Lesson Explainer: Hess’s Law Chemistry

In this explainer, we will learn how to define and use Hess’s law to calculate enthalpy changes.

For some chemical reactions, it can be difficult to find the enthalpy change for the reaction directly. In a laboratory, it may be difficult to control the reaction that is taking place. The reaction might occur very slowly and be too dangerous to carry out safely in an experiment. Alternatively, it may be difficult to take measurements from the reaction mixture using conventional calorimetry techniques.

As an example, we could try to find the enthalpy change for the reaction of carbon with oxygen to form carbon monoxide gas: C()+O()CO()s,graphitegg122

Firstly, it is very difficult for carbon, in the form of graphite, to react with oxygen. Graphite needs to be heated to a very high temperature since the reaction has a very high activation energy. Graphite does not spontaneously react with oxygen gas at room temperature. Under standard conditions, it would be very difficult to take any meaningful measurements for this reaction in a laboratory. The second problem that would be encountered is that it is almost impossible to ensure that carbon monoxide is the only product formed in this reaction. It is highly likely that carbon dioxide would also be produced.

Fortunately, enthalpy changes can be calculated by alternative means, using enthalpy data from other, separate, related reactions. The separate processes can be arranged so that their sum represents an alternative pathway for the desired reaction. We could view the alternative pathway as an indirect route to form the desired products. The outcome of the sequence of reactions would be the same as if the reaction had taken place in a single step. An example of this concept is shown in the following diagram.

In this diagram, reaction 1 represents the route for the desired reaction that is difficult to directly measure. The direct route can be achieved indirectly by moving through reaction 2 and then following the reverse of reaction 3.

Another way of viewing this situation is that reaction 2, which represents the formation of one mole of carbon dioxide directly from graphite and oxygen, can be achieved by following reaction 1 and then reaction 3. By following reaction 1 first and then reaction 3, we progress to the formation of one mole of carbon dioxide via one mole of carbon monoxide. In either case, the initial reactants and the final products are identical. They are also in the same states and in the same molar quantities.

We can then attach enthalpy change values to these processes, as in the diagram below.

Energetically, reaction 2 is the sum of reaction 1 and reaction 3: reactionreactionreaction2=1+3.

Reaction 1 is unknown and we shall denote it as π‘₯. By substituting in the values for each of the reactions, we can see that βˆ’393/=π‘₯+(βˆ’283/).kJmolkJmol

So, π‘₯=βˆ’393/+283/π‘₯=βˆ’110/.kJmolkJmolkJmol

We have therefore determined the enthalpy change of the reaction 1 to be βˆ’110 kJ/mol.

In completing this energy cycle, we have used the concept of Hess’s law.

Definition: Hess’s Law

Hess’s law states that the enthalpy change for a reaction is independent of the path taken.

Hess’s law is also known as Hess’s law of constant heat summation, and it is an example of the law of conservation of energy. Hess’s law will always apply to an energy cycle, provided that the initial and final conditions are the same for each pathway taken.

In a similar way, it is difficult to measure the enthalpy change for the decomposition of calcium carbonate into calcium oxide and carbon dioxide directly. The reaction takes place at a high temperature and it would be hard to establish if the reaction were complete: CaCO()CaO()+CO()32ssg

We can use the idea that both calcium carbonate and calcium oxide react with aqueous hydrochloric acid to create an alternative reaction cycle. Both substances are bases, producing calcium chloride and water in separate neutralization reactions.

The cycle could be established as in the diagram below.

Ξ”π»οŠ§ represents the enthalpy change for the decomposition reaction of calcium carbonate into calcium oxide and carbon dioxide.

Ξ”π»οŠ¨ represents the enthalpy change for the reaction of calcium carbonate with excess hydrochloric acid.

Ξ”π»οŠ© represents the enthalpy change for the reaction of calcium oxide with excess hydrochloric acid.

Both Ξ”π»οŠ¨ and Ξ”π»οŠ© can be established quite easily by using simple calorimetry apparatus. We can see from the cycle created above that Δ𝐻=Δ𝐻+Ξ”π»οŠ¨οŠ§οŠ©. Rearranging this equation to make Ξ”π»οŠ§ the subject gives Δ𝐻=Ξ”π»βˆ’Ξ”π»οŠ§οŠ¨οŠ©.

During the cycle, we follow the direction of the arrow for Ξ”π»οŠ¨ but go in the opposite direction of the arrow for Ξ”π»οŠ©. As a result, the sign for the value Ξ”π»οŠ© changes and becomes βˆ’Ξ”π»οŠ©.

In fact, since energy can be neither created nor destroyed, if we were to make a journey around this cycle in either direction and arrive back at our starting point, all of the enthalpy values would sum to zero. This is a useful approach that can be taken for solving problems involving energy cycles.

In the diagram above, the cycle is completed in a clockwise direction. Ξ”π»οŠ§ and Ξ”π»οŠ© are completed first by following the respective arrows. These values are added together, retaining their original signs. Ξ”π»οŠ¨ is completed in the reverse direction, so the enthalpy value for this step has a change of sign. Since we arrive back at the start, all the values must sum to zero since we cannot create nor destroy heat energy in this thermodynamic cycle.

Hence, we have Δ𝐻+Ξ”π»βˆ’Ξ”π»=0.

Example 1: Using Hess’s Law to Find an Equivalent Enthalpy Change

Which of the following values is equivalent to Ξ”π»οŠ©?

  1. βˆ’Ξ”π»+Δ𝐻+Ξ”π»οŠ§οŠ¨οŠͺ
  2. βˆ’Ξ”π»βˆ’Ξ”π»βˆ’Ξ”π»οŠ§οŠ¨οŠͺ
  3. Ξ”π»βˆ’Ξ”π»βˆ’Ξ”π»οŠ§οŠ¨οŠͺ
  4. Δ𝐻+Ξ”π»βˆ’Ξ”π»οŠ§οŠ¨οŠͺ
  5. Δ𝐻+Δ𝐻+Ξ”π»οŠ§οŠ¨οŠͺ

Answer

Ξ”π»οŠ© in the diagram above represents the enthalpy change for the direct reaction that converts E into C. In this reaction, E represents the reactants and C represents the products.

We can find an alternative pathway from E to C, moving indirectly around the energy cycle in a clockwise direction. This involves three separate reactions, taking place one after the other.

Firstly, we would move from E to FG+. This is the reverse of Δ𝐻οŠͺ. Energetically, this step would equate to βˆ’Ξ”π»οŠͺ.

The second step in the alternative route would be the reaction from FG+ to AB+. This is the reverse of Ξ”π»οŠ¨, so energetically this step equates to βˆ’Ξ”π»οŠ¨.

Finally, we would move from AB+ to C. This step completes the alternative enthalpy cycle. Energetically, this equates to Ξ”π»οŠ§.

Since Ξ”π»οŠ© represents the sum of the three alternative steps, we have Δ𝐻=(βˆ’Ξ”π»)+(βˆ’Ξ”π»)+Δ𝐻.οŠͺ

These quantities may be added together in any order, provided the signs are observed. Rearranging and retaining the original signs, we have Δ𝐻=Ξ”π»βˆ’Ξ”π»βˆ’Ξ”π».οŠͺ

The correct answer is therefore answer C.

Example 2: Applying Hess’s Law to an Energy Cycle

Considering the diagram, which of the following equations would Hess’s law predict to be true?

  1. Δ𝐻=Ξ”π»βˆ’Ξ”π»οŠ§οŠ¨οŠ©
  2. Δ𝐻=Ξ”π»βˆ’Ξ”π»οŠ¨οŠ©οŠ§
  3. Δ𝐻=Δ𝐻+Ξ”π»οŠ§οŠ¨οŠ©
  4. Δ𝐻=Δ𝐻+Ξ”π»οŠ¨οŠ©οŠ§
  5. Δ𝐻=Δ𝐻+Ξ”π»οŠ©οŠ§οŠ¨

Answer

In the reaction cycle seen above, the enthalpy change for the conversion of AB+ to C is equivalent to Ξ”π»οŠ§. This reaction can be achieved indirectly by converting AB+ to E in the first instance and then by converting E into C. Since these two steps are equivalent to Ξ”π»οŠ¨ and Ξ”π»οŠ©, respectively, Ξ”π»οŠ§ is the sum of Ξ”π»οŠ¨ and Ξ”π»οŠ©.

We have Δ𝐻=Δ𝐻+Ξ”π»οŠ§οŠ¨οŠ©.

Rearranging this equation, we have Δ𝐻=Ξ”π»βˆ’Ξ”π»οŠ¨οŠ§οŠ© and Δ𝐻=Ξ”π»βˆ’Ξ”π»οŠ©οŠ§οŠ¨.

The only response that indicates a correct calculation for either Ξ”π»οŠ§, Ξ”π»οŠ¨, or Ξ”π»οŠ© is C.

Enthalpy changes can be measured for very specific types of reactions. When a combustion reaction takes place, oxidation products are formed. If a hydrocarbon fuel is burned completely in air, containing excess oxygen, the products will consist of carbon dioxide and water. An example of this is when methane burns in oxygen: CH()+2O()CO()+2HO()4222gggl

To measure the enthalpy change for this reaction, the reaction would be conducted under standard conditions. We would measure the heat energy transferred in the reaction under constant pressure. Chemists determined that the standard pressure for chemical reactions should be 1 atm. A pressure of 1 bar is sometimes used instead because 1 bar is essentially equivalent to 1 atm (1=1.01325atmbar).

A stated temperature of 25∘C (298 K) is used when quoting enthalpy values under standard conditions. It is important that the reactants and products are in their standard states under these conditions. For example, water is in liquid form at 25∘C and 1 atm. Any water vapor produced in a combustion reaction must be condensed to liquid water when enthalpy change measurements are made.

The enthalpy change for a combustion reaction, when measured under standard conditions, is called the standard enthalpy change of combustion.

Definition: Standard Enthalpy of Combustion (Δ𝐻c⦡)

The standard enthalpy of combustion is the enthalpy change when one mole of substance burns completely in oxygen under standard conditions and standard states.

Combustion reactions are exothermic. The combustion of methane to produce carbon dioxide and water releases heat to the surroundings.

Another example of a specific type of enthalpy change is when 1 mole of a compound is formed from its elements. We could form one mole of methane gas as in the reaction below: C()+2H()CH()sgg24

If the elements are in their standard states and standard conditions apply, the measured enthalpy change for this reaction would be known as the standard enthalpy change of formation. It must be noted that the enthalpy change of formation for an element in its standard state is zero.

Definition: Standard Enthalpy of Formation (Δ𝐻f⦡)

The standard enthalpy of formation is the enthalpy change when one mole of substance is formed from its constituent elements in their standard states and under standard conditions.

It would be difficult to measure the standard enthalpy of formation for 1 mole of methane directly, since carbon and hydrogen do not react simply with each other to form methane under standard conditions. We can construct an energy cycle using standard enthalpies of combustion to find the standard enthalpy of formation for methane. This energy cycle is seen below.

In reaction 1, one mole of methane is formed from its elements in their standard states under standard conditions. This is the enthalpy of formation ο€ΊΞ”π»ο†βŠ–f for methane.

Reaction 2 shows the combustion of one mole of carbon and the combustion of two moles of hydrogen gas. The complete combustion products, one mole of carbon dioxide and two moles of water, are seen written below. Energetically, reaction 2 would equate to the sum of these combustion enthalpies: Δ𝐻(2)=ο€Ί1×Δ𝐻+ο€Ί2×Δ𝐻.βŠ–βŠ–βŠ–reactionC()H()cc2sg

In reaction 3, we see the combustion of one mole of methane. This gives us the same products as reaction 2, so we have closed the energy cycle. Energetically, reaction 3 is equal to the enthalpy of combustion of methane: Δ𝐻(3)=1×Δ𝐻.βŠ–βŠ–reactionCH()c4g

By applying Hess’s law, we can see that the enthalpy change for reaction 1 is equal to the enthalpy change for reaction 2 minus the enthalpy change for reaction 3: Δ𝐻(1)=Δ𝐻(2)βˆ’Ξ”π»(3)Δ𝐻(1)=1×Δ𝐻+2Γ—Ξ”π»οžβˆ’ο’1Γ—Ξ”π»οž.βŠ–βŠ–βŠ–βŠ–βŠ–βŠ–βŠ–ffcc2c4reactionreactionreactionreactionC()H()CH()sgg

In this example, we have used enthalpies of combustion to find an enthalpy of formation. With the correct energy cycle, it is possible to use enthalpies of formation to find the enthalpy change for a given reaction.

Example 3: Using Standard Enthalpies of Formation to Find Δ𝐻⦡ in an Energy Cycle

Using the diagram and the standard enthalpies of formation in the given table, what is the standard enthalpy of reaction ο€ΊΞ”π»ο†βŠ–οŽ for the decomposition of phosphorus pentachloride?

ChemicalPCl()5gPCl()3g
Standard Enthalpy of Formation (kJβ‹…molβˆ’1)βˆ’399βˆ’306

Answer

In this question, we are required to calculate the enthalpy change for the conversion of one mole of PCl5 into one mole of PCl3 and one mole of Cl2. This can be found indirectly by using the enthalpies of formation provided and creating a Hess’s cycle. Let’s label each reaction in the diagram, as shown below.

At the bottom of the energy cycle, we see the elements that are required to form the reactants and the products.

In reaction 2, one mole of PCl5 is formed from its constituent atoms. The table shows that the enthalpy change for this reaction is βˆ’399 kJβ‹…molβˆ’1.

Step 3 corresponds to the sum of the standard enthalpies of formation for PCl3 and Cl2 from their constituent atoms. However, the enthalpy change for the formation of chlorine is 0 kJ/mol, because all elements in their standard states have enthalpies of formation equal to zero. The enthalpy change for this step is therefore (βˆ’306/+0/)kJmolkJmol, which is equal to βˆ’306 kJ/mol.

To find the enthalpy change of reaction 1, we can apply Hess’s law and state that Δ𝐻=Δ𝐻(1)=βˆ’Ξ”π»(2)+Δ𝐻(3).βŠ–οŽreactionreactionreaction

We must first change the sign of the enthalpy of formation for PCl5, since we are following this reaction in the opposite direction. Substituting in the values for the enthalpies of reactions 2 and 3 gives Δ𝐻=+399+(βˆ’306+0)Δ𝐻=(+399βˆ’306).βŠ–οŽβŠ–οŽ

The correct answer is Δ𝐻=+93β‹…βŠ–οŽοŠ±οŠ§kJmol.

Example 4: Using Enthalpies of Formation to Calculate the Enthalpy of Combustion

When calculating the enthalpy of combustion of ethanol using Hess’s law, what value is used for Z?

  1. +3Γ— the enthalpy of formation of water
  2. +3Γ— the enthalpy of combustion of hydrogen
  3. +1Γ— the enthalpy of formation of water
  4. βˆ’3Γ— the enthalpy of combustion of hydrogen
  5. βˆ’3Γ— the enthalpy of formation of water

Answer

To calculate the enthalpy of combustion of ethanol, we can use the enthalpies of formation of the reactants and products to create an energy cycle. The reactants and products are formed from their respective elements, as seen at the bottom of the cycle diagram.

To complete the energy cycle, we must know the enthalpies of formation for 1 mole of ethanol, 3 moles of oxygen gas, 2 moles of carbon dioxide gas, and 3 moles of gaseous water. The water, as presented in the equation here, is not in its standard state and we must use the enthalpy of formation for gaseous water. Since there are 3 moles of gaseous water formed, we must multiply the enthalpy of formation by 3.

If the enthalpy change for reaction 1 is to be determined, we can equate this to βˆ’(Δ𝐻2)+(Δ𝐻3).reactionreaction

The sign for Δ𝐻 reaction 3 is not changed as we are moving in the direction of the reaction processes.

Since reaction Z is part of the enthalpy change for reaction 3, we would use +3Γ— the enthalpy of formation of gaseous water.

Response A is therefore the correct answer.

Enthalpies of reaction can also be calculated using mean bond enthalpies. The energy required to break a specific bond will depend upon the location of the bond within a certain molecule. For example, the strength of a CC bond will vary in different molecules. The CC bond strength in ethane will be slightly different from the CC bond strength in ethanol. Mean bond enthalpies are calculated by taking the average bond strength from a range of different molecules. They are displayed in data tables.

Calculations for enthalpy changes based upon mean bond enthalpies may differ slightly from calculations for similar reactions based upon enthalpies of formation, using a Hess’s cycle. Enthalpies of formation and combustion are very specific to the compounds concerned and the data is precise.

Let’s summarize how we define and use Hess’s law to calculate enthalpy changes.

Key Points

  • An energy cycle can be used to find the enthalpy change for a reaction that may be difficult to determine experimentally or directly.
  • Standard enthalpy changes are measured under standard conditions. The βŠ– symbol is used to denote that standard conditions apply. These are 298 K and 100 atm.
  • Hess’s law states that the enthalpy change for a reaction is independent of the route taken. It is an application of the law of conservation of energy.
  • In an energy cycle, if a process occurs in the opposite direction to the stated arrow, the sign of the enthalpy change must be changed.
  • In a complete energy cycle, the sum of all the enthalpy changes must be zero.
  • The standard enthalpy of formation ο€ΊΞ”π»ο†βŠ–f is the enthalpy change when one mole of substance is formed from its constituent elements in their standard states under standard conditions. By definition, it is zero for all elements.
  • The standard enthalpy of combustion ο€ΊΞ”π»ο†βŠ–c is the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions and standard states.
  • Mean bond enthalpies can be used to calculate an enthalpy change for a reaction, but they are not as precise when compared to enthalpies of formation for specified compounds.

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