Created by S. Colgan;
Modified by K. Slater
Resources: http://lincoln.pps.k12.or.us/lscheffler/Energetics.ppt#269,36,Sta ndard Enthalpy Changes
1

IB Topic 5: Energetics
5.1: Exothermic and Endothermic Reactions




5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction
( Δ H o ).
5.1.2 State that combustion and neutralization are exothermic processes.
5.1.3 Apply the relationship between temperature change, enthalpy change and the classification of a reaction as endothermic or exothermic.
5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of enthalpy change for the reaction.
2

 Heat is energy that is transferred from one object to another due to a difference in temperature
 Temperature is a measure of the average kinetic energy of a body
 Heat is always transferred from objects at a higher temperature to those at a lower temperature
3
3

5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction ( Δ H o ).
4

5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction ( Δ H o ).
Standard Enthalpy Change of Reaction ( ∆H): The heat energy exchanged with the surroundings when a reaction happens under standard conditions (NOT STP… see below).
Since the enthalpy change for any given reaction will vary with the conditions, esp. concentration of chemicals, Δ H are measured under standard conditions:



 pressure = 101.3 kPa temperature = 25ºC = 298 K
Concentrations of 1 mol dm-3
The most thermodynamically stable allotrope (which in the case of carbon is graphite)
Only Δ H can be measured, not H for the initial or final state of a system.
5

5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction ( Δ H o ).
Pseudonyms (other names) for

H


Heat of Reaction:

H rxn heat produced in a chemical reaction
Heat of Combustion:

H comb heat produced by a combustion reaction






Heat of Neutralization: heat produced in a neutralization reaction
(when an acid and base are mixed to get water, pH = 7)
Heat of solution:

H sol
Heat of Fusion:

H fus heat produced by when something dissolves heat produced when something melts
Heat of Vaporization:

H vap heat produced when something evaporates
Heat of Sublimation:

H sub heat produced when something sublimes
Heat of formation:

H f change in enthalpy that accompanies the formation of 1 mole of compound from it’s elements (this has special uses in chemistry…)
6

5.1.2 State that combustion and neutralization are exothermic processes.
Combustion
Exothermic reaction
General Combustion Reaction Formula:
CH
Compound (usually hydrocarbon) + O
 CO
4
2
+ 2O
+ H
2
O + energy
2
 CO
2
+ 2H
2
O + 890kJ
2
∆H = -890kJ
Neutralization
Exothermic reaction
Acid + Base  Salt + Water + energy
HCl + NaOH  NaCl + H
2
O + 57.3 kJ
∆H = -57.3kJ
7

5.1.3 Apply the relationship between temperature change, enthalpy change and the classification of a reaction as endothermic or exothermic.
Exothermic
Heat flows out of the system
Surroundings heat up
Heat change ( Δ H) < 0
(negative)
C
8
H
18
+ 12½O
2
9H
2
 8CO
2
+
O ΔH = -5512 kJ mol -1
H
2
+ ½O
2
 H
2
O
ΔH = -286 kJ mol -1
Endothermic
Heat flows into the system
Surroundings cool down
Heat change ( Δ H) > 0
(positive)
H
2
O(s)  H
2
O(l)
ΔH = +6.01 kJ mol -1
½N
2
+ O
2
 NO
2
ΔH = +33.9 kJ mol -1
8

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
2H
2
( g ) + O
2
( g ) 2H
2
O ( l ) + energy
H
2
O ( g ) H
2
O ( l ) + energy
Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO ( s ) energy + H
2
O ( s )
2Hg ( l ) + O
2
( g )
H
2
O ( l )
9 6.2

10 6.5

5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of enthalpy change for the reaction.
Exothermic Reactions
Products more stable than reactants
(lower energy).
ΔH = H products
– H reactants
Since the products have less energy than the reactants, the ΔH value is negative.
Endothermic Reactions
Products less stable than reactants
(higher energy)
ΔH = H products
– H reactants
Since the products have more energy than the reactants, the ΔH value is positive.
11

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction at constant pressure
H products
< H

H < 0 reactants
H products
> H

H > 0 reactants
12 6.4

5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of enthalpy change for the reaction.
13

Definition
Examples (2)
Change in Temperature
∆H value
Direction of heat flow
Stability of reactants
Stability of products
Bonding
Endothermic Exothermic
14

Endothermic Exothermic
Definition
Examples (2)
A process that absorbs heat from the surroundings
A process that releases heat into the surroundings
Combustion &
Neutralization reactions
Increases Change in
Temperature
∆H value
Decreases
Positive Negative
Direction of heat flow From surroundings into system
Stability of reactants More stable
Stability of products Less stable
Bonding Bond breaking
From system into surroundings
Less stable
More stable
Bond making 15

IB Topic 5: Energetics
5.2: Calculation of Enthalpy Changes




5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.
5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.
5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water
5.2.4 Evaluate the results of experiments to determine enthalpy changes.
16

 The amount of heat contained by an object depends primarily on three factors:
 The mass of material
 The temperature
 The kind of material and its ability to absorb or retain heat.
17
17



The heat required to raise the temperature of
1.00 g of water 1 o C is known as a calorie
Calorie (with a capital “C”) : dietary measurement of heat. Food has potential energy stored in the chemical bonds of food.
1 Cal = 1 kcal = 1000 cal
 The SI unit for heat is the joule . It is based on the mechanical energy requirements.
 1.00 calorie = 4.184 Joules
18
18

5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.
Heat Energy Change q = m x c x ΔT q = heat (joules or calories) m = mass (g) c = specific heat (J g -1 o C -1 )
The amount of heat required to raise the temperature of 1 g of a substance 1 o C.
ΔT = change in temperature
19

5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.
How much heat in joules will be absorbed when 32.0 g of water is heated from 25.0 o C to 80.0 o C?
When 435 J of heat is added to 3.4 g of olive oil at 21 o C, the temperature increases to 85 o C. What is the specific heat of olive oil?
q = m x c x ΔT q = ?
m = 32.0 g c = 4.18 J g -1 o C -1
ΔT = 80.0-25.0 = 55.0 o C q = m x c x ΔT q = 435 J m = 3.4 g c = ?
ΔT = 85-21 = 64 o C q = 32.0 x 4.18 x 55.0 = 7,360
J
435 = 3.4 x c x 64 = 2.0 J g -1 o C -1
20

Calculate the heat that would be required an aluminum cooking pan whose mass is 400 grams, from 20 o C to
200 o C. The specific heat of aluminum is 0.902 J g -1 o C -1 .
Solution

Q = mC

T
= (400 g) (0.902 J g -1 o C -1 )(200 o C – 20 o C )
= 64,944 J
21
21

5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.
How much heat in joules is required to raise the temperature of 250 g of mercury 52 o C?
A 1.55 g piece of stainless steel absorbs 141 J of heat when its temperature increases by 178 o C.
What is the specific heat of the stainless steel?
22

5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed.
How much heat in joules is required to raise the temperature of 250 g of mercury 52 o C?
A 1.55 g piece of stainless steel absorbs
141 J of heat when its temperature increases by 178 o C. What is the specific heat of the stainless steel?
q = m x c x ΔT q = ?
m = 250 g c = 0.14 J g -1 o C -1 (Table
11.2)
ΔT = 52 o C q = m x c x ΔT q = 141 J m = 1.55 g c = ?
ΔT = 178 o C q = 250 x 0.14 x 52 = 1800
J
141 = 1.55 x c x 178 = 0.511 J g -1 o C -1
23

5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions.

Calorimeter: Reactions used to heat up an external source of water.
Temperature change of water, mass of material and mass of water are measured.
Use q = m x c x ΔT to solve for q then find the heat of reaction in kJ/mol of reacting substance.
24

 Calorimetry involves the measurement of heat changes that occur in chemical processes or reactions.
 The heat change that occurs when a substance absorbs or releases energy is really a function of three quantities:


The mass
The temperature change
 The heat capacity of the material
25
25

 The ability of a substance to absorb or retain heat varies widely.
 The heat capacity depends on the nature of the material.
 The specific heat of a material is the amount of heat required to raise the temperature of 1 gram of a substance 1 o C (or Kelvin)
26
26

What is the final temperature when 50 grams of water at
20 o C is added to 80 grams water at 60 o C? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 o C -1
Solution:

Q
(Cold)
=

Q (
Let T = final temperature hot
) mC

T= mC

T
(50 g) (4.184 J g -1 o C -1 )(T- 20 o C)
= (80 g) (4.184 J g -1 o C -1 )(60 o C- T)
(50 g)(T- 20 o C) = (80 g)(60 o C- T)
50T -1000 = 4800 – 80T
130T =5800
T = 44.6 o C
27
27

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.
Using Thermochemical Equations
Calcium oxide combines with water to produce calcium hydroxide and heat
(exothermic reaction).


CaO(s) + H
2
O(l)  Ca(OH)
2
(s) + 65.2 kJ OR
CaO(s) + H
2
O(l)  Ca(OH)
2
(s)  H = -65.2 kJ
How many kJ of heat are produced when 7.23 g of CaO react?
1) Balance equation
2)
3)
Moles CaO: 7.23 g/56.01 g/mol = 0.129 mol
Ratio using heat: x kJ = 65.2 kJ
0.129 CaO 1 CaO
8.41 kJ
28

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.
Using Thermochemical Equations
Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to sodium carbonate, water, and carbon dioxide (endothermic reaction).


2NaHCO
3
(s) + 129 kJ  Na
2
2NaHCO
3
(s)  Na
2
CO
3
(s) + H
CO
3
2
(s) + H
2
O(g) + CO
O(g) + CO
2
(g) OR
2
(g)  H = 129 kJ
How many kJ of heat are needed to decompose 2.24 mol NaHCO
3
(s)?
1) Balanced equation
2)
3)
Moles NaHCO
3
(s) = 2.24 mol
Ratio: x kJ
2.24 NaHCO
3
(s)
= 129 kJ 144 kJ
2 mol
Do sample problems 13 & 14 on page 304
Pg. 306: 15,16,18
29

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.
Using Experimental Data
In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup o calorimeter. At the start, the solutions and the calorimeter are all at 25.0
C. During the reaction, the highest temperature observed is 32.0 o C.
Calculate the heat (in kJ) released during this reaction. Assume the densities of the solutions are 1.00 g mL -1 .
Use q = m x c x ΔT m = mass of solution = 50.0 mL x 1.00 g mL -1 = 50.0 g
C = 4.18 j g -1 o C -1
ΔT = 32.0 – 25.0 = 7.0 o C q = 50.0 g x 4.18 J g -1 o C -1 x 7.0 o C = 1463 J = 1.5 kJ
Pg. 302: 11 & 12
30

31

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.




Using Thermodynamic Quantities (Standard Heats of Formation)
Heat of reaction can be found by: sum the heats of formation of all the products – sum of heats of formation of all the reactants

H rxn
=

H f products
– 
H f reactants

H
 f
= standard enthalpy of formation. Energy required to form a compound from its elements.
“standard” is a term used a lot in chemistry. It usually means that the values are experimentally determined and compared to an agreed upon reference value
Since the

H f is given per mole, we must multiply by coefficients
32

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.
Using Thermodynamic Quantities (Standard Heats of
Formation)
Using the table of thermodynamic quantities, calculate the heat of reaction for 2SO
2
(g) + O
2
(g)  2SO
3
(g)
Heat of reaction = H f
Heat of products: H f
(SO
3 products – H f of reactants
) = -395.2 kJ/mol x 2 = -790.4 kJ
Heat of reactants = H f
(SO
2
) + H f
(O
2
)
(-296.9 kJ/mol x 2) + (0) = -593.8 KJ
Heat of reaction = -790.4 kJ – (-593.8 kJ) = -196.6 kJ
33

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.
N
2
O
4
+ 3 CO  N
2
O + 3CO
2
Reactants

H
 f
N
2
O
4
9.7 kJ mol -1
CO -110 kJ mol -1
Products
N
2
O
CO
2

H
 f
81 kJ mol -1
-393 kJ mol -1

H f products

H f reactants

H f products
= 1(81 kJ mol -1 ) + 3(-393 kJ mol -1 ) = -1098 kJ/mol
= 1(-9.7 kJ mol
– 
H f reactants
-1 )+ 3(-110 kJ mol
= (-1098 kJ mol -1 )
-1 ) = -320.3 kJ mol
– (-320.3 kJ mol -1 )
-1
= -778 kJ mol -1

H rxn
= -778 kJ mol -1
Therefore it is exothermic
34

5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water.
Ca(OH)
2
(s) + CO
2
(g)  H
2
O(g) + CaCO
3
(s)
Reactants

H
 f
Products

H
 f
Ca(OH)
2
-986.1 kJ mol -1 H
2
O -241.8 kJ mol -1
CO
2
-393.5 kJ mol -1 CaCO
3
-1206.9 kJ mol -1

H f products

H f reactants

H f products
= 1(-241.8 kJ mol
= 1(-986.1 kJ mol
– 
H f reactants
-1
-1
) + 1(-1206.9 kJ mol
)+ 1(-393.5 kJ mol
= (-1448.7 kJ mol -1 )
-1
-1 ) = -1448.7 kJ/mol
) = -1379.6 kJ mol
– (-1379.6 kJ mol -1 )
-1
= -69.1 kJ mol -1

H rxn
= -69.1 kJ mol -1
Therefore it is exothermic
35

IB Topic 5: Energetics
5.3 Hess’s Law & 5.4 Bond Enthalpies



5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
5.4.1 Define the term average bond enthalpy
5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic
36

5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
Hess’s Law: Reactions can be added together in order to determine heats of reactions that can’t be measured in the lab.
C(diamond)  C(graphite)
This reaction is too slow to be measured in the lab. Two reactions can be used that can be measured in the lab: a) C(graph) + O
2
(g)  CO
2
(g)  H = -393.5 kJ b) C(diam) + O
2
(g)  CO
2
(g)  H = -395.4 kJ
Since C(graphite) is a product, write equation a) in reverse to give: c) CO
2
(g)  C(graph) + O
2
(g)  H = 393.5 kJ
Now add equations b) and c) together:
C(diam) + O
2
(g) + CO
2
(g)  C(graph) + O
2
(g) + CO
2
(g)
 H = -395.4 kJ + 393.5 kJ = -1.9 kJ
Final equation: C(diamond)  C(graphite)  H = - 1.9 kJ
37

5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction for:
C
2
H
4
(g) + H
2
O(l)  C
2
H
5
OH(l) a) C
2
H
5
OH(l) + 3O
2
(g)  2CO
2
(g) + 3H
2
O(l)  H = -1367 kJ b) C
2
H
4
(g) + 3O
2
(g)  2CO
2
(g) + 2H
2
O(l)  H = -1411 kJ
38

5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction for
C
2
H
4
(g) + H
2
O(l)  C
2
H
5
OH(l) a) C
2
H
5
OH(l) + 3O
2
(g)  2CO
2
(g) + 3H
2
O(l)  H = -1367 kJ b) C
2
H
4
(g) + 3O
2
(g)  2CO
2
(g) + 2H
2
O(l)  H = -1411 kJ
Since C
2
H
5
OH(l) is a product, write equation a) in reverse order: c) 2CO
2
(g) + 2H
2
O(l)  C
2
H
5
OH(l) + 3O
2
(g)  H = 1367 kJ
Add equations b) & c) together, cancelling out substances on opposite sides of the arrow. Add the heat values to obtain the heat of reaction.
C
2
H
4
(g) + H
2
O(l)  C
2
H
5
OH(l)  H = -44 kJ
39

5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction for:
C(s) + 2H
2
(g)  CH
4
(g) a) C(s) + O
2
(g)  CO
2
(g)  H = -393 kJ mol -1 b) H
2
(g) + ½ O
2
(g)  H
2
O(l)  H = -286 kJ mol c) CH
4
(g) + 2O
2
(g)  CO
2
(g) + 2H
2
O(l)  H = -890 kJ mol -1
-1
40

5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes.
Given the following thermochemical equations, calculate the heat of reaction for:
C(s) + 2H
2
(g)  CH
4
(g)
C(s) + O
2
(g)  CO
2
(g)  H = -393 kJ
2(H
2
(g) + ½ O
2
(g)  H
2
O(l))
2H
2
(g) + O
2
(g)  2H
2
O(l))


H = 2(-286 kJ mol
H = -572 kJ
CH
4
(g) + 2O
2
(g)  CO
2
(g) + 2H
2
O(l)  H = -890 kJ mol -1
CO
2
(g) + 2H
2
O(l)  CH
4
(g) + 2O
2
(g)  H = 890
C(s) + 2H
2
(g)  CH
4
(g)  H = -75 kJ
-1 )
41

5.4.1 Define the term average bond enthalpy
Enthalpy changes of reactions are the result of bonds breaking and new bonds being formed.


Breaking bonds requires energy
Forming new bonds releases energy
Bond enthalpy is the energy change required to break 1 mol of bonds or form 1 mol of bonds between two covalently bonded atoms in the gaseous state.
Bond enthalpies for unlike atoms will be affected by surrounding bonds and will be slightly different in different compounds so average
bond enthalpies are used.
42





43
43

 One approach to determining an enthalpy change for a chemical reaction is to compute the difference in bond enthalpies between reactants and products
 The energy to required to break a covalent bond in the gaseous phase is called a bond enthalpy.
 Bond enthalpy tables give the average energy to break a chemical bond. Actually there are slight variations depending on the environment in which the chemical bond is located
44
44



45
45

The average bond enthalpies for several types of chemical bonds are shown in the table below:
46
46





Bond enthalpies can be used to calculate the enthalpy change for a chemical reaction .
Energy is required to break chemical bonds .
Therefore when a chemical bond is broken its enthalpy change carries a positive sign .
Energy is released when chemical bonds form . When a chemical bond is formed its enthalpy change is expressed as a negative value
By combining the enthalpy required and the enthalpy released for the breaking and forming chemical bonds, one can calculate the enthalpy change for a chemical reaction
47
47

 Exothermic processes release energy
C
3
H
8
(g) + 5 O
2
(g)  3 CO
2
(g) + 4H
2
O (g)
+ 2043 kJ
 Endothermic processes absorb energy
C(s) + H
2
O (g) +113 kJ  CO(g) + H
2
(g)
48
48

In an endothermic reaction there is more energy required to break bonds than is released when bonds are formed.
The opposite is true in an exothermic reaction.
49
49

5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic
If the amount of energy required to break the bonds in the reactants is greater than the amount of energy released when bonds are formed in the products, the reaction is endothermic. average bond enthalpy reactants > average bond enthalpy products
If the amount of energy required to break the bonds in the reactants is less than the amount of energy released when bonds are formed in the products, the reaction is exothermic. average bond enthalpy reactants < average bond enthalpy products
50

5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic
Use the following average bond enthalpies (kJ mol -1 ) to determine the heat of reaction for 3F
2
+ NH
3
 3HF + NF
3
F-F = 158; N-H = 388; H-F = 562; N-F = 272
Energy in (kJ mol -1 )
3F-F is 3(158) + 3N-H is 3(388) = 1638 kJ
Energy out (kJ mol-1)
3H-F is 3(562) + 3N-F is 3(272) = 2502 kJ
Since Energy in < Energy out, reaction is exothermic
Heat of reaction is -864 kJ mol -1
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5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic
Using the Bond Enthalpy Table, determine the heat of reaction for:
CO(g) + 2H
2
(g)  CH
3
OH
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5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic
Using the Bond Enthalpy Table, determine the heat of reaction for:
CO(g) + 2H
2
(g)  CH
3
OH
Energy in (kJ)
1CO (triple bond) 1074 + 2H-H is 2(435) = 1944 kJ mol -1
Energy out (kJ)
3C-H is 3(393) + 1C-O is 356 + 1 O-H is 464 = 1999 kJ mol -1
Since Energy in < Energy out, reaction is exothermic
Heat of reaction is -55 kJ mol -1
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2
2

3
Bonds broken
1 N = N: = 945
3 H-H: 3(435) = 1305
Total = 2250 kJ
Bonds formed
2x3 = 6 N-H: 6 (390) = - 2340 kJ
Net enthalpy change
= + 2250 - 2340 = - 90 kJ
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5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic
In the 1960s NASA considered the relative merits of using hydrogen and oxygen compared with hydrogen and fluorine as rocket fuels.
Assuming all the reactants and products are in the gaseous state, use bond enthalpies to calculate the enthalpy change of reaction
(in kJ mol -1 of product) for both fuels. As mass is more important than amount in the choice of rocket fuels, which reaction would give more energy per kilogram of fuel?
Bond enthalpies (kJ mol-1):
H-H: 435; O O: 496; H-O: 464 kJ; F-F: 158; H-F: 562
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 Endothermic
 Exothermic
 Temperature
 Heat
 Average bond enthalpy
 Standard enthalpy of formation


Enthalpy of combustion
Hess’ Law
 Standard conditions
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