Thermochemistry

Chapter 6Thermochemistry

Thermochemistry

Energy

• The ability to do work or transfer heat.Work: Energy used to cause an object that

has mass to move.Heat: Energy used to cause the

temperature of an object to rise.

Thermochemistry

Potential Energy

Energy an object possesses by virtue of its position or chemical composition.

PE = mgh

Thermochemistry

Kinetic Energy

Energy an object possesses by virtue of its motion.

12

KE = mv2

Thermochemistry

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

1 J = 1 kg m2

s2

Thermochemistry

System and Surroundings

• The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).

• The surroundings are everything else (here, the cylinder and piston).

Thermochemistry

Work

• Energy used to move an object over some distance.

• w = F d,

where w is work, F is the force, and d is the distance over which the force is exerted.

Thermochemistry

Heat

• Energy can also be transferred as heat.

• Heat flows from warmer objects to cooler objects.

Thermochemistry

Transferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

Thermochemistry

Transferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

b) As the ball falls, its potential energy is converted to kinetic energy.

Thermochemistry

Transferal of Energy

a) The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

b) As the ball falls, its potential energy is converted to kinetic energy.

c) When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

Thermochemistry

First Law of Thermodynamics• Energy is neither created nor destroyed.• In other words, the total energy of the universe is

a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Use Fig. 5.5

Thermochemistry

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.

Use Fig. 5.5

Thermochemistry

Internal EnergyBy definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:

E = Efinal − Einitial

Use Fig. 5.5

Thermochemistry

Changes in Internal Energy

• If E > 0, Efinal > Einitial

Therefore, the system absorbed energy from the surroundings.

This energy change is called endergonic.

Thermochemistry

Changes in Internal Energy

• If E < 0, Efinal < Einitial

Therefore, the system released energy to the surroundings.

This energy change is called exergonic.

Thermochemistry

Changes in Internal Energy

• 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

E, q, w, and Their Signs

Thermochemistry

Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

Thermochemistry

Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

• When heat is released by the system to the surroundings, the process is exothermic.

Thermochemistry

State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

Thermochemistry

State Functions• However, we do know that the internal energy

of a system is independent of the path by which the system achieved that state. In the system below, the water could have reached

room temperature from either direction.

Thermochemistry

State Functions• Therefore, internal energy is a state function.• It depends only on the present state of the

system, not on the path by which the system arrived at that state.

• And so, E depends only on Einitial and Efinal.

Thermochemistry

State Functions

• However, q and w are not state functions.

• Whether the battery is shorted out or is discharged by running the fan, its E is the same.But q and w are different

in the two cases.

E = q + w

Thermochemistry

Work

When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

Thermochemistry

WorkWe can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.

w = −PV

Thermochemistry

Internal Energy Example• Calculate E for a system undergoing an endothermic

process in which 15.6 kJ of heat flows & where 1.4 kJ of work is done on the system.

Thermochemistry

PV Work Example• Calculate the work associated with the expansion of a

gas from 46 L to 64 L at a constant external pressure of 15 atm.

Thermochemistry

Internal Energy, Heat & Work Example

• A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.30 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate E for the process. To convert between L * atm and J, use 1 L * atm = 101.3 J.

Thermochemistry

Enthalpy

• If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.

• Enthalpy is the internal energy plus the product of pressure and volume:

H = E + PV

Thermochemistry

Enthalpy

• When the system changes at constant pressure, the change in enthalpy, H, is

H = (E + PV)• This can be written

H = E + PV

Thermochemistry

Enthalpy

• Since E = q + w and w = −PV, we can substitute these into the enthalpy expression:

H = E + PVH = (q+w) − w

H = q• So, at constant pressure the change in

enthalpy is the heat gained or lost.

Thermochemistry

Endothermicity and Exothermicity

• A process is endothermic, then, when H is positive.

Thermochemistry

Endothermicity and Exothermicity

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

Thermochemistry

Enthalpies of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Thermochemistry

Enthalpies of Reaction

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

Thermochemistry

The Truth about Enthalpy

1. Enthalpy is an extensive property.

2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.

3. H for a reaction depends on the state of the products and the state of the reactants.

Thermochemistry

Calorimetry

Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

Thermochemistry

Heat Capacity and Specific Heat

• The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.

• We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

Thermochemistry

Heat Capacity and Specific Heat

Specific heat, then, is

Specific heat =heat transferred

mass temperature change

s =q

m T

Thermochemistry

Constant Pressure Calorimetry

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

Thermochemistry

Constant Pressure Calorimetry

Because the specific heat for water is well known (4.184 J/mol-K), we can measure H for the reaction with this equation:

q = m s T

Thermochemistry

Calorimetry Problem• When a 3.25g sample of solid sodium hydroxide was dissolved in a

calorimeter in 100.0 g of water, the temperature rose from 23.9 C to 32.0 C. Calculate H (in kJ/mol NaOH) for the solution process:

NaOH (s) ----> Na+(aq) + OH (aq)Assume it's a perfect calorimeter and that the specific heat of the solution is the same as that of pure water.

Thermochemistry

Enthalpy Example• When 1 mole of methane is burned at constant pressure, 890

kJ of energy is released as heat. Calculate H for a process in which a 5.8 g sample of methane is burned at constant pressure.

Thermochemistry

Constant Pressure CalorimetryWhen 1.00 L of 1.00 M Ba(NO3)2 solution at 25.00C is mixed with 1.00 L of 1.00

M Na2SO4 solution at 25.00C in a calorimeter, the white solid BaSO4 forms, and the temperature of the mixture increases to 28.10C. Assuming that the calorimeter absorbs only a negligible amount of heat, that the specific heat capacity of the solution is 4.18 J/0C*g, & the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of BaSO4 formed.

Thermochemistry

Bomb Calorimetry

Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water.

Thermochemistry

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

Bomb Calorimetry

Thermochemistry

Constant Volume or Bomb CalorimetryIt has been suggested that H gas obtained by the decomposition of water might

be a substitute for natural gas (principally methane). To compare the energies of combustion of these fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/0C. When a 1.50 g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.30C. When a 1.15 g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.30C. Compare the energies of combustion per gram for H & CH4

Thermochemistry

Calorimetry Continued• A 110. g sample of copper (specific heat capacity = 0.20 J/0C*g)

is heated to 82.40C & then placed in a container of water at 22.30C. The final temperature of the water & the copper is 29.40C. What is the mass of the water assuming that all the heat lost by the copper is gained by the water?

Thermochemistry

Calorimetry Continued• Consider the dissolution of CaCl2 :

• An 11.0 g sample of CaCl2 is dissolved in 125 g of water, with both substances at 25.00C. Calculate the final temperature of the solution assuming no heat loss to the surroundings & assuming the solution has a specific heat capacity of 4.18 J/0C*g.

kJHClCaCaCl aqaqs 5.812 )(2

)()(2

Thermochemistry

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 H values that are published and the properties of enthalpy.

Thermochemistry

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.”

Thermochemistry

Hess’s Law

Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Thermochemistry

Hess’s Law I• Two forms of carbon are graphite & diamond. Using the enthalpies of

combustion for graphite (-394 kJ/mol) & diamond (-396 kJ/mol), calculate H for the conversion of graphite to diamond:

)()( sdiamondsgraphite CC

Thermochemistry

Hess’s Law II• Diborane (B2H6) is a highly reactive boron hydride that was once considered

as a possible rocket fuel for the US space program. Calculate H for the of synthesis of diborane from its elements according to the equation

• Using the following data)(22)(2)( 32 ggs HBHB

)(2)(2

)(2)(2)(2

)(2)(32)(2)(62

)(32)(2)(

)(2

1)(

33)(2

32)(

gl

lgg

gsgg

sgs

OHOHd

OHOHc

OHOBOHBb

OBOBa

H

-1273 kJ

-2035 kJ

-286 kJ

44 kJ

Thermochemistry

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.

Thermochemistry

Standard Enthalpies of Formation

Standard enthalpies of formation, Hf, are measured under standard conditions (25°C and 1.00 atm pressure).

Thermochemistry

Calculation of H

• Imagine this as occurringin 3 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

Calculation of H

• Imagine this as occurringin 3 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

Calculation of H

• Imagine this as occurringin 3 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

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

Calculation of H

We can use Hess’s law in this way:

H = nHf(products) - mHf(reactants)

where n and m are the stoichiometric coefficients.

Thermochemistry

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

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

Thermochemistry

Standard Enthalpies of Formation I• Using the standard enthalpies of formation in the table below, calculate the

standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide & water.

)(2)(2)(2)(3 6474 lggg OHNOONH Compound Hf0(kJ/mol)

NH3(g) -46

NO2(g) 34

H2O (l) -286

Al2O3 (s) -1676

Fe2O3 (s) -826

CO2(g) -394

CH3OH (l) -239

C8H18(l) -269

Thermochemistry

Standard Enthalpies of Formation II• Using the standard enthalpies of formation in the table below, calculate the

standard enthalpy change for the thermite reaction

• This reaction occure when a mixture of powdered

aluminum and iron (III) oxide is ignited with a

magnesium fuse.

)()(32)(32)( 22 ssss FeOAlOFeAl Compound Hf0(kJ/mol)

NH3(g) -46

NO2(g) 34

H2O (l) -286

Al2O3 (s) -1676

Fe2O3 (s) -826

CO2(g) -394

CH3OH (l) -239

C8H18(l) -269

Thermochemistry

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

Thermochemistry

Fuels

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