diamond) + O2(g)  CO2(g) ΔH = kJ
-the goal is to manipulate the equations so that you get an equation that matches the final equation
a. The C(s, diamond) is in the right place (reactant), so leave it alone
b. The C(s, graphite) is on the wrong side. It needs to be a product, so flip the entire equation around

24. CO2(g)  C(s, graphite) + O2(g)
-When you flip an equation (or manipulate it any other way) you have to match that change in the ΔH value
ΔH= kJ becomes ΔH = 393.5kJ
-Now you can look at the entire set up, with all changes
C(s, diamond) + O2(g)  CO2(g) ΔH = kJ
CO2(g) C(s, graphite) + O2(g) ΔH = 393.5kJ
C(s, diamond)  C(s, graphite)
ΔH =-1.9kJ
*add together, cancelling like terms that occur on each side of arrow

25. *you must cancel all “spectator” substances (those that do not appear in the final equation), therefore you will sometimes flip an equation, multiply an equation by a coefficient, or divide and equation by a coefficient. REMEMBER: What you do to the equation, you must do to the ΔH value
-you can end up with fractions in order to match coefficients!