Focus 4

Thermodynamics: The First Law

Chemistry 200

Thermodynamics, is the study of Work and Heat are Energy

EnergyWork Heat

James Joule, English physicist, showed that both heat and work are forms of energy; now know as the two fundamental concepts of thermodynamics.

Sadi Carnot, a French engineer, believed heat flowed to produce work, just as water flowed to turn a water wheel .

Heat was thought of as a fluid, flowing from a hot to cold substance.

Heat & Work

Thermodynamics studies how energy is transformed from one form into another and transferred. Electricity generated in a power station is used a great distance away.

The System The Surroundings (transformed) (transferred)

To keep track of changes in energy, we always separate these two components.

Systems verses the Surroundings

Everything around the system, such as the water bath in a reaction is immersed, is called the surroundings. The surroundings are where we make observations on energy transfer into or out of the system.

Systems verses the Surroundings

The system and the surroundings jointly make up the universe.

You can “Open” the System to the surroundings (P, V, T all change)

You can “Close” the System to the surroundings (put a lid on it-constant V).

You can “Isolate” the System by preventing temperature from escaping (constant Temp=isothermal).

Universe

Systems verses the Surroundings

A closed system has a fixed amount of matter, but it can exchange energy with the surroundings, i.e. a cold pack used for treating athletic injuries.

An open system can exchange both matter and energy with the surroundings, i.e. automobile engines and the human body.

Isolated sys tems have no contact with its surroundings. A sealed, insulated, rigid container; i.e. a thermos.

Systems verses the Surroundings

Work is motion against an oppos ing force.

Work = opposing force × distance moved

The SI unit for work (energy), is joule, J,

where 1 J = 1 kg·m2·s-2

Work & Energy

Work is done when:

• Raising a weight against the pull of gravity;

• A chemical reaction in a battery does work when it pushes an electric current through a circuit;

• A gas in a cylinder does work when it expands and pushes back a piston.

Work & Energy

In thermodynamics, the capacity of a system to do work is called its internal energy, U.

Internal energy (U): sum of the kinetic and potential energies.

Absolute internal energy is not measurable because it includes the energies of all the atoms, their electrons, and the components of their nuclei .

The best we can do is to measure changes in internal energy or ΔX:

ΔX = Xfinal - Xinitial

Internal energy, U

If a system does 15 J of work, it used up some of its store of energy, so we say its internal energy has fallen by 15 J and write ΔX = -15 J (when your body burns food).

A negative value of ΔX, or ΔU - 15 J, means X has decreased.

If we do work on the system, lets say 15 J of work, we say its internal energy has increased by 15 J and write ΔX = +15 J (winding a spring increases U of the system, or we eat a sandwich).

A positive value of ΔX, or ΔU + 15 J, means X has increased.

Internal energy, U

Expansion work is a change in volume of a system. • Expansion work is when a gas expands in a cylinder pushing back a

piston which pushes out against the atmosphere.

Nonexpansion work is the flow of electrical current.• chemical reactions do nonexpansive work by causing electrical current to

flow.

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Work: Expansive and nonexpansive

We now introduce two forms of Expansive Work (W = - PΔV):

1. Expansion at constant pressure;

2. Expansion at constant temperature with changing Pexternal.

EnergyWork Heat

Expansive Work

The opposing force is the external pressure pushing against the outer face of the piston.

How much expansion work is done is when the system expands, ΔV, against the external pressure Pex.

Work = opposing force × distance moved

Work is done when the gas expands, the piston pushes against the opposing force (atmospheric pressure).

Expansive Work

work = PexA × Δ d A × Δ d = ΔV

The system does expansive work by pushing against the opposing pressure (force), so the system losses energy as work: ΔU = “-.”

• so if ΔV is positive (an expansion), w is negative; • so if ΔV is negative (contraction), w is positive.

Volume

ΔU = “-”

Expansive Work

w = -Pex ΔV

“-” because it is assumed work will be done so internal energy will decrease.

External pressure is expressed in Pascal's:

1 Pa = 1 kg·m-1·s-2

Change in External Pressure, by a cubic meter (m3) is:

1 Pa · 1 m3

or 1 kg·m-1·s-2·m3 = 1 J

101.323 J = 1 L·atm,

SI units for Pressure

w = -Pex ΔV

If the external pressure is zero (a vacuum), than w = 0

No expansion work is done in a vacuum because there is no opposing force, there is nothing to push against.

Expansion against zero pressure is called free expansion.

Work: Free Expansion

Practice 1: Suppose a gas expands by 500. mL (0.500 L) against a pressure of 1.20 atm. How much work is done in the expansion in Joules?

w = -Pex ΔV

w = - 1.20 atm × 0.500 L = - 0.600 L·atm

Since 101.323 J = 1 L · atm

- 0.600 L · atm = - 60.8 J

w = -60.8 J

ΔU = -60.8 J

Work is a measure of P and V changes.

Measuring work where V changes in either direction as a "reversible" expansion must be done in "infinitesimal" changes.

Of course this means that pressure increases or decreases infinitesimally as well.

These reversible processes are of the greatest importance in thermodynamics, and we shall encounter them many times.

Changing external pressure

The simplest reversible change is an isothermal (constant temperature) expansion of an ideal gas.

The criteria for isothermal expansion is the temperature remains unchanged. The reason is that temperature implies change in motion of atoms so an "infinitesimal" change in P or V is impossible if the atoms speed up or slow down during the expansion (contraction).

An ice bath can maintain constant temperature.

Measuring isothermal work with changing external pressure

wiso = -nRT ln

Measuring isothermal work with changing external pressure

Reversible, isothermal work (constant temperature)

Practice 2: A piston confines 0.100 mol Ar (g) in a volume of 1.00 L at 25°C. Two experiments are performed. 1. The piston is allowed to expand through 1.00 L against a constant

pressure of 1.00 atm. 2. The piston is allowed to expand reversibly and isothermally to the

same final volume.

Which process does more work?

We know how to calculate the amount of work two ways:

1. For expansion against constant external pressure:

w = -Pex ΔV

2. For reversible, isothermal expansion: w = -nRT ln

1. we use w = -Pex ΔV = 1.00 atm × 1.00 L × = - 101 J

2. isothermal expansion: w = -nRT ln

= - 0.100 mol × 8.3145 J·K-1·mol-1 × 298 K × ln = -172 J

The reversible process does more work (yellow) than constant pressure (green).

Practice 2: A piston confines 0.100 mol Ar (g) in a volume of 1.00 L at 25°C. Two experiments are performed. In one, the piston is allowed to expand through 1.00 L against a constant pressure of 1.00 atm. In the second, it is allowed to expand reversibly and isothermally to the same final volume. Which process does more work?

Is there a difference between heat and temperature?

Heat and Temperature

Heat is what Physically happens when two objects at different temperatures touch.

Energetic molecules in higher temperature regions vigorously stimulate slower moving molecules in the lower temperature region into higher energetic states.

Heat: Flow of energy due to a T difference.

ColdWater

HotWater

This results in a decrease for higher energy molecules (they become cooler) and an increase the lower energetic molecules (they become warmer).

Temperature is measured with a thermometer, and it must come into contact with an object. Fluid inside the thermometer, often mercury or alcohol, reacts to changes in the energetic molecules it makes contact with.

Heat and Temperature

SurroundingsSurroundings

SystemSystem

EnergyEnergy

ExothermicEndothermic

U 0U 0

U = q + w

Energy flows into system via heat (endothermic): q = +x

Energy flows out of system via heat (exothermic): q = -x

System does work on surroundings: w = -x U < 0Surroundings do work on the system: w = +x U > 0

Heat & Work

So far, we have considered the transfer of energy as work or heat separately.

Many processes, involve changes in internal energy that are both work and heat.

When a spark ignites the mixture of gasoline vapor and air in the engine of a moving automobile, the vapor burns and expands, transferring energy to its surroundings as both work and heat.

ΔU = q + w

The first law of thermodynamics

Law of conservation of energy: energy of the universe is constant.

Practice 6: A system was heated by using 300. J of heat, yet it was found that its internal energy decreased by 150. J (the system loses it’s internal energy). Calculate was work done on the system or did the system do work?

ΔU = q + w

ΔU = -150. J lost

q = +300. J added to system

w = ΔU - q or -150. J - (+300. J) = - 450. J

The “-” means work lost by the system or the system did – 450 J of work.

Heat

units of heat: calorie (cal) English system

joule (J) SI system

1 cal = 4.184 J

Joule: Energy (heat) required to raise T of one gram of water by 1C.

Food energy is measured in Calories (note the capital C).

1 Cal = 1 kcal = 1000 cal

Amount of heat = specific heat capacity × mass × change in temperature

q = Cp × m × (Tfinal – Tinitial)

Tfinal = final temperature (ºC)

Tinitial = initial temperature(ºC)

Cp = Specific heat capacity (cal/g °C or J/g °C)

m = mass (g)

Heat

• Specific heat capacity is the energy required to change the temperature of a mass of one gram of a substance by one Celsius degree.

Heat

specific heat, Cp = where m = molar mass, and

molar heat capacity, Cpm = where n = moles.

For Water the specific heat is 4.18 J·(°C)-1·g-1 or 4.18 J·K-1·g-1 and the molar heat capacity is 75 J·K-1·mol-1 .

Also commonly used in the following form:

specific heat capacity: q = m Cp

molar heat capacity: q = n Cpm

Two forms of heat capacity

• A silver-gray metal weighing 15.0 g requires 133.5 J to raise the temperature by 10.°C. Find the heat capacity.

q = Cp x m x T

(133.5 J) = Cp x (15.0 g) x (10.°C)

Cp = 0.89 J/g°C

Can you determine the identity of the metal using

Table 10.1?

HeatPractice 3:

Al

q = Cp = n Cpm

q = 2.00 mol × 75 J·K-1·mol-1 × 20. K = + 3.0 kJ (“+” adding heat)

Common student error:

When working with , it’s important for the student to remember that

(°C) = (K).

In (a) above, = 20.°C, is the same as= 20. K

(°C) = (K)

0.0 °C 100.0 °C = 100.0 °C

273.0 K 373.0 K = 100.0 K

Practice 4: Calculating the heat needed to bring about a rise in temperature Calculate the heat necessary to increase the temperature of 2.00 mol H2O(l) by 20.°C from room temperature. (Cpm = 75 J.K-1.mol-1)

1. A calorimeter has a known Ccal (calorimeter constant).

2. When a sample is placed in the calorimeter, heat from the experiment goes into the calorimeter.

3. of the calorimeter, is the heat lost by the experiment into the calorimeter.

4. By knowing information about the calorimeter (1) and temperature changes (3) we can calculate,

q = Ccal , the heat form our experiment.

A calorimeter is a common device to measure energy transferred as heat.

Calorimeter

The sample generates heat. Then, that heat is absorbed the calorimeter (qcal = -qexp).

Calorimeters are calibrated with a known specific heat, usually water.

Calibration means finding the calorimeter constant Ccal.

36A more sophisticated version made of metal, called a bomb calorimeter.

Heat from the Experiment = Heat of the Calorimeter.

Calorimeter

Practice 5: A constant-volume calorimeter was calibrated by carrying out a reaction known to release 1.78 kJ of heat in 0.100 L of solution in the calorimeter, resulting in a tempera ture rise of 3.65°C. Next, 50. mL of 0.20 M HCl(aq) and 50. mL of 0.20 M NaOH(aq) were mixed in the same calorimeter and the temperature rose by 1.26°C. What is the change in the internal energy of the neutralization reaction?

First, we calibrate or find the amount of heat absorbed by the calorimeter.

qcal = Ccal or Ccal = plug in our data,

Ccal = = 0.488

Second, we find the heat the sample generated, we do this by measuring the amount of heat was absorbed by the calorimeter.

Here we notice the calorimeter’s temperature rose by 1.26°C

qcal = Ccal plugging in our data, 0.488 = 0.614 kJ

For our reaction, heat is leaving the reaction so q = ΔU = - 0.614 kJ

The first law of thermodynamics: The internal energy of an isolated system is constant.

If we were to prepare a second system consisting of exactly the same amount of sub stance in exactly the same state as the first, and inspect that system, we would find that the second system has the same internal energy as the first system.

A property that depends only on the current state of the system and is independent of how that state was prepared, is called a State Function.

State functions

Pressure, volume, temperature, mass and altitude of a system are also state functions, and so is internal energy.

Work is not a state function, the physical exertion of a climber climbing path A or B requires a wildly different amount (and kind) of “work” by the climber; work = P∆V

State functions

If in one step we raised its temperature, its internal energy would change by a certain amount.

In multiple steps of raising and lowering the temperature, we use a different amount of energy, but the change is the same.

The net change (A to B) in the internal energy in each step is exactly the same.

In the next example internal energy, ΔU = q + w, is calculated by two different paths, both over the same range.

1. One path calculates internal energy only reversibly,

2. The other path is a combination of both reversible and irreversible.

The biggest difference between the two is for a reversible process the temperature remains constant (isothermal).

Also remember that all isothermal expansions are the most efficient (though not common).

State functions

Practice 7: Suppose that 1.00 mol of ideal gas molecules at 292 K at 3.00 atm expands from 8.00 L to 20.00 L and has a final pressure of 1.20 atm. Find the amount of work by two different paths.

(a) Path A is an isothermal, reversible expansion.

(b) Path B has two parts. In step 1, the gas is cooled at constant volume until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed to expand against a constant pressure of 1.20 atm until its volume is 20.00 L and T = 292 K. Determine for each path the work done, the heat transferred, and the change in internal energy (w, q, and ΔU).

Isothermal: w = -nRT ln

Practice 7: Path B has two parts. In step 1, the gas is cooled at constant volume until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed to expand against a constant pressure of 1.20 atm until its volume is 20.00 L and T = 292 K. Determine for each path the work done, the heat transferred, and the change in internal energy (w, q, and ΔU).

Step 2, let V change V↑T↑,keeping P constant

Step 1, cool, keeping V constant. P↓T↓

Practice 7: Suppose that 1.00 mol of ideal gas molecules at 292 K and 3.00 atm expands from 8.00 L to 20.00 L and a final pressure of 1.20 atm by two different paths. (a) Path A is an isothermal, reversible expansion.

Part A: ΔU = q + w

For reversible, isothermal expansion:

w = -nRT ln 1.00 mol × 8.3145 J·K-1·mol-1 × 292 K × ln = -2.22 kJ

ΔU = q + w, note for an isothermal expansion ΔU = 0 (this is just a gas expanding, so it is not a chemical reaction losing energy).

0 = q + w or q = -w so here q = - (-2.22 kJ)

= +2.22 kJ

Practice 7: Path B has two parts. In step 1, the gas is cooled at constant volume until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed to expand against a constant pressure of 1.20 atm until its volume is 20.00 L and T = 292 K. Determine for each path the work done, the heat transferred, and the change in internal energy (w, q, and ΔU).

Part B: ΔU = q + w

Step 1 work: cooling at constant V, no work or w = 0

Step 2 work: w = -Pex ΔV, ΔV = (20.00 – 8.00 )L = 12.00 L

w = -Pex ΔV = 1.20 atm × 12.00 L × = - 1.46 kJ

Step 1 + Step 2 = 0 + -1.46 kJ = - 1.46 kJ

ΔU = q + w, note for an isothermal expansion ΔU = 0 (same as (a)).

0 = q + w or q = -w so here q = - (- 1.46 kJ) = +1.46 kJ

1

2

Practice 7: Suppose that 1.00 mol of ideal gas molecules at 292 K and 3.00 atm expands from 8.00 L to 20.00 L and a final pressure of 1.20 atm by two different paths. (a) Path A is an isothermal, reversible expansion. (b) Path B has two parts. In step 1, the gas is cooled at constant volume until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed to expand against a constant pressure of 1.20 atm until its volume is 20.00 L and T = 292 K. Determine for each path the work done, the heat transferred, and the change in internal energy (w, q, and ΔU).

Isothermal, reversible does more work, +2.22 kJ.

Nonreversible does less work, +1.46 kJ.

ΔU = 0 for both

In a chemistry lab, many reactions are done in open contain ers, under nonexpansive work (ΔV = 0) conditions at a constant pressure of 1 atm.

ΔU = q + w , and for nonexpansion work w = 0

So at constant volume: ΔU = q

In this case energy supplied is strictly heat, q only.

Enthalpy

For a reaction in constant pressure,the change of enthalpy is equal to energy that flows as heat.

Hp = heatConstant pressure

Enthalpy (Thermochemistry): heat of chemical reactions.

Heat Transfers at Constant Pressure:

The new state function is called enthalpy, H:

H = U + PV

U, P, and V are the internal energy, pressure, and volume of the system.

Enthalpy is a state function because U (from the first law), P, and V are also state functions.

Constant pressure means the system is open, and the work is pushing against the atmosphere; that is the constant pressure is the constant pressure of the atmosphere.

Enthalpy

It follows from enthalpy, H = U + PV, that changes at constant pressure: ΔH = ΔU + PΔV

We substitute for ΔU = q + w , to get, ΔH = q + w + PΔV

A system doing expansion work only, we substitute for w = -Pex ΔV

to get:

Enthalpy

Now, if we leave our system open to the atmosphere, the pressure of the system is the same as the external pressure; so Pex = P, and ΔH = q + -PΔV + PΔV, the last two terms cancel to give us:

ΔH = q (constant pressure)

Previously: ΔU = q (constant volume)

ΔH = q + -Pex ΔV + PΔV

Surroundings

System

Energy

Exothermic

Exothermic: H is negative.(heat flows out of the system).

Surroundings

System

Energy

Endothermic

Endothermic: H is positive.(heat flows into the system).

Enthalpy

Remember that ΔH is a constant pressure (open to the atmosphere) scenario.

(PE)

Reactant

Product

Energy released to the surroundings as heat

SurroundingExothermic (burning)

Exothermic = exit!

The enthalpy of a system is like a measure of the height of water in a reservoir. When a reaction releases 208 kJ of heat at constant pressure, the "reservoir" falls by 208 kJ and ΔH = -208 kJ.

Enthalpy

Practice 8: In an endothermic reaction at constant pressure, 30. kJ of energy entered the system as heat. The products took up less volume than the reactants, and 40. kJ of energy entered the system as work as the outside atmosphere pressed down on it. What are the values of (a) ΔH and (b) ΔU for this process?

(a) ΔH = q at constant pressure

The reaction absorbs energy so ΔH = + 30.0 kJ

(b) ΔU = q + w, where w = -Pex ΔV ,

since ΔV is Vfinal – V initial, and here Vfinal < V initial so ΔV is “-”

then w = - 40.0 kJ × “-” = + 40.0 kJ

ΔU = + 30.0 kJ + (+ 40.0 kJ) = + 70.0 kJ (an increase in U)

Phase changes approximate how far apart molecules are.

Vaporization, requires adding energy to break and separate intermolecular attractions, an endothermic process.

Freezing, is exothermic because energy is given off to allow new intermolecular attractions to form between molecules.

Phase changes usually take place at constant pressure, the heat transfer is due to changes in enthalpy.

The Enthalpy of Physical Change

Phase change: liquid to a gas or vaporizing is called the enthalpy of vaporization, ΔHvap:

Enthalpy of vaporization, ΔHvap = Hm(vapor) - Hm(liquid)

Where ΔHm is the molar heat

For water at its boiling point, 100°C, ΔHvap = 40.7 kJ·mol-1, and at 25°C the value is ΔHvap = 44.0 kJ·mol-1.

This means to vaporize 1.00 mol H2O(l) (18.02 g of water) at 25°C and constant pressure, requires 44.0 kJ of energy as heat.

Enthalpy of vaporization

All enthalpies of vaporization are positive (endo). Stronger intermolecular forces, such as hydrogen bonds, have the highest enthalpies of vaporization.

Practice 9: Heating a sample of ethanol, C2H5OH, of mass 23 g to its boiling point, it was found that 22 kJ was required to vaporize all the ethanol. What is the enthalpy of vaporization of ethanol at its boiling point?

Enthalpy of vaporization, ΔHvap = Hm(vapor)

Where Hm is the molar heat, so we need to find kJ·mole-1.

46.069 g·mol-1 C2H5OH.

= 0.50 mole C2H5OH

= 44 kJmol-1 per 1 mole C2H5OH

Phase change: solid to a liquid or melting (fusion) is called the enthalpy of fusion,

ΔHfus = Hm(liquid) - Hm(solid)

The enthalpy of fusion of water at 0.0 °C is 6.0 kJ·mole-1: to melt 1.0 mol H2O(s) (18 g of ice) at 0.0 °C, we have to sup ply 6.0 kJ of heat.

Vaporizing water takes much more energy (more than 40 kJ) because to vaporize a gas, its molecules are separated completely, and KE increases dramatically. In melting, the molecules stay close together, and so the forces of attraction and repulsion are nearly as strong as in the solid.

Enthalpy of fusion

Freezing is the change from liquid to solid.

Enthalpy is a state property:

ΔHreverse process = - ΔHforward process

Since the enthalpy of fusion of water at 0.0°C is 6.0 kJ·mole-1 the enthalpy of freezing for water is at 0.0°C is - 6.0 kJ·mole-1

Condensation is the change from gas to liquid. So this process is the reverse of ΔHvap. For water at boiling, 100°C, ΔHvap = 40.7 kJ·mol-1 , therefore

ΔHcondensation = - 40.7 kJ·mol-1

Enthalpy of freezing

Enthalpy of condensation

Sublimation is the direct conversion of a solid into its vapor.

ΔHsublimation = Hm(vapor) - Hm(solid)

Frost disappears on a cold, dry morning as the ice sublimes directly into water vapor. Solid carbon dioxide also sublimes, which is why it is called dry ice.

Deposition is the reverse of sublimation a vapor into its solid.

ΔHdeposition = Hm(solid) - Hm(vapor)

Enthalpy of sublimation

Enthalpy of deposition

A heating curve shows the variation in the temperature of a heated sample.

As a sample of ice is heated the temperature rises steadily.

1

2

1. In a solid the molecules are still locked together, as they oscillate vigorously around their mean positions.

2. As the temperature rises reaching the melting point, the molecules have enough energy to move past one another, overcoming the attractive forces between molecules.

Heating Curves

3. At this temperature, all the added energy is used break all the attractive forces, the temperature remains constant at the melting point until all the ice has melted.

1

2 34

4. Only then does the temperature rise again, and the rise continues right up to the boiling point.

5. At the boiling point, the tempera ture rise again comes to a halt. 6. Now the water molecules have enough energy to escape into the vapor state, and all the heat supplied is used to form the vapor.

5 6

Heating Curves

A. Steeper slopes mean lower the heat capacity. Both the solid and vapor phases have steeper slopes so the liquid has a greater heat capacity.

B. The liquids high heat capacity is due largely to loose intermolecular hydrogen bonds that can take up more energy than stiff chemical bonds between atoms found in the solid.

7. After the sample has evaporated and heating continues, the temperature of the vapor rises again.

5 67

B

A

A

Heating Curves

Heat supplied

T

empe

ratu

re (°

C)

-7.2

58.78

70.0

ΔHfus

ΔHvap

Practice 10: Use the following information to construct a heating curve for bromine from -7.2°C to 70.0°C. The molar heat capacity of liquid bromine is 75.69 kJ·mole-1 and that of bromine vapor is 36.02 kJ·mole-1. For bromine the specific heat of vaporization 0.225 J·g-1·°C-1 and of liquid is 0.473 J·g-1·°C-1. Bromine melts at -7.2°C and boils at 58.78°C. Also calculate the energy to melt 10.0 g of bromine, Br2.

Hg(gas)

Hg(liquid)

Htotal = ΔHfus+Hg(liq) + ΔHvap+ Hg(gas)

Some H’s are in mole, so 0.0626 mol Br2

Step 1 (melt/fus): 0.0626 mol × × = 4740 J

Step 2 (heat liq): gCΔT: 10.0 g × 0.473 J·g-1·°C-1 × 66.0 °C = 312 J

Step 3 (vap): 0.0626 mol × × = 2250 J

Step 4 (heat gas): 10.0 g × 0.225 J·g-1·°C-1 × 11.2 °C = 25.2 J

ΔHfus 75.69 kJ·mole-1

Cg(liq) 0.473 J·g-1·°C-1

ΔHvap 36.02 kJ·mole-1

Cg(vap) 0.225 J·g-1·°C-1

q = C

Practice 10: Use the following information to construct a heating curve for bromine from -7.2°C to 70.0°C. The molar heat capacity of liquid bromine is 75.69 kJ·mole-1 and that of bromine vapor is 36.02 kJ·mole-1. For bromine the specific heat of vaporization 0.225 J·g-1·K-1 and of liquid is 0.473 J·g-1·K-1. Bromine melts at -7.2°C and boils at 58.78°C. Also calculate the energy to melt 10.0 g of bromine, Br2.

Total: 4740 J + 312 J + 2250 J + 25.2 J = 7330 J

Heat supplied

T

empe

ratu

re (°

C)

-7.2

58.78

70.0

ΔHfus

ΔHvap Hg(gas)

Hg(liquid)

Htotal = ΔHfus+Hg(liq) + ΔHvap+ Hg(gas)

Knowing enthalpies of chemical change help us determine which are good fuels, or to study of biochemical processes.

Every chemical reaction is accompanied by energy transfer as heat.

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(I) ΔH = - 890. kJ

Enthalpy of chemical change (reaction)

The stoichiometric coefficients corresponding to an enthalpy change.

1 mol CH4 (g) produces 890. kJ heat (energy).

S(s) + O2(g) SO2(g) ΔH = –296 kJ

• Calculate the quantity of heat released when 2.10 g of sulfur is burned in oxygen at constant pressure.

Use the H value like a conversion factor.

2.10 g S = 0.0655 mol Sx1 mol S

32.26 g S

0.0655 mol S = – 19.4 kJx– 296 kJ

1 mol S

Enthalpy

Practice 11:

1 mol S(s) produces 296 kJ heat (energy).

Practice 12: When 0.113 g of benzene, C6H6 (78.12 g·mole-1) burns in excess O2 in a calibrated calorimeter with a heat capacity of 551 J·(°C)-1, the temperature of the calorimeter rises by 8.60°C. Write the thermochemical equation and determine the reaction enthalpy.

2 C6H6(g) + 15 O2(g) 12 CO2(g) + 6 H2O(I) H = ?

Total heat from the calorimeter using qrxn = - qcal = CcalΔT

551 J·(°C)-1 × 8.60°C = - 4.74 kJ

This is from 0.113 g, we need to know per mole:

=

- 3.28 × 103 kJ per mole

Our final answer is kJ per mol, though we only write it as kJ, mole is implied.

Hess’s Law

State function: a property of system that changes independently of its pathways.

Enthalpy is a state function.

In a chemical reaction, change of enthalpy is the same whether the reaction takes place in one step or in a series of steps.

N2(g) + 2O2(g) 2NO2(g) H1 = 68 kJ1 Step

N2(g) + O2(g) 2NO(g) H2 = 180 kJ2NO(g) + O2(g) 2NO2(g) H3 = -112 kJ

N2(g) + 2O2(g) 2NO2(g) H2 + H3 = 68 kJ

2 Steps

Two rules about enthalpy

1. If a reaction is reversed, the sign of H is also reversed.

N2(g) + 2O2(g) 2NO2(g) H1 = 68 kJ

2NO2(g) N2(g) + 2O2(g) H1 = -68 kJ

2. If the coefficients in a balanced reaction are multiplied by an integer, the value of H is also multiplied by the same integer.

N2(g) + 2O2(g) 2NO2(g) H1 = 68 kJ

2N2(g) + 4O2(g) 4NO2(g) H1 = 2 68 kJ = 136 kJ 2

Hess's law is a way to find unknown enthalpies.

• Step 1: Select one of the reactants in the overall reaction and write down a chemical equation in which it also appears as a reactant;

• Step 2: Select one of the products in the overall reaction and write down a chemical equation in which it also appears as a product;

• Step 3: Cancel unwanted species in the sum;

• Step 4: Once the sequence is complete, combine the standard reaction enthalpies;

Note: Reversing the equation, changes the sign of enthalpy;

Note: Mul tiplying by a factor and so multiply the reaction enthalpy by the same factor.

Combining Reaction Enthalpies: Hess's Law

Practice 13: Using Hess's law, calculate the standard enthalpy for 3 C(gr) + 4 H2(g) C3H8(g) from the following experimental data: (a) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ΔH° = -2220. kJ

(b) C(gr) + O2(g) CO2(g) ΔH° = -394 kJ

(c) H2(g) + ½ O2(g) H2O(l) ΔH° = -286 kJ

Step 1: Treat carbon as a reactant, (b) and multiply it through by 3.

3 C(gr) + 3 O2(g) 3 CO2(g) ΔH° = 3(-394) kJ = -1182 kJ

Step 2: Reverse equation (a), changing the sign of H.

3 CO2(g) + 4 H2O(l) C3H8(g) + 5 O2(g) ΔH° = +2220. kJ

Simplify by combining the two previous equations.

3 C(gr) + 4 H2O(l) C3H8(g) + 2 O2(g) ΔH° = +1038. kJ

Continued: Overall reaction 3 C(gr) + 4 H2(g) C3H8(g)

(a) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ΔH° = -2220. kJ

(b) C(gr) + O2(g) CO2(g) ΔH° = -394 kJ

(c) H2(g) + ½ O2(g) H2O(l) ΔH° = -286 kJ

Previous:

3 C(gr) + 4 H2O(l) C3H8(g) + 2 O2(g) ΔH° = +1038. kJ

Step 3: To cancel unwanted H2O and O2 , multiply equation (c) by 4.

4 H2(g) + 2 O2(g) 4 H2O(l) ΔH° = 4(-286) kJ = -1144 kJ

Step 4: Combine and simplify

3 C(gr) + 4 H2(g) C3H8(g) ΔH° = -106 kJ

For reactions in which no gas is generated or consumed, little expansion work is done the difference between ΔH and ΔU is negligible; so we can set ΔH = ΔU.

However, if a gas is formed in the reaction and expansion work is done the difference can be significant.

We can use the ideal gas law to relate the values of ΔH and ΔU for gases that behave ideally.

The Relation Between ΔH and ΔU

The relation between ΔH and ΔU, is H = U + PV, or H = U + nRT

ΔH = ΔU + ΔngasRT

Previous: ΔH = ΔU + PΔV

Practice 14: constant-volume calorimeter showed that the heat generated by the combustion of 1.000 mol glucose molecules in the reaction C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) is 2559 kJ at 298K, and so ΔU = -2559 kJ. What is the change in enthalpy for the same reaction?

ΔH = ΔU + ΔngasRT

We need Δngas = 12 mol final – 6 mol initial = + 6 mol

R = 8.314 J·K-1·mol-1

ΔH = -2559 kJ + ( 6 mol × 8.314 J·K-1·mol-1 × 298 K) = -2559 kJ + 14.9 kJ = - 2544 kJ

Notice that ΔH is less negative (more positive) than ΔU for reactions that generate gases: less energy is obtained because some of the energy is used to expand to make room for the reaction products.

The standard reaction enthalpy, ΔH°, is the reaction enthalpy when reactants are in their standard states.

Standard Reaction Enthalpies

Standard state: pure form of a substance at exactly 1 bar.

First calculate the reaction enthalpy for the formation of all products.

Second, calculate the reaction enthalpy for all reactants.

The difference between these two totals is the standard enthalpy of the reaction.

n n

n is the amounts in moles, Σ (sigma) means sum.

Combining standard enthalpies of formation to calculate a standard reaction enthalpy.

Practice 15: You have an inspiration: maybe diamonds would make a great fuel! Calculate the standard enthalpy of combustion of diamonds.

C(s, dia) + O2(g) CO2(g) (kJ·mol-1) = 1.895 0.0 -393.51

n n

= (1 mol ) (CO2(g)) - [(1 mol) (C(s, dia)) + (1 mol ) (O2(g)]

= (1 mol ) (-393.51 kJ·mol-1) – [(1 mol) (1.895 kJ·mol-1) + (1 mol) (0)]

= -393.51 - 1.895

= -395.41 kJ (a little better than ethanol made from corn)

In focus 2 we calculated energy changes in an ionic system to find lattice energy.

The difference in molar enthalpy of solid and gas is called the lattice enthalpy of the solid, ΔHL.

The lattice enthalpy of a solid cannot be measured directly.

Instead we measure it indirectly by using an application of Hess's law.

The procedure uses a Born-Haber cycle, a closed path of steps.

The Born-Haber Cycle

In a Born-Haber cycle, we

a) break apart the bulk elements into atoms,

b) ionize the atoms,

c) combine the gaseous ions to form the ionic solid,

d) then form the elements again from the ionic solid,

a

b

c

d

The sum of a complete Born-Haber cycle is zero, because the enthalpy of the system must be the same at the start and finish.

Note: Lattice energies are “-”, though when calculated with the Born-Haber cycle the value will be positive.

The Born-Haber Cycle

Practice 16: Devise and use a Born-Haber cycle to calculate the lattice enthalpy of potassium chloride.

1. form gaseous atoms, 2. ionize the atoms to form gaseous ions, 3. allow the ions to form an ionic solid, 4. and then convert the solid back into the

pure elements.

2

1

4 3

1

2

(s)

Begin with the pure elements K(s) + ½ Cl2(g)

1. atomize them to form gaseous atoms,

K(s) K(g) = + 89 kJ·mol-1

½ Cl2(g) Cl(g) = + 122 kJ·mol-1

2. ionize the atoms to form gaseous ions,

K(g) K+ (g) (1st ionization) = + 418 kJ·mol-1

Cl(g) Cl- (g) (electron affinity) = - 349 kJ·mol-1

3. allow the ions to form an ionic solid,

K+ (g) + Cl- (g) KCl(s) = -ΔHL

4. then convert the solid back into the pure elements.

K(s) + ½ Cl2(g) = -(-437 kJ·mol-1)

note sign is “-” because the reaction is reversed.

All the steps together sum to 0.

+ 89 kJ·mol-1 + 122 kJ·mol-1 + 418 kJ·mol-1 - 349 kJ·mol-1 + ΔHL -(-437 kJ·mol-1) = 0

Solve for -ΔHL = + 717 kJ·mol-1

81

(s)