Lesson 2: Chemical Reactions and Enthalpy Change
Part b: Hess's Law
Part a:
Enthalpy Change
Part b: Hess's Law
Part c:
Heat of Formation
Part d:
Bond Enthalpy and ∆H
Part e:
Thermal Stoichiometry
The Big Idea
Hess’s Law reveals that the energy change for a chemical reaction is independent of the path taken - whether the reaction happens in one step or many. By rearranging, reversing, and summing known thermochemical equations, we can predict the overall enthalpy change (ΔH) of a target reaction.
Three Rules of Thermochemical Equations
We introduced thermochemical equation in Lesson 2a. They are balanced chemical equations that are accompanied by information about the enthalpy change for the reaction as stated. Here’s two versions of thermochemical equations for the combustion of methane.
CH4(g) + 2 O2(g) → 2 H2O(l) + CO2(g) ∆H = -890 kJ
CH4(g) + 2 O2(g) → 2 H2O(l) + CO2(g) + 890 kJ
In the first version, the enthalpy change is listed as a ∆H value to the right of the equation. In the second version, the enthalpy change is added into the equation as a heat term. The heat term is a reactant for endothermic reactions and a product for exothermic reactions.
In this part of Lesson 2, we will learn three rules regarding thermochemical equations and learn how to use them to determine the enthalpy change of chemical reactions. The three rules are:
- The ∆H is directly proportional to the amount of reactants and products.
We have referred to enthalpy change as a stoichiometric quantity. The kJ of heat absorbed or released by a specific reaction is dependent on how much reactant or product is involved. It can be expressed as a ratio. For the methane combustion reaction (above), the ratio is -890 kJ/1 mol CH4 or -890 kJ/2 mol O2. If the coefficients of the balanced equation are doubled or halved, the ∆H value will double or halve.
- The ∆H value for a reaction is equal and opposite to the ∆H value for the reverse reaction.
The synthesis of liquid water from its elements is exothermic and has a ∆H value of -286 kJ/mol. The reverse process - the decomposition of water into its elements - is endothermic and has a ∆H of +286 kJ/mol. This is illustrated by the two thermochemical equations below.
H2(g) + ½ O2(g) → H2O(l) ∆H = -286 kJ
H2O(l) → H2(g) + ½ O2(g) ∆H = +286 kJ
The ∆H value for a reaction is independent of how many steps are taken between reactants and products.
This ↑↑↑↑↑. This is the topic of Lesson 2b. Understanding this third rule and being able to apply it with Rules 1 and 2 is the goal of Lesson 2b. This third rule suggests that enthalpy is a state function. That will take some explaining. Read on!
What Does "Enthalpy is a State Function" Mean?
Altitude is a property that we refer to as a
state function. While it has little to do with Chemistry, it is a useful property for explaining the idea of a state function. Suppose that you are in a very tall building with an elevator that has floors positioned 5-meters apart. You start on Ground level - an altitude of 0 meters. You ride the elevator up to the 10
th floor at an altitude of 50 meters. Here’s an easy question - by how much did your altitude change? Easy answer - 50 meters.
Now suppose you start at Ground level and ride the elevator to the 15
th floor, down to the 5
th floor, and finally back up to the 10
th floor. The ride ends. You’ve reached the same final destination in three steps instead of one step. What is your altitude change? It’s a bit more difficult question. But the answer is still 50 meters. In the initial state, you had an altitude of 0 meters. In the final state you had an altitude of 50 meters. The change in altitude is still 50 meters.
This illustrates the concept of a state function. A change in its value from the initial to the final state is not dependent upon the path. The altitude change only depends upon the two states - the altitude at Ground floor and the altitude on the 10
th floor. The change in altitude for a trip from the Ground floor to the 10
th floor is always 50 meters no matter how you get there. Altitude is a path-independent quantity. This makes it a state function.
Many chemical properties are what we refer to as state functions. Enthalpy is a state function. As we have discussed, reactants have a certain amount of enthalpy. Products have a different amount of enthalpy. If generic reactants A and B change into generic products C and D, then the enthalpy change is simply the difference between the enthalpy of products (the final state) and the enthalpy of reactants (the initial state).
∆H = Hproducts - Hreactants
And more importantly, because enthalpy is a state function, the enthalpy change will be the same no matter how the overall change occurs. The change could occur in one step, two steps, three steps, or more steps. But however it does occur, the change in enthalpy will not be affected by the pathway or number of steps.
Just as there are many ways to use an elevator and get from the Ground floor to the 10
th floor, there are many ways to get from reactants A + B to products C + D. And no matter how one gets from the intial state (reactants) to the final state (products), the enthalpy change will be the same because enthalpy is a state function.
Hess’s Law of Constant Heat Summation
Hess’s Law of Constant Heat Summation, more affectionately known as
Hess’s Law, states that the enthalpy change for a multi-step process is the sum of the enthalpy change of the individual steps. The diagram below illustrates this idea for the generic reaction of A + B → C + D:
Observe how the enthalpy change between the initial state (the reactants A + B) and the final state (the products C + D) is -400 kJ regardless of the number of steps. Also observe how the enthalpy change for the overall process is the sum of the enthalpy changes of the individuals steps in the process.
Adding Equations to Achieve the Target Equation
We’re getting close to being equipped to accomplish our goal - using the three rules of thermochemical equations to determine the enthalpy change of a reaction. There’s one more skill that you will need - the skill of adding reactions together to obtain an overall target reaction. The typical Hess’s law problem will provide two or three reactions with given ∆H values. The problem will require that you arrange and “add the equations together” to achieve the
Target Equation and to determine the ∆H of the
Target Equation. An example of such a problem is shown at the right.
The concept of adding chemical equations is a difficult one for many students. For those who have added algebraic equations in math with formulas including x, y, and z (or the like), it may help to think about that process since adding chemical equations works the same way. (And for those who have not done this in math or regret having done it in math, forget the previous sentence.)
When you add equations in chemistry, you are adding all the reactants together with their coefficients and all the products together with their coefficients. Consider Equations 1 and 2 below.
Equation 1: J + 3M → 2L + N
Equation 2: 3J + 2L → 2M + N
The sum of these two equations leads to Equation 3:
Equation 3: J + 3M + 3J + 2L → 2L + N + 2M + N
Note how both original equations had two reactants each and two products each. So, the sum of the two equations has four reactants and four products as expected.
Once the two (or more) equations are added, you will need to
cancel and group formulae. Formulae (plural for formula) will cancel if the same amount of that formula is present on opposite sides of the chemical equation (the
summed one). In the above example (Equation 3), the 2L will cancel since it is present on both the reactant and product side. Formulae will partially cancel if there are unequal amounts of that formula present on opposite sides of the chemical equation. For instance, there are 3M on the reactant side and 2M on the product side; this will result in partial cancellation. 2M can be cancelled from each side, leaving 1M on the reactant side. The equation becomes three formulas shorter.
Equation 3 (shortened): J + M + 3J → N + N
Grouping formulas means that if there are two instances of a formula on the same side of the equation, then they can be added together. For instance, there is a J and a 3J on the reactant side of the above equation. Those would add to 4J. And similarly, there is an N and a second N on the product side. Those would add to 2N. Implementing these two groupings would simplify the chemical equation to three terms:
Equation 3 (very shortened): M + 4J → 2N
And if you’re wondering, it doesn't matter the order in which you write these reactants as long as you keep them as reactants. Wrriting 4J + M means the same thing as writing M + 4J.
Developing the Skill of Adding Equations
In our J-L-M-N example, you were given the equations to add. In a typical Hess’s Law problem, you will be given equations but will have to decide how to manipulate them (reverse one, multiply another by two, etc.) in order for them to add up to the Target Equation. The task of arranging and adding the equations so as to achieve the Target Equation is the more difficult task. It is a type of task that is
better caught than taught. In other words, you will need to see several worked-out examples and try some on your own. You must be patient with the process of arranging and adding. There is no single way to solve this problem of adding two equations to arrive at the Target Equations. It is a process that requires trial and error. Commit to patiently working the process.
How to Solve a Hess's Law Problem with Examples
Let’s begin with our first example -
the problem listed above. As we proceed through the solution, we will depend upon these four steps. While the process will always involve trial-and-error, these steps will efficiently guide each iteration of the trial-and-error process.
Example 1 - Using Hess's Law to Determine ∆H
If given the thermochemical equations …
Equation 1: N
2(g) + O
2(g) → 2 NO
(g) ∆H = +182 kJ
Equation 2: N
2(g) + 2 O
2(g) → 2 NO
2(g) ∆H = +66 kJ
Determine the ∆H of the
Target Equation:
Target Eq’n: 2 NO
2(g) → 2 NO
(g) + O
2(g) ∆H = ??
Solution:
Step 1: Manipulate the Given Equations
We want to begin by manipulating and adding Equations 1 and 2 in order to achieve the Target Equation. Remember the first two rules of thermochemical equations - they involve multiplying an equation through by a coefficient and reversing and equation. This is how you will “manipulate” the equations before adding them.
Notice that the target Equation 2 has NO
2(g) on the reactant side. Only Equation 2 has NO
2 in it. But because NO
2 is on the product side of Equation 2, we will reverse Equation 2. When we do reverse it, we will have to change the sign on ∆H from + to -.
Equation 1: N
2(g) + O
2(g) → 2 NO
(g) ∆H = +182 kJ
Equation 2 (reversed): 2 NO
2(g) → N
2(g) + 2 O
2(g) ∆H = -66 kJ
Step 2: Add the Manipulated Versions of the Given Equations
Now we will add Equations 1 and 2 to get Equation 3 below:
Equation 3: N
2(g) + O
2(g) +
2 NO2(g) →
2 NO(g) + N
2(g) + 2 O
2(g)
Observe how Equation 3 has two of the formulas we need in our Target Equation -
2 NO2(g) on the reactant side and
2 NO(g) on the product side. That means we’re getting close to the Target Equation.
Step 3: Cancel and Group Formulae
Now we need to do some Cancelling and Grouping. There are the same number of N
2(g) on reactant and product side of Equation 3. They will cancel. Equation 3 becomes:
Equation 3 (shortened): O
2(g) + 2 NO
2(g) → 2 NO
(g) + 2 O
2(g)
And there is one O
2(g) on the reactant side and two O
2(g) on the product side. We can cancel one O
2(g) from each side, leading to a more shortened version:
Equation 3 (very shortened): 2 NO
2(g) → 2 NO
(g) + O
2(g)
Eureka! Equation 3 is now identical to the Target Equation.
Let’s review what was done:
First, we manipulated Equation 2, reversing it so as to get the 2 NO
2(g) formula on the reactant side. Second, we added the two equations together. Third, we did some Cancelling and Grouping and were fortunate enough to have produced the Target Equation as a result.
Step 4: Use Hess’s Law to Calculate the ∆H
Our fourth and final step can now be done - use Hess’s Law to calculate ∆H of the Target Equation.
We produced the Target Equation by adding Equation 1 (as is) and a reversed form of Equation 2. So, we will need to add the ∆H of Equation 1 to the ∆H of the reversed form of Equation 2.
∆HTarget Equation = ∆HEquation 1 + ∆H Equation 2 (reversed)
∆HTarget Equation = +182 kJ + (-66 kJ)
∆HTarget Equation = +116 kJ
Our four-step process, combined with Hess’s Law, has led to the following thermochemical equation:
2 NO2(g) → 2 NO(g) + O2(g) ∆H = +116 kJ
We have two more worked-out examples. We recommend that you attempt them on your own using these four steps as a guide to your trial-and-error solution. Then tap on the
View Answer button to check your answer or to view the complete solution.
Example 2 - Using Hess's Law to Determine ∆H
If given the thermochemical equations …
Equation 1: 2 NO
(g) + O
2(g) → 2 NO
2(g) ∆H = -116 kJ
Equation 2: N
2O
4(g) → 2 NO
2(g) ∆H = +55 kJ
Determine the ∆H of the
Target Equation:
Target Eq’n: 2 NO
2(g) + O
2(g) → N
2O
4(g) ∆H = ??
View Answer
Solution
Every trial-and-error iteration will begin with three steps - manipulate the given equations, add the equations, and then cancel and group formulae. If successful, then you can proceed with the fourth step.
Step 1: Manipulate the Given Equations
When you compare the two given equations to the Target Equation, you will notice that the Target Equation has N
2O
4(g) formula on the product side. It is on the reactant side of Equation 2. So, we will reverse Equation 2 and change its ∆H from + to -. This gives:
Equation 1: 2 NO
(g) + O
2(g) → 2 NO
2(g) ∆H = -116 kJ
Equation 2 (reversed): 2 NO
2(g) → N
2O
4(g) ∆H = -55 kJ
Step 2: Add the Manipulated Versions of the Given Equations
Now we will add Equation 1 and the reversed form of Equation 2 to yield Equation 3:
Equation 3: 2 NO
(g) + O
2(g) + 2 NO
2(g) → 2 NO
(g) + N
2O
4(g)
Step 3: Cancel and Group Formulae
The grouping and cancelling step can be used to shorten Equation 3. The 2 NO
(g) formula appears on both reactant and product side; it can be cancelled and Equation 3 can be shortened.
Equation 3 (shortened): O
2(g) + 2 NO
2(g) → N
2O
4(g)
Say it loud now! Eureka! We have produced the Target Equation. We did so by adding Equation 1 to the reversed form of Equation 2.
Step 4: Use Hess’s Law to Calculate the ∆H
Using Hess’s Law, we can determine the ∆H of the Target Equation by adding the ∆H of Equation 1 to the ∆H of the reversed form of Equation 2.
∆HTarget Equation = ∆HEquation 1 + ∆H Equation 2 (reversed)
∆HTarget Equation = -116 kJ + (-55 kJ)
∆HTarget Equation = -171 kJ
We can now write the thermochemical equation for the Target Equation:
2 NO2(g) + O2(g) → N2O4(g) ∆H = -171 kJ
Example 3 - Using Hess's Law to Determine ∆H
If given the thermochemical equations …
Equation 1: N
2(g) + 3 H
2(g) → 2 NH
3(g) ∆H = -92 kJ
Equation 2: 2 H
2(g) + O
2(g) → 2 H
2O
(l) ∆H = -572 kJ
Equation 3: N
2(g) + 2 O
2(g) → 2 NO
2(g) ∆H = +66 kJ
Determine the ∆H of the
target equation:
Target Eq’n: 2 NH
3(g) + 4 H
2O
(l) → 2 NO
2(g) + 7 H
2(g) ∆H = ??
View Answer
Solution
Once again, every trial-and-error iteration will begin with the three steps - manipulate the given equations, add the equations, and then cancel and group formulae. If successful, then you can proceed with the fourth step.
Step 1: Manipulate the Given Equations
When you compare the given equations to the Target Equation, there should be two noteworthy observations. First, the formula 2 NH
3(g) on the reactant side of the Target Equation is only present in Equation 1. And it is on the product side of Equation 1. So, Equation 1 will need to be reversed. Second the formula 4 H
2O
(l) on the reactant side of the Target Equation is present on the product side of Equation 2 as 2 H
2O
(l). So, Equation 2 will need to be reversed and
doubled. Multiplying all coefficients of Equation 2 by two will produce the 4 H
2O
(l). The ∆H values of both Equations 1 and 2 will need to be modified to be consistent with these two manipulations. Here is the result of reversing both Equations 1 and 2 and
doubling Equation 2:
Equation 1 (reversed): 2 NH
3(g) → N
2(g) + 3 H
2(g) ∆H = +92 kJ
Equation 2 (reversed, X2): 4 H
2O
(l) → 4 H
2(g) + 2 O
2(g) ∆H = +1144 kJ
Equation 3: N
2(g) + 2 O
2(g) → 2 NO
2(g) ∆H = +66 kJ
Step 2: Add the Manipulated Versions of the Given Equations
Now we will add all three equations to yield Equation 4:
Equation 4:
2 NH
3(g) + 4 H
2O
(l) + N
2(g) + 2 O
2(g) → N
2(g) + 3 H
2(g) + 4 H
2(g) + 2 O
2(g) + 2 NO
2(g)
Step 3: Cancel and Group Formulae
Now it is time for the Grouping and Cancelling step in order to shorten Equation 4. The N
2(g) formula appears on both reactant and product side; it can be cancelled. And the formula 2 O
2(g) appears on both sides; it too can be cancelled. The result is …
Equation 4 (shortened):
2 NH
3(g) + 4 H
2O
(l) → 3 H
2(g) + 4 H
2(g) + 2 NO
2(g)
Now we can group the 3 H
2(g) and the 4 H
2(g) on the product side to form 7 H
2(g). Equation 4 is shortened even more:
Equation 4 (very shortened):
2 NH
3(g) + 4 H
2O
(l) → 7 H
2(g) + 2 NO
2(g)
Now compare the latest version of Equation 4 to the Target Equation. Yell Eureka and tell yourself you’re a Chemistry Wizard! Equation 4 is the Target Equation. We formed the target equation by reversing Equations 1 and 2 and multiplying all terms in Equation 2 by two.
Step 4: Use Hess’s Law to Calculate the ∆H
We can use Hess’s Law to determine the ∆H of the Target Equation as …
∆HTarget Equation = ∆HEquation 1 (reversed) + ∆H Equation 2 (reversed, X2) + ∆H Equation 3
∆HTarget Equation = +92 kJ + 1144 kJ +66 kJ
∆HTarget Equation = 1302 kJ
The Target Equation can be written as the following thermochemical equation.
2 NH3(g) + 4 H2O(l) → 3 H2(g) + 4 H2(g) + 2 NO2(g) ∆H = +1302 kJ
Some Final Tips for Using Hess's Law
We will leave you with three tips for making smoother work of using Hess’s Law to solve problems. Then it is time for you to practice.
- Observe. Inspect. Gather Information.
Don't stall just because you don't know what to do. Instead, look at and compare the given equations and the target equation. Pay attention to details. Ask and answer ... What formula (if any) is present in only one of the given equations and shows up in the Target Equation? How do the coefficients in the given equation and the Target Equation compare for this formula? And what formulas are present in both of the given equations but doesn't show up in the Target Equations. For this formula, how do the coefficients in the given equations compare to each other? After you have gathered this information, you will be ready for the next two strategies.
- Look for a Formula Present in only One of the Given Equations
A formula that is present in only one of the given equations will show up in the Target Equation. This is an important starting point. If the formula is a reactant in the given equation but a product in the Target Equation, then you know for certain that the given equation will have to be reversed. And if the coefficients are different, then you also know that a multiplier will have to be used so that the coefficients are the same in both the manipulated version of the given equation and the Target Equation. Make the needed adjustments based on this observation. Often times doing so will be all you need to do before adding the given equations. If you need to do more then proceed to the next strategy ...
- Look for a Formula Present in Both of the Given Equations and Absent in the Target Equation
A formula that is present in both of the given equations but absent in the Target Equation is a formula that has cancelled. The only way that such a formula can cancel when added is if the formula is a reactant in one of the given equations and a product in the other given equation ... and if they have the same coefficient. This suggests two possible manipulations: First, one of the two given equations may have to be reversed so that the formula in focus is a reactant in one equation and a product in the other equation. And second, one of the equations may need a multiplier in order for the coefficient in front of the formula to be the same coefficient in both given equations. Make the needed adjustments based on this observation. Often times doing so will be all you need to do before adding the given equations.
Before You Leave - Practice and Reinforcement
Now that you've done the reading, take some time to strengthen your understanding and to put the ideas into practice. Here's some suggestions.
- Our Hess’s Law Concept Builder may be the best practice ever for manipulating and adding the given equations to form the Target Equation. You have to at least try it.
- 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. Try Problem Set TC11: Hess’s Law for more practice.
- The Check Your Understanding section below include questions with answers and explanations. It provides a great chance to self-assess your understanding.
- Download our Study Card on Hess’s Law. Save it to a safe location and use it as a review tool.
Check Your Understanding of Hess's Law
Use the following questions to practice the skill of using Hess's Law to determine an enthalpy change for a reaction. Tap the Check Answer buttons when ready.
1. Use the first two rules of thermochemical equations to answer the following questions:
- Given: N2(g) + 3 H2(g) → 2 NH3(g) ∆H = -92 kJ,
what is the ∆H of 3 N2(g) + 9 H2(g) → 6 NH3(g)?
Check Answer
Answer: -276 kJ
All coefficients in the given equation are multiplied by 3. So the ∆H of the given equation must be multiplied by 3.
- Given: N2(g) + 3 H2(g) → 2 NH3(g) ∆H = -92 kJ,
what is the ∆H of 2 NH3(g) → N2(g) + 3 H2(g)?
Check Answer
Answer: +92 kJ
The given equation has been reversed so the sign on ∆H must be changed from - to +.
- Given: N2(g) + 3 H2(g) → 2 NH3(g) ∆H = -92 kJ,
what is the ∆H of 4 NH3(g) → 2 N2(g) + 6 H2(g)?
Check Answer
Answer: +184 kJ
Two changes have been made to the given equation. First it has been reversed. So, the sign on ∆H must be changed from - to +. Second, all coefficients have been doubled. So, the value of ∆H must be doubled from the 92 kJ to 184 kJ.
- Given: N2(g) + 3 H2(g) → 2 NH3(g) ∆H = -92 kJ,
what is the ∆H of 10 NH3(g) → 5 N2(g) + 15 H2(g)?
Check Answer
Answer: +460 kJ
Two changes have been made to the given equation. First it has been reversed. So, the sign on ∆H must be changed from - to +. Second, all coefficients have been multiplied by 5. So, the value of ∆H must be multiplied by 5, changing the 92 kJ to 460 kJ.
2. Which of the following statements are true of a quantity that is a state function? Select all that are true.
- The value of the quantity is always the same.
- A change in the quantity between two different states will not depend upon how many steps were taken between those states.
- The value of the quantity depends upon the properties of a particular state and not on how the state was reached.
- The quantity is a path-dependent quantity.
- A change in the quantity will depend upon how many steps are taken from the initial value to the final value.
Check Answer
Answer: B and C are TRUE
Statement A is incorrect: the value of a state quantity can definitely change from one state to another state.
Statement D is incorrect: state quantities are path-independent quantities.
Statement E is incorrect: the change in the quantity will be the same regardless of whether it happens in a single step or in multiple steps.
3. Consider:
A + B → C ∆H = -800 kJ
The proccess is carried out in two steps as shown with D being an
intermediate.
Step 1: A → D ∆H = +200 kJ
Step 2: B + D → C ∆H = ???? kJ
What is ∆H of step 2?
Check Answer
Answer: -1000 kJ
∆Hoverall = ∆H1 + ∆H2
-800 kJ = +200 kJ + ∆H2
∆H2 must be -1000 kJ in order for the above statement to be true.
4. Consider the enthalpy diagram at the right for E changing to A, directly and by means of four individual steps. Given that ...
ΔH
1 = 50 kJ,
ΔH
2 = 75 kJ,
ΔH
4 = 150 kJ, and
ΔH
5 = 375 kJ,
... determine the value of ΔH
3?
Check Answer
Answer: 100 kJ
∆H5 is the enthalpy change for the direct, 1-step change from E to A.
∆H1 + ∆H2 + ∆H3 + ∆H4 is the sum of the four steps leading from E to A in step-wise fashion. And so,
∆H5 = ∆H1 + ∆H2 + ∆H3 + ∆H4
The numbers can be inserted into the equation.
375 kJ = 50 kJ + 75 kJ + ∆H3 + 150 kJ
The right side of this equation must add to 375 kJ. There are three known values (50 kJ, 75 kJ, and 150 kJ) and one unknown value (∆H3) on the right side. The three known values on the right side - 50 kJ + 75 kJ + 150 kJ - add to 275 kJ. So the unkown ∆H3 must be 100 kJ to make up the difference between the 375 kJ (left side) and the 275 kJ (known values on right side).
5. On Planet Exwizee, where the main elements are X, Y, and Z, it is known that compounds of elements X, Y, and Z will react in the following ways:
Equation 1: Y + X
2Z → XY + XZ
Equation 2: 2 X
2Z → X
2 + 2 XZ
The enthalpy changes for these two reactions are ...
∆H
1 = +208 kJ
∆H
2 = -173 kJ.
Use this information to determine the enthalpy change (∆H) of the Target Equation …
X
2 + 2 Y → 2 XY
Check Answer
Answer: +589 kJ
Each given reaction has the formula X2Z in it. But this formula does not show up in the Target Equation. So we know that X2Z must get cancelled through the addition process. This provides HUGE clues on how to manipulate the two given equations before adding them. We can also observe that Y is only listed in Equation 1 and it is also in the Target Equation. In the Target Equation, there is a coefficient of 2 in front of it, but there is a coefficient of 1 in front of it in Equation 1. Making these observations provides the clues we need to know how to manipulate the given equations before adding them.
Step 1: Manipulate the Given Equations
Let's multiply Equation 1 by 2. And let's reverse Equation 2. This gives:
Equation 1 (multiplied by 2): 2 Y + 2 X2Z → 2 XY + 2 XZ ∆H = +416 kJ
Equation 2 (Reversed): X2 + 2 XZ → 2 X2Z ∆H = +173 kJ
Step 2: Add the Manipulated Versions of the Given Equations
Now add the two manipulated forms of the equations to yield Equation 3. We will hope Equation 3 turns out to be the Target Equation.
Equation 3: 2 Y + 2 X2Z + X2 + 2 XZ → 2 XY + 2 XZ + 2 X2Z
Step 3: Cancel and Group Formulae
The formula 2 X2Z is present on both sides. These can be cancelled.
Equation 3 (shortened): 2 Y + X2 + 2 XZ → 2 XY + 2 XZ
The formula 2 XZ is present on both sides. These also can be cancelled.
Equation 3 (very shortened): 2 Y + X2 → 2 XY
There it is! Equation 3 now looks identical to the Target Equation. It's time to finish up by using Hess's Law to find the ∆H value.
Step 4: Use Hess’s Law to Calculate the ∆H
We doubled Equation 1; it's new ∆H is +416 kJ. And we reversed Equation 2; it's new ∆H becomes +173 kJ. Because these two equations were added, we must add their ∆H values. That's Hess's Law!
∆HTarget Equation = +416 kJ + (+173 kJ)
∆HTarget Equation = +589 kJ
6. Consider these balanced chemical equations and their corresponding enthalpy change values:
Equation 1: P
4(s) + 10 Cl
2(g) → 4 PCl
5(g) ∆H = -7736 kJ
Equation 2: PCl
3(g) + Cl
2(g) → PCl
5(g) ∆H = -562 kJ
Use this information and Hess's law to determine the enthalpy change for this Target Equation:
P
4(s) + 6 Cl
2(g) → 4 PCl
3(g)
Check Answer
Answer: -5488 kJ
Always begin by making observations. PCl5 is in both given equations but it is not in the Target Equation. So we know that PCl5 must get cancelled through the addition process. There are 4PCl5 in Equation 1 but only 1 PCl5 in Equation 2. So Equation 2 must be multiplied by 4. And these formulae are on the product side of both given equations. Thus, one of the equations will have to be reversed. If Equation 2 is reversed, then that places the formula PCl3 of Equation 2 on the product side. That is where it needs to be since PCl3 is on the product side of the Target Equation.
Step 1: Manipulate the Given Equations
Let's multiply Equation 2 by 4 and let's reverse Equation 2. That gives ...
Equation 1: P4(s) + 10 Cl2(g) → 4 PCl5(g) ∆H = -7736 kJ
Equation 2 (reversed and x4): 4 PCl5(g) → 4 PCl3(g) + 4 Cl2(g) ∆H = +2248 kJ
Step 2: Add the Manipulated Versions of the Given Equations
Now add the two equations to yield Equation 3. We will hope Equation 3 turns out to be the Target Equation.
Equation 3: P4(s) + 10 Cl2(g) + 4 PCl5(g) → 4 PCl5(g) + 4 PCl3(g) + 4 Cl2(g)
Step 3: Cancel and Group Formulae
The formula 4 PCl5(g) is present on both sides. These can be cancelled.
Equation 3 (shortened): P4(s) + 10 Cl2(g) → 4 PCl5(g) + 4 Cl2(g)
There are 4 Cl2(g) on the product side that will partially cancel the 10 Cl2(g) on the reactant side. Doing so results in ...
Equation 3 (shortened): P4(s) + 6 Cl2(g) → 4 PCl5(g)
Done! We have produced the Target Equation. It is time to apply Hess's Law and determine its ∆H value.
Step 4: Use Hess’s Law to Calculate the ∆H
We reversed Equation 2 and multiplied through by 4; it's new ∆H becomes +2248 kJ. Because thee two equations were added, we must add their ∆H values. That's Hess's Law!
∆HTarget Equation = -7736 kJ + (+2248 kJ)
∆HTarget Equation = -5488 kJ