Check Answer
Answer: ∆H_reaction = -2191.1 kJ

We will use the equation
∆H_reaction =  ∑ ∆H_f°_products  -  ∑ ∆H_f°_reactants
 to determine the ∆H. Heat of formation values (∆H_f°) are given in the table above:
 
To avoid issues associated with + and - signs, we will calculate ∑ ∆H_f°_products and ∑ ∆H_f°_reactants individually. While this is not the only way to approach the problem, it is the safest way.
 
∑ ∆H_f°_products = 3 mol•(-178.4 kJ/mol)  +  6 mol•(-183.4 kJ/mol) = -1635.6 kJ
 
∑ ∆H_f°_reactants = 5 mol•(+140.7 kJ/mol)  +  2 mol•(-74.0 kJ/mol) = +555.5 kJ
 
Now we can calculate the ∆H.
 
∆H_reaction = ∑ ∆H_f°_products  -  ∑ ∆H_f°_reactants = -1635.6 - (555.5 kJ kJ)
 
∆H_reaction = -2191.1 kJ

 

Lesson 2: Chemical Reactions and Enthalpy Change

Part c: Heat of Formation

Part a: Enthalpy Change
Part b: Hess's Law
Part c: Heat of Formation
Part d: Bond Enthalpy and ∆H
Part e: Thermal Stoichiometry

 

Standard Heat of Formation Values

Scientists have collected a boat load of thermodynamic data on nearly every substance in existence. Some of that data can be found in our Reference section. See the Standard Enthalpy of Formation page as an example. The table at the right provides a small sample of data from that page for a variety of the more common substances. What exactly do all these numbers mean? How will I use these numbers in Chemistry class? And does a student of Chemistry have to memorize these values?
 
We will answer the last question first. Does a student of Chemistry have to memorize these values? No! A student of Chemistry does not have to memorize these values. But how cool would you be if you did! For instance, just imagine mentioning to your friends at the lunch table that the standard heat of formation of liquid water is -285.8 kJ/mole. Now that would impress! And if it doesn’t, you need new friends. But otherwise … no! … you probably don’t need to memorize these values. But you do need to know what they mean and how to use them. Read on!
 
These values are standard heat of formation values or standard enthalpy of formation values. Every substance has its own value. For some substances (elements), the value is 0 kJ/mol. Each value indicates the enthalpy change for the formation of 1 mole of that substance from its elements in the standard state. It is the ∆H value for a specific type of reaction … a reaction in which there is 1 mole of the substance as the product. And the reactants are the elements that are in the substance as they exist naturally at a pressure of 1 atm and a temperature of 25°C. It’s a simple idea with a lot of details.
 
Here's a summary of the details:

• It is the ∆H for a formation (or synthesis) reaction.
• The coefficient in front of the product is 1 (for 1 mole).
• Reactants are the elements … as they exist naturally.
• The values apply at standard pressure and temperature - 1 atm and 25°C.
• The state symbol is important. The value is different for (l) as it is for (g).

 

Let’s consider five substances from the list - H₂O(l), H₂O(g), CO₂(g), NH₃(g), and Al₂O₃(s) - and one not on the list - O₂(g). The values in the table refer to the enthalpy change for the following formation or synthesis reactions occurring at 1 atm and 25°C.
 
Eq’n 1:            H₂(g)  +  ½ O₂(g)  →  H₂O(l)                         ∆H_f° = -285.8 kJ/mol
 
Eq’n 2:            H₂(g)  +  ½ O₂(g)  →  H₂O(g)                        ∆H_f° = -241.8 kJ/mol
 
Eq’n 3:            C(s)  +  O₂(g)  →  CO₂(g)                              ∆H_f° = -393.5 kJ/mol
 
Eq’n 4:            ½ N₂(g)  +  ³/₂ H₂(g)  →  NH₃(s)                   ∆H_f° = -45.9 kJ/mol
 
Eq’n 5:            2 Al(s)  + ³/₂ O₂(g)  →  Al₂O₃(s)                    ∆H_f° = -1675.7 kJ/mol
 
Eq’n 6:            O₂(g)  →  O₂(g)                                              ∆H_f° = 0.0 kJ/mol
 
 
You should notice the following about every one of these equations:

• They are balanced equations.
• There is only 1 product. It is the substance for which the heat of formation is given.
• The product coefficient is always 1. The values are based on 1 mole.
• Fractions may be needed for the reactant coefficients.
• The reactants are elements … as they exist naturally. Remember your diatomic elements!
• The formula and the state symbol for reactants match how those reactants would exist at 1 atm and 25°C.

 
And for the last equation (Equation 6), the enthalpy change is 0.0 kJ/mol. That makes sense because it takes zero energy to form O₂(g) from its elements since oxygen gas is made of O₂(g). The ∆H_f° for any element in its natural state is 0.0 kJ/mol.
 
That completes the answer to the question What exactly do all these numbers mean? Now let’s begin the discussion of How will I use these numbers in Chemistry class?  
 
 

Formation versus Decomposition

A formation reaction is a synthesis reaction. We discussed synthesis reactions in Chapter 8 of our Chemistry Tutorial. We also discussed decomposition reactions. Decomposition reactions are the opposite of synthesis or formation reactions. Synthesis reactions involve starting with elements and putting them together to form compounds. Decomposition reactions involve starting with compounds and pulling them apart to form their elements. Because a decomposition reaction is the opposite of a formation reaction, the ∆H of a decomposition reaction is the opposite of the corresponding formation reaction. In other words, if formation of 1 mole of ammonia gas releases 45.9 kJ:
   
Then it can be said that the decomposing of 1 mole of ammonia gas requires 45.9 kJ:
   
This is a natural extension of the second rule of thermochemical reactions that we discussed in Lesson 2b. Whenever we know a heat of formation value for a compound, we also know the enthalpy change for the corresponding decomposition reaction; it is a ∆H value that is equal in magnitude and opposite in sign.
 
 
 

Thinking Stepwise

Hess’s Law was the topic on the previous page of this Lesson. Because of Hess’s Law, we can think of reactions as occurring in steps and use heat of formation values to determine the enthalpy changes for a reaction. Let’s do some stepwise thinking about the combustion of methane. The balanced chemical equation is:
   
Let’s think about performing this reaction in steps, beginning with the reactants - CH₄(g)  +  2 O₂(g). The first step would be to decompose CH₄(g) into its elements - C(s) and H₂(g). This is the opposite of formation so the ∆H for this step is the opposite of the ∆H_f° of CH₄(g).  We can write …
   
Once this step is done, we would have the other reactant - 2 O₂(g) - plus the products of the CH₄ decomposition - C(s)  +  2 H₂(g).  We now have all elements in their natural state. Now we can start forming the products in the equation. Let’s start with the formation of 2 H₂O(l) from H₂(g) and O₂(g). Since we are making two moles of H₂O(g), the enthalpy change would be twice the heat of formation value of H₂O(l). We can write …
   
Now for the final step - the formation of 1 mole of CO₂(g) from the C(s) and the remaining O₂(g). The ∆H of this third step is the ∆H_f° of CO₂(g). We can write …
   
According to Hess’s Law, the enthalpy change for the entire process is the sum of the ∆H values of the individual steps. That is, …
   
Our 3-step approach to this reaction can be represented by the following schematic. We have arranged it roughly to scale on an enthalpy graph.
   
 
 

 

The ∆H Equation

The principle of Hess’s Law is that the enthalpy change of a multi-step process is equal to the sum of the enthalpy changes of the individual steps. We can think of any reaction as involving the following two steps.
 

• Reactant compounds are decomposed into their elements. The ∆H for this process is the opposite or the negative of the ∆H_f° values for those compounds.
• Products compounds are formed from their elements. The ∆H for this process is equal to the ∆H_f° values for those compounds.

 
Applying Hess’s Law, the enthalpy change of the reaction is the sum of all the ∆H_f° values for product compounds plus the negative of the ∆H_f° values of the reactant compounds. Put another way, we could say that …  
When using this equation, be sure to …

• Multiply ∆H_f° values by the coefficients in the balanced chemical equation. The ∆H_f° values are expressed on a per mole basis. A multiplier must be used if there is more than one mole listed in the balanced chemical equation.
• Reactants or products that are elements existing in their standard state can be disregarded. They have a ∆H_f° value of 0.
• Subtract the total or sum (∑) for the reactants from the total or sum (∑) of the products.
• Be very careful about + and - signs. Know that in some instances you will be subtracting negative enthalpy values. The equation is not a rocket science equation; but it is an equation that requires attention to details.

 
We will demonstrate the use of this equation in the following two examples.  
 
 

Example 1

Hydrazine (N₂H₄)  is used as a fuel in liquid-fueled rockets.  Dinitrogen tetroxide can be used as an oxidizing agent.  The balanced equation for the reaction is:
   
Use heat of formation values in the table above to determine the enthalpy change for this reaction.
 
 

Solution

We will use the equation
to determine the ∆H. Heat of formation values (∆H_f°) are given in the table above:
 
N₂H₄(l):  -50.6 kJ/mol                                   N₂O₄(l):  -19.5 kJ/mol          
N₂(g) :  0.0 kJ/mol (element)                       H₂O(l):  -285.8 kJ/mol
 
To avoid issues associated with + and - signs, we will calculate ∑∆H_f°_products and ∑∆H_f°_reactants individually.
 
∑∆H_f°_products = 3 mol•(0.0 kJ/mol)  +  4 mol•(-285.8 kJ/mol) = -1143.2 kJ
 
∑∆H_f°_reactants = 2 mol•(-50.6 kJ/mol)  +  1 mol•(-19.5 kJ/mol) = -120.7 kJ
 
Now we can calculate the ∆H.
   
 
 

Example 2

Solid fuel booster rockets sometimes rely on a mixture of aluminum and ammonium perchlorate as a rocket fuel.  The reaction is:
   
Use heat of formation values in the table above to determine the enthalpy change for this reaction.
 
 

Solution

We will use the equation
 
∆H_reaction =  ∑ ∆H_f°_products  -  ∑ ∆H_f°_reactants
 
to determine the ∆H. Heat of formation values (∆H_f°) are given in the table above:
 
3 Al(s):  0.0 kJ/mol (element)                      NH₄ClO₄(s):  -295.8 kJ/mol
Al₂O₃(s):  -1676 kJ/mol                                AlCl₃(s):  -704 kJ/mol
NO(g):  +91.3 kJ/mol                                    H₂O(l):  -285.8 kJ/mol
 
To avoid issues associated with + and - signs, we will calculate ∑∆H_f°_products and ∑∆H_f°_reactants individually. While this is not the only way to approach the problem, it is the safest way.
 
∑∆H_f°_products = 1 mol•(-1676 kJ/mol)  +  1 mol•(-704 kJ/mol) + 3 mol•(+91.3 kJ/mol)  +  6 mol•(-285.8 kJ/mol) = -3820.9 kJ
 
∑∆H_f°_reactants = 3 mol•(0.0 kJ/mol)  +  3 mol•(-295.8 kJ/mol) = -887.4 kJ
 
Now we can calculate the ∆H.
   
 

Hopefully the answer to the question How will I use these numbers in Chemistry class? is now clear. At this point, it is important that you put the ideas to practice by utilizing one or more of the suggestions in the Before You Leave section.
 
 
 

Before You Leave

• Download our Study Card on Heat of Formation. Save it to a safe location and use it as a review tool. (Coming Soon.)
• Our Heat of Formation Concept Builder provides a developmental approach to this topic. It emphasizes the stepwise approach used in this Tutorial.
• Our Calculator Pad section is the go-to location to practice solving problems. You’ll find plenty of practice problems on our Thermal Chemistry page. Here’s a good follow-up: Problem Set TC13: Heat of Formation and Enthalpy Change.
• The Check Your Understanding section below include questions with answers and explanations. It provides a great chance to self-assess your understanding.

 
 
 

 
 

Check Your Understanding

Use the following questions to assess your understanding. Tap the Check Answer buttons when ready.
 
1. The ∆H_f° value of hydrogen - H₂(g) - is not listed in the table of values. Yet it is a pretty common substance. What is the ∆H_f° value of H₂(g)?
 

 


2. The ∆H_f° value of AlCl₃(s) is -704 kJ/mol. Write the thermochemical equation for the formation of this AlCl₃(s) with the heat term in the equation.
 
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3. The ∆H_f° value of NO₂(g) is +33.8 kJ/mol. Write the thermochemical equation for the formation of this NO₂(g) with the heat term in the equation.
 
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4. Write the equations in Questions #2 and #3 as decomposition reactions. Include the heat term in the equation.
 
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5. Using a graphical schematic (like the one shown at the end of the Thinking Stepwise section), show the enthalpy change for the combustion of ethanol

C₂H₅OH(l)  +  3 O₂(g)  →  3 H₂O(l)  +  2 CO₂(g)

Use the ∆H_f° values provided in the table above.
 

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6. Consider the following balanced chemical equation for the reaction of fictional compounds A and B to produce fictional compounds C and D.

5 A + 2 B → 3 C + 6 D

The standard enthalpy of formation (ΔH_f°) for these compounds are:
A: +140.7 kJ/mol
B: -74.0 kJ/mol
C: -178.4 kJ/mol
D: -183.4 kJ/mol
 
Use these values to calculate the enthalpy change (∆H) for the given chemical reaction.
 

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7. Welding torches commonly use the combustion of acetylene gas (C₂H₂) to produce the heat required to cut through steel and other metals. The reaction is …

2 C₂H_2(g) + 5 O_2(g) → 4 CO_2(g) + 2 H₂O_(l)

Use this table of heat of formation values to determine the enthalpy change of this reaction. Enter a negative answer if appropriate.
 

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