Thermochemistry

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Chapter 5Thermochemistry

John D. BookstaverSt. Charles Community College

Edited by Debbie AmusoCamden High School

Chemistry, The Central Science, 11th editionTheodore L. Brown; H. Eugene LeMay, Jr.;

and Bruce E. Bursten

Thermochemistry

Thermochemistry

Thermodynamics = study of energy and its transformations

Thermochemistry = study of chemical reactions involving changes in heat energy

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Thermochemistry

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Energy

Energy = the ability to do work or transfer heat energy.

Work = energy used to cause an object with mass to move (w = f x d)

Heat = energy used to cause the temperature of an object to increase

Thermochemistry

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Major Types of Energy

Potential energy = energy an object possesses by virtue of its position or chemical composition.

Kinetic energy = energy an object possesses by virtue of its motion.

Thermochemistry

Kinetic Energy

m = mass in kilograms (kg)

v = velocity in meters per second (m/s)

KE = kinetic energy in joules (J)

1 Joule = 1 kg-m2/s2

A mass of 2 kg moving at a speed of one meter per second possesses a kinetic energy of 1 Joule.

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12

KE = m v 2

Thermochemistry

Potential Energy

m = mass in kilograms (kg)

g = acceleration due to gravity (9.8 m/s2)

h = height (m)

PE = potential energy in joules (J)

1 Joule = 1 kg-m2/s2

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PE = m g h

Thermochemistry

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Units of Energy

• The SI unit of energy is the joule (J).

• An older, non-SI unit is still in widespread use: the calorie (cal).

1 cal = 4.184 J

• 1000 calories = one nutritional Calorie

Thermochemistry

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First Law of Thermodynamics• Energy is neither created nor destroyed, but it can undergo a transformation from

one type to another. (Law of Conservation of Energy)

• The total energy of the universe is a constant.

• The energy lost by a system must equal the energy gained by its surroundings, and vice versa.

Thermochemistry

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System and Surroundings

System = the molecules we want to study (here, the hydrogen and oxygen molecules).

Surroundings = everything else (here, the cylinder and piston).

Thermochemistry

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Internal EnergyThe internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

E = Efinal − Einitial (It’s a state function)

• If E is positive, the system absorbed energy from the surroundings.

• If E is negative, the system released energy to the surroundings.

Thermochemistry

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E = q + w

• When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).

• That is, E = q + w.

Thermochemistry

q, w, and their signs

+ q = system gains or takes in heat

- q = system loses or gives off heat

+ w = work is done on the system by the surroundings (piston pushed in)

- w = work is done by the system on its surroundings (piston moves out)

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Thermochemistry

ExampleAs hydrogen and oxygen gas are ignited in a cylinder,

the system loses 550 J of heat to its surroundings. The expanding gases move a pistion to do 240 J of work on its surroundings. E for system = ?

Answer:

E = q + w

E = (-550 J) + (-240 J)

E = - 790 J

What does it mean?

The system gave off 790 J of energy to its surroundings

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Thermochemistry

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Enthalpy & H

• The symbol for enthalpy is H.• Enthalpy is the internal energy plus the product of

pressure and volume:

• At constant pressure:

H =E = q • So at constant pressure,

heat lost or gained by the system.

H = E + PV

Thermochemistry

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Recall: Endothermic• When heat is absorbed (taken in) by the system from the surroundings, the

process is endothermic.

H = Hfinal − Hinitial

H = Hproducts − Hreactants

H = positive value for endothermic

Thermochemistry

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Recall: Exothermic When heat is released (given off) by the system into the surroundings, the process is

exothermic.

H = Hfinal − Hinitial

H = Hproducts − Hreactants

H = negative value for exothermic

Thermochemistry

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Enthalpy of Reaction

1. This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

2. Enthalpy is an extensive property.

3. Reverse Rx Negate H value.

4. Phase (state) matters.

Thermochemistry

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Enthalpies of Formation

An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

Example:

2 Al (s) + 3 O2 (g) 2 Al2O3 (s)

Hf = -3340 kJ

Thermochemistry

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Standard Enthalpies of Formation

Standard enthalpies of formation, Hf°, is defined as the enthalpy change for the reaction in which one mole of a compound is made from its constituent elements in their standard states at 25o C (298 K) and 1.00 atm pressure.

Example:

2 Al (s) + 3/2 O2 (g) Al2O3 (s)

Hf°= - 1670 kJ/mol

Thermochemistry

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Calculation of H

H = nHf°products – mHf° reactants

where n and m are the coefficients.

Hf°Values are listed in Appendix C of your Textbook.

Thermochemistry

Calculation of H

H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)]= (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ

H = nHf products – mHf° reactants

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C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Thermochemistry

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Calorimetry is Another Way to Measure H values

We measure H through calorimetry, which measures heat flow.

But first, some definitions.

Thermochemistry

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Heat Capacity The amount of energy (joules) required to raise the temperature of a substance by 1 K (1C) is its heat capacity. Units are J/K or J/ C

q = C * T or C = q / T

q = heat in Joules

C = heat capacity in J/K

T = change in temperature

Thermochemistry

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Specific Heat

Specific heat capacity (or simply specific heat) is the amount of energy (joules) required to raise the temperature of 1 g of a substance by 1 K. Units are J/g-K

Thermochemistry

Molar Heat Capacity

• Molar heat capacity is the amount of energy (joules) required to raise the temperature of 1 mole of a substance by 1 K. Units are J/mol-K

Also helpful: # mol = mass/gfm

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Thermochemistry

Use Units

What is the molar heat capacity of CH4 (g) if its specific heat capacity is 2.20 J/g-K?

Answer:

2.20 J x 16.0 g =

g – K 1 mol

35.2 J/mol-K

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Thermochemistry

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Constant Pressure Calorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the H for the system by measuring the heat change for the water in the calorimeter.

q = m sh T

Thermochemistry

Calorimetry

q = m * sh * T

q = Joules of heat

m = total mass in calorimeter

(If a solution, use m = D * V to obtain the mass)

Water has a density of 1.00 g/mL

Most water solutions have a density of 1.02 g/mL

sh = specific heat (for water, 4.18 J/g-K)

T = Tf - Ti

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Thermochemistry

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Calorimetry Σ q = 0

In other words, the heat released (or absorbed) by the reaction of interest = the sum of

the heat gained (or released) by the resulting solution & calorimeter.

qcalorimeter + qreaction+ qresulting solution = 0

If we assume qcalorimeter is negligible the equation reduces to the following:

qreaction = - qsolution = -m*sh*ΔT

where T = Tf - Ti and m = D * V

H = qreaction /# moles

Thermochemistry

Specific Heat of a MetalHow to determine specific heat of a metal:

1) Mass metal

2) Put metal in HOT water & measure initial temp of hot metal

3) Measure temp of 100.0 mL (100.0 g) of COLD water

4) Put hot metal in cold water

5) Record temp of water with metal in it (that temp is the final temp for both the metal & water)

6) Calculate the sh of the metal

(m * sh * T)metal = (m * 4.18 * T)water

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Thermochemistry

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Bomb Calorimetry

• Reactions can be carried out in a sealed “bomb” such as this one.

• The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.

Thermochemistry

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Bomb Calorimetry

• Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.

• For most reactions, the difference is very small.

Thermochemistry

H via Heat Capacity

q = C * T

q = heat in Joules

C = heat capacity in J/K

T = change in temperature

where T = Tf – Ti

H = q / # moles

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Thermochemistry

Stoichiometric Determination of

Given the H for a reaction, you can use mole ratios to determine H values of

different sample sizes.

See examples.

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Thermochemistry

Sample Problem #1

H2 (g) + I2 (g) 2 HI (g) H = +53.0 kJ

How many joules of heat are required to form 5 moles of HI (g)?

5 mol HI __x___

2 mol HI = +53.0 kJ

x = +133 kJ

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Thermochemistry

Sample Problem #2

H2 (g) + I2 (g) 2 HI (g) H = +53.0 kJ

How many moles of HI (g) are formed by the expenditure of 235 kJ of heat energy?

__x___ +235 kJ

2 mol HI = +53.0 kJ

x = 8.87 mol HI

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Thermochemistry

Sample Problem #3

H2 (g) + I2 (g) 2 HI (g) H = +53.0 kJ

How many joules of heat are required to form 109 grams of HI?

# mol HI = mass/gfm

# mol HI = 109 g / 128 g/mol = 0.852 mol HI

0.852 mol HI __x___

2 mol HI = +53.0 kJ

x = + 22.6 kJ

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Thermochemistry

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Hess’s Law

H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.

However, we can estimate H using published H values and the properties of enthalpy.

Thermochemistry

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Hess’s Law

Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

H = H1 + H2 + H3

Thermochemistry

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Calculation of H

• Imagine this as occurringin three steps:

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry

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Calculation of H

• Imagine this as occurringin three steps:

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry

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Calculation of H

• Imagine this as occurringin three steps:

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry

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C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Calculation of H

• The sum of these equations is:

Thermochemistry

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Energy in FoodsMost of the fuel in the food we eat comes from carbohydrates and fats.

Thermochemistry

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Energy in Fuels

The vast majority of the energy consumed in this country comes from fossil fuels.

Thermochemistry

Summary: Ways to Calculate H

H = nHf°products – mHf° reactants

Using Hfo values from Appendix C

Or

By Using a Calorimeter

qreaction = - qsolution = - m*sh*ΔT

where T = Tf - Ti and m = D * V

H = qreaction /# moles

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Thermochemistry

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Summary: Ways to Calculate HUsing Heat Capacity

q = C * T

q = heat in Joules

C = heat capacity in J/K

T = change in temperature

where T = Tf – Ti

H = q / # moles

Thermochemistry

Summary: Ways to Calculate H

Hess’s Law

H = H1 + H2 + H3

Or

By Stoichiometric Determination

Given the H for a reaction, you can use mole ratios to determine H values of different sample sizes.

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