UNIT 6

Thermodynamics

Explore how energy is transferred during physical and chemical processes and learn how heat, enthalpy, phase changes, and chemical bonds relate to energy.

Topics

Heat Transfer and Calorimetry

Distinguish heat from temperature and calculate energy transfer using specific heat, calorimetry, and thermal equilibrium.

Phase Changes and Heating Curves

Analyze heating curves and calculate the energy involved when substances change temperature or physical state.

Enthalpy Changes

Calculate reaction enthalpy using energy diagrams, bond enthalpies, standard enthalpies of formation, and Hess’s law.

Heat Transfer and Calorimetry

Thermodynamics examines the transfer of energy between a system and its surroundings. In chemistry, the system is the reaction or process being studied, while the surroundings include everything outside that system.

Heat and Temperature

Heat and temperature are related, but they do not mean the same thing.

Quantity	Meaning
Temperature	A measure related to the average kinetic energy of the particles in a substance
Heat	Energy transferred between objects because of a temperature difference

Heat naturally transfers from an object at a higher temperature to an object at a lower temperature until the objects reach thermal equilibrium.

Thermal equilibrium: The condition reached when objects in contact have the same temperature and no net heat transfer occurs between them.

Endothermic and Exothermic Processes

The sign of heat, q, depends on whether energy enters or leaves the system.

Process	Direction of Heat Transfer	Sign of q for the System
Endothermic	Heat enters the system	Positive
Exothermic	Heat leaves the system	Negative

If the surroundings become warmer, the system released heat and the process is exothermic.

If the surroundings become cooler, the system absorbed heat and the process is endothermic.

Specific Heat Capacity

Specific heat capacity is the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius.

q = mcΔT

In this equation:

q is the heat absorbed or released.
m is the mass of the substance.
c is its specific heat capacity.
ΔT is the change in temperature.

ΔT = T_final − T_initial

A substance with a large specific heat capacity requires more energy to produce a given temperature change than the same mass of a substance with a smaller specific heat capacity.

Example: Heating a Metal

A 75.0 g sample of a metal has a specific heat capacity of 0.385 J/(g·°C). How much heat is absorbed when its temperature increases from 22.0°C to 68.0°C?

First, calculate the temperature change:

ΔT = 68.0°C − 22.0°C = 46.0°C

Substitute the values into the heat equation:

q = (75.0 g)(0.385 J/(g·°C))(46.0°C)

q = 1.33 × 10³ J

The metal absorbs 1.33 kJ of heat. The value is positive because the metal gains energy.

Heat Capacity

Heat capacity describes the energy needed to increase the temperature of an entire object or calorimeter by one degree. Unlike specific heat capacity, it applies to the complete object rather than one gram of material.

q = CΔT

In this equation, C is the object’s heat capacity, commonly measured in J/°C.

Conservation of Energy

Energy cannot be created or destroyed. Therefore, heat lost by one part of an isolated system must be gained by another part.

q_system + q_surroundings = 0

q_system = −q_surroundings

For example, if a hot metal sample loses 500 J of heat, the cooler water surrounding it must gain 500 J, assuming no energy escapes to the outside environment.

Thermal Equilibrium Calculations

Example: Mixing Water at Different Temperatures

A 50.0 g sample of water at 80.0°C is mixed with 100.0 g of water at 20.0°C. Assume no heat escapes and determine the final temperature.

Heat lost by the warmer water equals heat gained by the cooler water:

q_hot + q_cold = 0

(50.0 g)c(T_f − 80.0°C) + (100.0 g)c(T_f − 20.0°C) = 0

Because both samples are water, the same specific heat capacity appears in both terms and can be canceled.

50.0(T_f − 80.0) + 100.0(T_f − 20.0) = 0

150.0T_f = 6000

T_f = 40.0°C

The final temperature must fall between the two initial temperatures, which supports the calculated result.

Calorimetry

Calorimetry is the experimental measurement of heat transferred during a physical or chemical process. A calorimeter reduces energy exchange with the outside environment.

In a coffee-cup calorimeter, a reaction occurs in a solution. The temperature change of the solution is used to determine how much heat the reaction released or absorbed.

q_reaction = −q_solution

q_solution = mcΔT

If the solution’s temperature increases, the solution absorbed heat. That heat came from the reaction, so the reaction’s q is negative.

Example: Coffee-Cup Calorimetry

A reaction occurs in 100.0 g of solution. The temperature increases from 22.5°C to 28.7°C. Assume the solution has a specific heat capacity of 4.184 J/(g·°C).

Calculate the heat of the reaction. Then calculate the energy released per mole if 0.0500 mol of reactant was consumed.

First, calculate the temperature change:

ΔT = 28.7°C − 22.5°C = 6.2°C

Calculate the heat absorbed by the solution:

q_solution = (100.0 g)(4.184 J/(g·°C))(6.2°C)

q_solution = 2.59 × 10³ J = 2.59 kJ

The solution absorbed heat, so the reaction released the same amount:

q_reaction = −2.59 kJ

Calculate the energy released per mole:

Energy per mole = −2.59 kJ 0.0500 mol

Energy per mole = −51.8 kJ/mol

The negative sign shows that the reaction is exothermic.

Accounting for the Calorimeter

Some problems provide the heat capacity of the calorimeter. In that case, both the solution and the calorimeter absorb energy.

q_calorimeter = C_calorimeterΔT

q_reaction = −(q_solution + q_calorimeter)

Always identify which objects absorb heat. If the calorimeter has a provided heat capacity, its absorbed energy should not be ignored.

Common Calorimetry Assumptions

No heat escapes into the outside environment.
The solution has the same specific heat capacity as water when the problem tells you to make that assumption.
The solution’s mass may be approximated by adding the masses of the combined solutions.
The reaction and solution exchange equal amounts of heat with opposite signs.

Practice Questions

Question 1: Calculating Heat

How much heat is required to increase the temperature of 125 g of water from 18.5°C to 32.0°C? Use a specific heat capacity of 4.184 J/(g·°C).

Show solution

First, calculate the temperature change:

ΔT = 32.0°C − 18.5°C = 13.5°C

Use the specific-heat equation:

q = mcΔT

q = (125 g)(4.184 J/(g·°C))(13.5°C)

q = 7.06 × 10³ J = 7.06 kJ

The water absorbs 7.06 kJ of heat.

Question 2: Finding Specific Heat Capacity

A 50.0 g metal sample absorbs 1.80 kJ of heat as its temperature increases from 20.0°C to 95.0°C.

Calculate the metal’s specific heat capacity.

Show solution

Convert the heat from kilojoules to joules:

1.80 kJ = 1800 J

Calculate the temperature change:

ΔT = 95.0°C − 20.0°C = 75.0°C

Rearrange the heat equation to solve for c:

c = q mΔT

c = 1800 J (50.0 g)(75.0°C)

c = 0.480 J/(g·°C)

Question 3: Conservation of Energy

A warm metal sample is placed into 100.0 g of water. The water’s temperature increases from 22.0°C to 25.0°C. Assume no heat escapes.

Calculate the heat absorbed by the water and the heat released by the metal.

Show solution

The water’s temperature change is:

ΔT = 25.0°C − 22.0°C = 3.0°C

q_water = (100.0 g)(4.184 J/(g·°C))(3.0°C)

q_water = 1.26 × 10³ J = 1.26 kJ

The water gains heat, so its q is positive. The metal loses the same amount of energy.

q_metal = −q_water

q_metal = −1.26 kJ

Question 4: Reaction Calorimetry

A reaction occurs in 150.0 g of solution. The temperature increases from 21.3°C to 26.8°C. Assume the solution has a specific heat capacity of 4.184 J/(g·°C).

If 0.0750 mol of reactant is consumed, calculate the heat released per mole of reactant.

Show solution

Calculate the temperature change:

ΔT = 26.8°C − 21.3°C = 5.5°C

Calculate the heat absorbed by the solution:

q_solution = (150.0 g)(4.184 J/(g·°C))(5.5°C)

q_solution = 3.45 × 10³ J = 3.45 kJ

The solution absorbs heat, so the reaction releases the same amount.

q_reaction = −3.45 kJ

Calculate the heat released per mole:

Energy per mole = −3.45 kJ 0.0750 mol = −46.0 kJ/mol

The negative result indicates an exothermic reaction.

Question 5: Including the Calorimeter

A reaction occurs in 80.0 g of solution inside a calorimeter with a heat capacity of 35.0 J/°C. Both the solution and calorimeter increase in temperature by 4.0°C.

Assume the solution has a specific heat capacity of 4.184 J/(g·°C). Calculate the heat of the reaction.

Show solution

Calculate the heat absorbed by the solution:

q_solution = (80.0 g)(4.184 J/(g·°C))(4.0°C)

q_solution = 1.34 × 10³ J

Calculate the heat absorbed by the calorimeter:

q_calorimeter = (35.0 J/°C)(4.0°C) = 140 J

Add the energy absorbed by both surroundings:

q_surroundings = 1340 J + 140 J = 1480 J

The reaction has the opposite sign:

q_reaction = −1.48 kJ

Question 6: Calorimetry Error Analysis

A student uses a poorly insulated coffee-cup calorimeter to measure the energy released by an exothermic reaction. Some heat escapes into the room.

How will this affect the measured temperature change and the calculated magnitude of the reaction’s heat?

Show solution

Because some heat escapes into the room, less heat is absorbed by the solution. The measured temperature increase will therefore be too small.

Because q is calculated using ΔT, the calculated amount of heat released will also have a magnitude that is too small.

The calculated value for an exothermic process would be less negative than the value obtained under perfectly insulated conditions.

VIDEO EXPLANATION

Watch: Calorimetry

Watch a worked example showing how mass, specific heat, temperature change, and energy conservation are used in calorimetry calculations.

Watch on TikTok →

Phase Changes and Heating Curves

Energy can increase a substance’s temperature or cause it to change physical state. A heating curve shows how temperature changes as energy is continuously added to a substance.

Common Phase Changes

Phase Change	Direction	Energy Change
Melting	Solid → Liquid	Energy absorbed
Freezing	Liquid → Solid	Energy released
Vaporization	Liquid → Gas	Energy absorbed
Condensation	Gas → Liquid	Energy released
Sublimation	Solid → Gas	Energy absorbed
Deposition	Gas → Solid	Energy released

Moving toward a less organized phase—solid to liquid to gas—requires energy.

Moving toward a more organized phase—gas to liquid to solid—releases energy.

Temperature Changes Within One Phase

When a substance remains in one physical state, added energy increases the average kinetic energy of its particles. The temperature therefore increases.

q = mcΔT

Different phases of the same substance may have different specific heat capacities. A complete calculation may therefore require a different value of c for the solid, liquid, and gas.

Energy During a Phase Change

During a phase change, the temperature remains constant even though energy continues to enter or leave the substance.

The transferred energy changes the arrangement and attractions between particles rather than changing their average kinetic energy.

During a phase change:

• Temperature remains constant

• Average kinetic energy remains constant

• Potential energy changes

• More than one physical phase is present

Enthalpy of Fusion

The molar enthalpy of fusion, ΔH_fus, is the energy required to melt one mole of a substance at its melting point.

q = nΔH_fus

Melting has a positive energy change. Freezing has the same magnitude but the opposite sign.

Example: Melting Ice

Calculate the energy required to melt 25.0 g of ice at 0°C. The molar enthalpy of fusion of water is 6.01 kJ/mol.

First, convert the mass of water to moles:

25.0 g H₂O × 1 mol H₂O 18.02 g H₂O = 1.39 mol H₂O

q = (1.39 mol)(6.01 kJ/mol)

q = 8.34 kJ

Melting the ice requires 8.34 kJ of energy.

Enthalpy of Vaporization

The molar enthalpy of vaporization, ΔH_vap, is the energy required to vaporize one mole of a substance at its boiling point.

q = nΔH_vap

Vaporization has a positive energy change. Condensation has the same magnitude but the opposite sign.

For a given substance, ΔH_vap is generally larger than ΔH_fus. Separating particles into the gas phase requires overcoming more intermolecular attraction than converting a solid into a liquid.

Reading a Heating Curve

Heating curve of water showing solid warming, melting, liquid warming, boiling, and gas warming — Sloped sections represent temperature changes within one phase. Horizontal sections represent phase changes at constant temperature.

Segment	Process	Main Energy Change
A	Solid warming	Kinetic energy increases
B	Melting	Potential energy increases
C	Liquid warming	Kinetic energy increases
D	Boiling	Potential energy increases
E	Gas warming	Kinetic energy increases

Sloped Sections of a Heating Curve

On a sloped section, the substance remains in one phase and its temperature changes. Use:

q = mcΔT

A steeper slope means that a smaller amount of energy is required for a given temperature increase. For equal masses, this corresponds to a smaller specific heat capacity.

Horizontal Sections of a Heating Curve

On a horizontal section, the substance changes phase while its temperature remains constant.

Use one of the following equations:

q = nΔH_fus

q = nΔH_vap

A longer horizontal section represents a phase change requiring a greater amount of energy under the conditions shown.

Calculations Involving Multiple Segments

If a substance undergoes both temperature changes and phase changes, calculate the heat for each separate segment and then add the results.

q_total = q₁ + q₂ + q₃ + ...

Example: Heating and Melting

A 10.0 g sample of ice is heated from −10.0°C to 20.0°C. This process contains three separate energy calculations:

Warm the solid ice from −10.0°C to 0°C.
Melt the ice at 0°C.
Warm the liquid water from 0°C to 20.0°C.

q_total = m(c_ice)ΔT + nΔH_fus + m(c_water)ΔT

Each part must be calculated separately because the appropriate equation or specific heat capacity changes between segments.

Practice Questions

Question 1: Reading a Heating Curve

Refer to the labeled heating curve. For each segment, identify whether kinetic energy or potential energy primarily increases.

Show solution

Segment A: Kinetic energy increases as the solid warms.
Segment B: Potential energy increases as the solid melts.
Segment C: Kinetic energy increases as the liquid warms.
Segment D: Potential energy increases as the liquid boils.
Segment E: Kinetic energy increases as the gas warms.

Temperature changes during sloped sections, so average kinetic energy changes. Temperature remains constant during horizontal sections, while particle attractions are overcome and potential energy changes.

Question 2: Enthalpy of Fusion

Calculate the energy required to melt 45.0 g of ice at 0°C. Use a molar mass of 18.02 g/mol and ΔH_fus = 6.01 kJ/mol.

Show solution

First, convert the mass to moles:

45.0 g H₂O × 1 mol H₂O 18.02 g H₂O = 2.50 mol H₂O

Calculate the energy required for melting:

q = nΔH_fus

q = (2.50 mol)(6.01 kJ/mol) = 15.0 kJ

Melting the ice requires 15.0 kJ of energy.

Question 3: Condensation

A 1.50 mol sample of water vapor condenses at its condensation point. Calculate the energy change if ΔH_vap = 40.7 kJ/mol.

Show solution

Vaporization absorbs energy, but condensation is the reverse process and releases the same amount of energy per mole.

ΔH_condensation = −40.7 kJ/mol

q = (1.50 mol)(−40.7 kJ/mol)

q = −61.1 kJ

The water releases 61.1 kJ. The negative sign indicates that energy leaves the water.

Question 4: Temperature During a Phase Change

A sample is boiling while energy is continuously added. A student claims that the added energy must be increasing the average kinetic energy of the particles.

Is the student correct? Explain.

Show solution

The student is not correct. During boiling, the temperature remains constant, so the particles’ average kinetic energy also remains constant.

The added energy increases potential energy by separating particles and overcoming intermolecular attractions as the liquid becomes a gas.

Question 5: A Multistep Heating Calculation

Calculate the total energy required to convert 18.0 g of ice at −15.0°C into liquid water at 25.0°C.

Use the following values:

Specific heat of ice: 2.09 J/(g·°C)
Specific heat of liquid water: 4.184 J/(g·°C)
Molar enthalpy of fusion: 6.01 kJ/mol
Molar mass of water: 18.02 g/mol

Show solution

This process requires three separate calculations.

Step 1: Warm the Ice to 0°C

q₁ = (18.0 g)(2.09 J/(g·°C))(15.0°C)

q₁ = 564 J = 0.564 kJ

Step 2: Melt the Ice

18.0 g × 1 mol 18.02 g = 0.999 mol

q₂ = (0.999 mol)(6.01 kJ/mol) = 6.00 kJ

Step 3: Warm the Liquid Water

q₃ = (18.0 g)(4.184 J/(g·°C))(25.0°C)

q₃ = 1880 J = 1.88 kJ

Add the Three Energy Changes

q_total = 0.564 kJ + 6.00 kJ + 1.88 kJ

q_total = 8.44 kJ

The complete process requires approximately 8.44 kJ of energy.

Question 6: Comparing Heating-Curve Plateaus

On a heating curve with heat added on the horizontal axis, the boiling plateau is much longer than the melting plateau. What does this indicate?

Show solution

The longer boiling plateau indicates that vaporization requires more energy than melting for the amount of substance represented.

During vaporization, particles must become much more widely separated. Overcoming enough intermolecular attraction to form a gas generally requires more energy than rearranging a solid into a liquid.

Therefore, ΔH_vap is generally greater than ΔH_fus.

VIDEO EXPLANATION

Watch: Heating Curves and Phase Changes

Watch a worked example showing how to interpret a heating curve and calculate energy across temperature changes and phase changes.

Watch on TikTok →

Enthalpy Changes

Enthalpy, represented by H, describes the energy of a system under constant-pressure conditions. Chemical reactions are commonly analyzed using the enthalpy change, ΔH.

ΔH = H_products − H_reactants

Exothermic and Endothermic Reactions

Reaction Type	Energy Transfer	Sign of ΔH	Relative Energy
Exothermic	Energy leaves the system	Negative	Products are lower than reactants
Endothermic	Energy enters the system	Positive	Products are higher than reactants

ΔH describes the difference between the initial and final energy levels. It does not depend on the pathway taken between the reactants and products.

Enthalpy Is a State Function

A state function depends only on the initial and final states, not on the individual steps used to reach the final state.

Therefore, the total enthalpy change for a reaction is the same whether the reaction occurs in one step or through several intermediate steps.

Because enthalpy is a state function, enthalpy changes can be added together. This principle is the basis of Hess’s law.

Bond Enthalpy

Bond enthalpy is the energy required to break one mole of a particular type of bond in the gas phase.

Breaking bonds requires energy.

Forming bonds releases energy.

An approximate reaction enthalpy can be calculated by comparing the energy required to break reactant bonds with the energy released when product bonds form.

ΔH_reaction ≈ Energy of bonds broken − Energy of bonds formed

Example: Calculating ΔH from Bond Enthalpies

H₂(g) + Cl₂(g) → 2HCl(g)

Use the following average bond enthalpies:

H—H: 436 kJ/mol
Cl—Cl: 243 kJ/mol
H—Cl: 431 kJ/mol

One H—H bond and one Cl—Cl bond must be broken.

Energy absorbed = 436 kJ/mol + 243 kJ/mol = 679 kJ/mol

Two H—Cl bonds are formed.

Energy released = 2(431 kJ/mol) = 862 kJ/mol

ΔH = 679 kJ/mol − 862 kJ/mol

ΔH = −183 kJ/mol

The negative result indicates that the reaction is exothermic.

Bond enthalpies are average values obtained from many substances, so calculations using them produce approximate reaction enthalpies.

Standard Enthalpy of Formation

The standard enthalpy of formation, ΔH_f^°, is the enthalpy change when one mole of a substance forms from its elements in their standard states.

The standard enthalpy of formation of an element in its standard state is zero.

Examples include O₂(g), H₂(g), N₂(g), and C(s, graphite).

Standard reaction enthalpy can be calculated using formation enthalpies:

ΔH_reaction^° = Sum of ΔH_f^° for products − Sum of ΔH_f^° for reactants

Each formation enthalpy must be multiplied by the corresponding coefficient in the balanced equation.

Example: Enthalpy of Combustion

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Use these formation enthalpies:

Substance	ΔH_f^° (kJ/mol)
CH₄(g)	−74.8
O₂(g)	0
CO₂(g)	−393.5
H₂O(l)	−285.8

Calculate the sum for the products:

Products = (−393.5) + 2(−285.8) = −965.1 kJ/mol

Calculate the sum for the reactants:

Reactants = (−74.8) + 2(0) = −74.8 kJ/mol

ΔH_reaction^° = −965.1 − (−74.8)

ΔH_reaction^° = −890.3 kJ/mol

Hess’s Law

Hess’s law states that if several reactions add together to form a target reaction, their enthalpy changes also add together to produce the enthalpy change of the target reaction.

When manipulating a chemical equation:

• Reverse the equation → reverse the sign of ΔH

• Multiply the equation → multiply ΔH by the same factor

• Add equations → add their ΔH values

Example: Using Hess’s Law

Determine ΔH for:

C(s) + 1 2 O₂(g) → CO(g)

Use the following equations:

C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ

CO(g) + 1 2 O₂(g) → CO₂(g) ΔH = −283.0 kJ

Reverse the second equation so that CO appears as a product. The sign of ΔH also reverses.

CO₂(g) → CO(g) + 1 2 O₂(g) ΔH = +283.0 kJ

Add the equations. CO₂ cancels, and part of the O₂ also cancels.

ΔH = −393.5 kJ + 283.0 kJ

ΔH = −110.5 kJ

Comparing the Methods

Method	Information Used	Important Detail
Calorimetry	Measured temperature change	Based on experimental heat transfer
Bond enthalpies	Bonds broken and formed	Produces an approximate value
Formation enthalpies	Tabulated ΔH_f^° values	Products minus reactants
Hess’s law	Enthalpies of related reactions	Equations and ΔH values are combined

Practice Questions

Question 1: Interpreting an Enthalpy Diagram

An energy diagram shows reactants at 120 kJ/mol and products at 75 kJ/mol.

Calculate ΔH and determine whether the reaction is exothermic or endothermic.

Show solution

ΔH = H_products − H_reactants

ΔH = 75 kJ/mol − 120 kJ/mol

ΔH = −45 kJ/mol

The products are lower in energy than the reactants, so the reaction is exothermic.

Question 2: Bond Enthalpies

Consider the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Use the following average bond enthalpies:

N≡N: 945 kJ/mol
H—H: 436 kJ/mol
N—H: 391 kJ/mol

Estimate ΔH for the reaction.

Show solution

One N≡N bond and three H—H bonds are broken.

Energy of bonds broken = 945 + 3(436)

Energy of bonds broken = 2253 kJ/mol

Each NH₃ molecule contains three N—H bonds. Two NH₃ molecules contain six N—H bonds.

Energy of bonds formed = 6(391) = 2346 kJ/mol

ΔH ≈ Energy of bonds broken − Energy of bonds formed

ΔH ≈ 2253 − 2346

ΔH ≈ −93 kJ/mol

The estimated reaction enthalpy is −93 kJ/mol, so the reaction is exothermic.

Question 3: Standard Enthalpies of Formation

Consider the decomposition reaction:

CaCO₃(s) → CaO(s) + CO₂(g)

Use the following standard formation enthalpies:

Substance	ΔH_f^° (kJ/mol)
CaCO₃(s)	−1206.9
CaO(s)	−635.1
CO₂(g)	−393.5

Calculate ΔH_reaction^°.

Show solution

Calculate the total for the products:

Products = −635.1 + (−393.5) = −1028.6 kJ/mol

Calculate the total for the reactants:

Reactants = −1206.9 kJ/mol

ΔH_reaction^° = Products − Reactants

ΔH_reaction^° = −1028.6 − (−1206.9)

ΔH_reaction^° = +178.3 kJ/mol

The positive value indicates that the decomposition is endothermic.

Question 4: Hess’s Law

Calculate ΔH for the target reaction:

2S(s) + 3O₂(g) → 2SO₃(g)

Use the following equations:

S(s) + O₂(g) → SO₂(g) ΔH = −296.8 kJ

2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = −198.2 kJ

Show solution

Multiply the first equation by two so that it produces two moles of SO₂.

2S(s) + 2O₂(g) → 2SO₂(g)

ΔH = 2(−296.8 kJ) = −593.6 kJ

Add the second equation:

2SO₂(g) + O₂(g) → 2SO₃(g)

ΔH = −198.2 kJ

The 2SO₂ molecules cancel when the equations are added.

ΔH_total = −593.6 kJ + (−198.2 kJ)

ΔH_total = −791.8 kJ

Question 5: Manipulating a Thermochemical Equation

Consider the thermochemical equation:

2H₂(g) + O₂(g) → 2H₂O(l) ΔH = −571.6 kJ

Determine ΔH for the following reaction:

H₂O(l) → H₂(g) + 1 2 O₂(g)

Show solution

First, reverse the original equation. Reversing the reaction reverses the sign of ΔH.

2H₂O(l) → 2H₂(g) + O₂(g)

ΔH = +571.6 kJ

Next, divide the entire equation and its ΔH value by two.

ΔH = +571.6 kJ 2 = +285.8 kJ

The target reaction has an enthalpy change of +285.8 kJ.

Question 6: Comparing Calculation Methods

A student uses average bond enthalpies to calculate ΔH and obtains a value that is slightly different from a measured calorimetry value.

Does this difference necessarily mean that the student made a mistake? Explain.

Show solution

No. Average bond enthalpies are averages obtained from bonds in many different chemical environments. The exact energy of a particular bond depends on the molecule containing it.

A bond-enthalpy calculation therefore produces an approximate reaction enthalpy. A small difference from an experimental calorimetry result does not necessarily indicate an error.

VIDEO EXPLANATION

Watch: Calculating Reaction Enthalpy

Watch a worked example using bond enthalpies, standard enthalpies of formation, or Hess’s law to determine the energy change of a reaction.

Watch on TikTok →