DOR2

MDQ2 How much energy does it take to break bonds, and how much is released when bonds are formed?

TEXT REFERENCE: Chapter 15

2.1 explain the enthalpy changes in a reaction in terms of breaking and reforming bonds, and relate this to the law of conservation of energy.

In any chemical reaction bonds within the reactants must be broken and new bonds formed to create the products. Energy is required to break chemical bonds and energy is released when bonds are formed. Hence a chemical reaction is a transfer of energy.

Heat is often required to initiate a chemical reaction. Particles are always in motion and reactant particles may collide with one another. However, if they do not have sufficient energy, they will merely bounce off one another. Adding heat increases the kinetic energy of the particles. This means they move more quickly and have greater momentum when they collide. When colliding particles have sufficient energy they may interact to form a new substance.

Energy is required to break chemical bonds and energy is released when new bonds are created.

We can calculate the net gain or loss of energy in a chemical reaction by subtracting the energy produced when the bonds are formed in the product molecules from the energy required to break the bonds in reactant molecules.

Or: ΔE = ΣE_{bonds broken} – ΣE_{bonds formed}

Different elements have different bond strengths both between their own atoms and when they form bonds with other atoms. We can rank these to describe the strength of a chemical bond. We can also use them to determine the net gain or loss of energy when new substances are formed. NB Don’t forget this energy is exchanged between the system and the surroundings according to the Law of Conservation of Energy.

When metallic or ionic bonds are involved in chemical reactions, we are not breaking bonds in the same way as covalent substances. For these reactions, we evaluate the energy required to dissociate one mole of the substance into gaseous ions. In this way, scientists have determined that Na-O bonds are stronger than Na-S or Na-Cl bonds. Large amounts of energy are released when metal and non-metal ions react to form ionic bonds.

Review Tx1 p478 Latent heat of fusion of ice

VIEW videos:

Breaking Bonds [1.17] https://www.youtube.com/watch?v=dvJaBUxaYuk
DoR#6 Bond breaking and making [6.24] https://www.youtube.com/watch?v=NVQZ-RyBhbo&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=2

TASK 2.1.1

1. The following table shows various bond energies (the energy required or produced when the bond is broken or formed). Consider the combustion of one mole of methane: CH_4(g) + 2O_2(g) ---- CO_2(g) + 2H₂O_(g)

This equation can also be shown as:

a) Calculate the total energy required to break the bonds in the reactants.

b) Calculate the total energy released when the products are formed.

c) Using your answer to a) and b), calculate the change in energy for this reaction.

TASK 2.1.2

Complete

Tx2 p170 WS4.8 Literacy review—comparing key terms

Tx2 p164 WS4.4 Breaking and forming bonds

TASK 2.1.3

Complete Txt1

p482 15.1 Review, Key Qs
p485 15.2 Review, Key Qs

2.2 investigate Hess’ Law in quantifying the enthalpy change for a stepped reaction using standard enthalpy change data and bond energy data, eg: carbon reacting with oxygen to form carbon dioxide via carbon monoxide

Not all chemical reactions are simple processes. Some products can be reached by more than one chemical path or step. This is particularly so for biochemical processes such as photosynthesis and respiration. The Law of Conservation of Energy has an important application in the form of Hess’ Law. Hess’ Law states:

If a reaction is carried out in a series of steps, ΔH for the overall reaction will equal the sum of the enthalpy changes for the individual steps. (Brown, et al., 2010, p. 143)

Another way to say this is:

In going from particular reactants to particular products, the total enthalpy change for the reaction is independent of the path of reaction, ie it is the same whether the reaction occurs in one step or multiple steps.

VIEW videos:

DoR#7 Hess’ Law [6.28] https://www.youtube.com/watch?v=9TIGS3t1QJM&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=1

TASK 2.2.1

1. Explain the diagram below in terms of Hess’s Law. (Hegarty, 2018, p. 62)

Task 2.2.2

Draw a Hess' Law cycle to represent the pathways in the diagram below.

Task 2.2.3

Complete from Tx1:

p482 15.1 Review, Key Qs
p485 15.2 Review, Key Qs

2.3 apply Hess’s Law to simple energy cycles and solve problems to quantify enthalpy changes within reactions, including:

a. heat of combustion

Hess’ Law can be used to calculate the enthalpy change for a reaction from known ΔH values for other reactions. This can be particularly useful if the ΔH value of the desired reaction cannot be measured directly. Remember to apply these rules: (Smith & Davis, 2017, p. 346)

1. If a reaction is reversed, the sign of the ΔH value must be changed.

2. If a reaction is doubled, multiplied or divided by any number, then the ΔH value must likewise be doubled, multiplied or divided by the same number.

View videos:

DoR#8 Applying Hess’ Law [6.28] https://www.youtube.com/watch?v=As6JobWhYoE&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=10

MORE WORKED EXAMPLES

WORK ALONG WITH THESE

TASK 2.3.1

1. Calculate ∆H for the following reaction:

NiO_(s) + 2HCl_(g) → NiCl_2(s) + H₂O_(g)

Use the following enthalpy changes:

Reaction ∆H (kJ.mol^-1)

SOCl_2(g) + NiO_(s) → SO_2(g) +NiCl_2(s)-150

SOCl_2(g) + H₂O_(g) → SO_2(g) +2HCl_(g)-27

2. Calculate ∆H for the hydrogenation of ethene to ethane:

C₂H_4(g) + H_2(g) → C₂H_6(g)

Use the following enthalpy changes:

Reaction ∆H (kJ.mol^-1)

2H_2(g) + O_2(g) → 2H₂O_(l)-571.6

C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(l)-1411

2C₂H_6(g) + 7O_2(g) → 4CO_2(g) + 6H₂O_(l)-3120

3. Calculate ∆H for the formation of nitrous acid:

HCl_(g) + NaNO_2(s) → HNO_2(l) + NaCl_(s)

Use the following enthalpy changes:

Reaction ∆H (kJ.mol^-1)

2NaCl_(s) + H₂O_(l) → 2HCl_(g) + Na₂O_(s)507

NO_(g) + NO_2(g) + Na₂O_(s) → 2NaNO_2(s)-427

NO_(g) + NO_2(g) → N₂O_(g) + O_2(g)-43

2HNO_2(l) → N₂O_(g) + O_2(g) + H₂O_(l)34

Task 2.3.2

1. Using the following standard enthalpy of reaction data, calculate the heat of reaction for the combustion of ethane. (Hegarty, 2018, p. 68)

2C_(s) + 3H_2(g) → C₂H_6(g) ΔH = -84.68 kJ.mol^-1

C_(s) + O_2(g) → CO_2(g) ΔH = -394 kJ.mol^-1

H_2(g) + ½O_2(g) → H₂O_(l) ΔH = -286 kJ.mol^-1

2.3 apply Hess’s Law to simple energy cycles and solve problems to quantify enthalpy changes within reactions, including:

b. enthalpy changes involved in photosynthesis

Probably the most critical biochemical process, which forever changed both the Earth’s atmosphere and the nature of its biology, is photosynthesis. Photosynthesis is the process by which plants transform light energy into chemical energy. Photosynthesis is a photochemical process which is both endothermic (requiring a net input of energy) and involves the reduction of carbon.

The word equation for photosynthesis is:

Carbon dioxide + Water → Glucose + Oxygen

The chemical equation is:

sunlight

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

chlorophyll

The raw materials required for this process are carbon dioxide and water. Plants obtain their carbon dioxide through diffusion and absorb water from the soil through their roots. The glucose molecules are either used in respiration or stored as starch molecules in the leaf.

We can use Hess’ Law to calculate the enthalpy change for photosynthesis by using the enthalpies of formation of water, carbon dioxide and glucose.

NB The enthalpy of formation of oxygen in its standard state = 0 kJ as oxygen is an element in its standard state.

The standard enthalpy of formation (or heat of formation), Δ_fH^Ө, is the increase in enthalpy when one mole of a compound in its standard state is formed from its elements in their standard states.

(Smith & Davis, 2017, p. 349)

The enthalpy change for a chemical reaction at standard state conditions can be calculated from its standard enthalpies of formation using the following expression (Woollett, et al., 2018, p. 490):

Δ_fH^Ө_(reaction) = ∑Δ_fH^Ө_(products) - ∑Δ_fH^Ө_(reactants)

Enthalpy of formation of water:

Equ 1: 2H_2(g) + O_2(g) → 2H₂O_(l) Δ_fH^Ө = −571.6 kJ.mol^-1

Enthalpy of formation of carbon dioxide:

Equ 2: C_(s) + O_2(g) → CO_2(g) Δ_fH^Ө = −393.5 kJ.mol^-1

Enthalpy of formation of glucose:

Equ 3: 6C_(s) + 6H_2(g) + 3O_2(g) → C₆H₁₂O_6(s) Δ_fH^Ө = −1271 kJ.mol^-1

To work out the enthalpy change for photosynthesis, we reverse Equ 1:

−Equ 1: 2H₂O_(l) → 2H_2(g) + O_2(g)Δ_fH^Ө = +571.6 kJ.mol^-1

We have 6 water molecules, so we need to multiply this equation by 3

3 x −Equ 1: 6H₂O_(l) → 6H_2(g) + 3O_2(g)Δ_fH^Ө = +1714.8 kJ.mol^-1

Next we reverse equ 2:

−Equ 2: CO_2(g) → C_(s) + O_2(g) Δ_fH^Ө = +393.5 kJ.mol^-1

We have 6 CO₂ molecules, so we need to multiply the equation by 6

6 x −Equ 2: 6CO_2(g) → 6C_(s) + 6O_2(g) Δ_fH^Ө = +2361 kJ.mol^-1

Then we can use Equ 3 as written:

Equ 3: 6C_(s) + 6H_2(g) + 3O_2(g) → C₆H₁₂O_6(s) Δ_fH^Ө = −1271 kJ.mol^-1

Applying Hess’ Law, we add these equations together and cancel unchanged terms:

6H₂O_(l) → 6H_2(g) + 3O_2(g)Δ_fH = +1714.8 kJ.mol^-1

6CO_2(g) → 6C_(s) + 6O₂_(g) Δ_fH = +2361 kJ.mol^-1

6C_(s) + 6H_2(g) + 3O_2(g) → C₆H₁₂O₆_(s) Δ_fH = −1271 kJ.mol^-1

We can cancel as shown, to end up with a final equation of:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ ΔH = 1714.8 + 2361 − 1271 kJ.mol^-1

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ΔH = 2804.8 kJ.mol^-1

Hence the enthalpy change for photosynthesis is +2805 kJ/mol of glucose formed. The positive sign confirms that the process is endothermic.

View videos:

in an understanding of the general equation vs the light and carbon-fixing (Calvin) stages of photosynthesis, DoR#9 Hess’ Law and photosynthesis [7.34] https://www.youtube.com/watch?v=vkMhZAN4UZA&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=9

TASK 2.3.3

1. Write a chemical equation to represent photosynthesis.

2. Describe the process used in the calculation of the enthalpy change for photosynthesis. Why is it helpful to use the given values for heats of formation? Why is it not quite correct to use glucose in the solid state?

3. Distinguish between the following terms, providing an example for each:

a) Standard heat of formation of an element

b) Standard heat of formation of a compound

2.3 apply Hess’s Law to simple energy cycles and solve problems to quantify enthalpy changes within reactions, including:

c. enthalpy changes involved in respiration

Aerobic respiration, which is carried out by most plants and animals, is in chemical terms the reverse of photosynthesis.

We could estimate the enthalpy change for aerobic respiration based on our previous calculation of photosynthesis to be −2805 kJ .mol^-1.

As the equation is reversed, the sign is reversed and this means respiration is an exothermic process. This is exactly what we would expect as respiration is the process by which energy is made available to the cells in the form of an energy-rich molecule called ATP (adenosine triphosphate).

However, there are also several forms of anaerobic respiration, one of which is fermentation and this chemical process is a useful one for us to use for practice in the application of Hess’ Law, just as we did for photosynthesis.

Fermentation equation:

C₆H₁₂O_6(s) → 2C₂H₅OH_(l) + 2CO_2(g)

Enthalpy of formation of ethanol:

Equ 4: 2C_(s) + 3H_2(g) + ½O_2(g) → C₂H₅OH_(l) Δ_fH^Ө = −277.7 kJ.mol^-1

Enthalpy of formation of carbon dioxide:

Equ 2: C_(s) + O_2(g) → CO_2(g) Δ_fH^Ө = −393.5 kJ.mol^-1

Enthalpy of formation of glucose:

Equ 3: 6C_(s) + 6H_2(g) + 3O_2(g) → C₆H₁₂O_6(s) Δ_fH^Ө = −1271 kJ.mol^-1

To work out the enthalpy change for fermentation, we need to multiply this equation by 2

2 x Equ 4: 4C_(s) + 6H_2(g) + O_2(g) → 2C₂H₅OH_(l)Δ_fH^Ө = −555.4 kJ.mol^-1

Next we multiply equ 2 by 2:

2 x Equ 2: 2C_(s) + 2O_2(g) → 2CO_2(g) Δ_fH^Ө = −787 kJ.mol^-1

Then we reverse Equ 3:

Equ 3: C₆H₁₂O_6(s) → 6C_(s) + 6H_2(g) + 3O_2(g) Δ_fH^Ө = +1271 kJ.mol^-1

Applying Hess’ Law, we add these equations together:

4C_(s) + 6H_2(g) + O_2(g) → 2C₂H₅OH_(l)Δ_fH = −555.4 kJ.mol^-1

2C_(s) + 2O_2(g) → 2CO₂_(g) Δ_fH = −787 kJ.mol^-1

C₆H₁₂O₆_(s) → 6C_(s) + 6H_2(g) + 3O_2(g) Δ_fH = +1271 kJ.mol^-1

We cancel to end up with a final equation of:

C₆H₁₂O_6(s) → 2C₂H₅OH_(l) + 2CO_2(g) ΔH = −555.4 − 787 + 1271 kJ

C₆H₁₂O_6(s) → 2C₂H₅OH_(l) + 2CO_2(g) ΔH = −71.4 kJ

The enthalpy change for fermentation is −71 kJ/mol of glucose. The negative sign confirms that the process is exothermic.

View videos:

in an understanding of the general equation vs the TCA cycle of respiration, DoR#10 Hess’ Law and Respiration [8.21] https://www.youtube.com/watch?v=1imCCMIpRVo&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=8

TASK 2.3.4

1. Anaerobic respiration can occur in human muscles when they are fatigued and low in oxygen. However the product is not ethanol but lactic acid, C₃H₆O₃.

The equation which represents this process is:

C₆H₁₂O_6(s) → 2C₃H₆O_3(s)

and the standard enthalpy of formation of lactic acid is:

3C_(s) + 3H_2(g) + O_2(g) → C₃H₆O_3(s) Δ_fH^Ө = −694.1 kJ

Calculate the enthalpy change associated with the production of lactic acid. (Woollett, et al., 2018, p. 494)

Other useful sites:

Libre texts https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Thermodynamics/Thermodynamic_Cycles/Hess's_Law

REVIEW

1. Complete Txt2 p166 WS4.5 Hess' Law

2. Complete Txt1 p496 15.3 Review, Key Qs

3. COMPLETE UNDER TEST CONDITIONS, THEN CHECK YOUR ANSWERS

TX1 p497-9 Chapter 15 Review

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