Chapter 5. Thermochemistry.

Chapter 5. Thermochemistry.

Thermochemistry deals with changes in energy that occur in chemical reactions. The study of energy and its transformations is known as thermodynamics.

5.1 The nature of energy.Different types of energy are kinetic energy and potential energy.

Different forms of energy are interconvertible:

Potential Energy Kinetic Energy Chemical Energy

Kinetic energy (Ek)

Ek = ½ mv2

Where m is the mass of the object in kg, and v is its speed in m/s (meters per second, also written as m.s-1).

Kinetic energy is important in chemistry because molecules are constantly in motion and so have kinetic energy.

Units of Energy:The unit of energy is the joule (J). This is the energy required to move an object weighing 2 kg at 1 m per second.

units of joules = kg x m2

s2

Example. What is the energy in J of an O2 molecule moving at 200 m/sec?

Mass of O2 molecule = 2 x 16.0 = 32.0 amu (1 kg = 1000 g)

One amu = 1.66 x 10-24 g = 1.66 x 10-24 g x _1 kg_ 1000 g

= 1.66 x 10-27 kg.

Mass of O2 molecule = 32.0 x 1.66 x 10-27 kg

= 5.31 x 10-26 kg

Energy in Joules is calculated as:

Ek = ½ mv2

Ek = ½ x 5.31 x 10-26 kg x (200 m/sec)2

= 1.06 x 10-21 J.

What is the kinetic energy of a person (50kg) movingat a speed of 1m/s?:

mass velocity

2

2

1mvEk

2

1)50(21

sm

kg

2

225

s

mkg J25

Weight of person

Walking speed

example

What is the kinetic energy in Joules of a 45 g golf-ball moving at 61 m/s?

Ek = ½ mv2

Note: m has units of kg, v of m/s

45 g = 45 g x 1 kg/1000 g = 0.045 kg.

Ek = ½ x 0.045 kg x 61 m x 61 m = 83.7 J. s s

What happens to this energy when the ball lands in a sand-trap? Ans. It is converted to heat.

unitscheckout

Units of energy

A joule is a very small amount of energy, and so one commonly uses the kJ. Energies for bonds are usually expressed in kJ/mol.

The calorie: Amount of energy required to raise temperature of 1 g of water by 1 ºC.

Use kcal.mol-1 (=kcal/mol) for bonds.

Nutritional Calorie (note upper case) = 1000 cal.

Types of energy

Potential energy: Object has this by virtue of its position.

Electrostatic energy (not covered in CHM 101). Potential energy due to electrostatic attraction or repulsion.

Chemical Energy: Due to arrangement of atoms, e.g. gasoline, glucose

Thermal Energy: Due to kinetic energy of molecules.

Calculation of work or potential energy:The potential energy equals the work done to raise the object to the height it is above the ground. e.g. a 5.4 kg bowling ball is raised to a height of 1.6 m above the ground. What is its potential energy? Note: The force is the gravitational constant is g = 9.8 m/s2.

Work = m x g x d= 5.4 kg x 9.8 m x 1.6 m

s2

= 85 kg.m2/s2 = 85 J

(check: J = units of kg.m2/s2 so ourcalculation produces the right units).

UnitsCheckout

System and surroundingsIn thinking about thermodynamics, we cannot think about the whole universe at one time. We have to think about the system of interest to us, which for chemistry is usually the contents of something the size of a beaker. Thermodynamics is the book-keeping of energy, and so we are concerned with how much heat goes in or out of the system from the surroundings.

an example of asystem – a beakerplus a solution

energyout

energyin

system

everything else= ‘surroundings’

A system is like a bank account –see below

The System and its Surroundings:

Energy can be transferred from the system to the surroundings

Energy can also be transferred from the surroundings to the system

Heat is transferred from the hotter to the colder object

… until their temperatures are equal

OR

Hot coffee Cold soda

heatout

heatin

The sign of the loss or gain of energy:

We are interested in how much energy goes in or out of a system because this is what causes a chemical reaction to take place. If energy is lost from the system into the surroundings the sign of the energy change is negative, and if energy is gained, it is positive.

system

energy out= negative

energy in= positive

the signs of the energychanges are rather like a bank account: +ve for moneyin, -ve for money out

A system is like your bank account. You only worry about what goes in or out of it, not what happens to the money in the surroundings, i.e. the rest of the world.

Work and Heat:

Energy can be transferred from one object to another either as work or as heat.

Energy used to make an object move against a force is called work.

w = F x d ( work = force x distance)

Heat is energy transferred from a hotter to a colder object.

Energy can also be transferred as work (w)

When an object is moved by a force, F, over a distance, d, energy (work) is transferred

dFw

Energy can be transferred as heat (q)

the soccer playeris doing work on the ball

energy

ENERGY IS THE CAPACITY TO DO WORK OR TO TRANSFER HEAT.

Heat is the non-ordered transfer of energy due to random collisions between particles, whereas work is the ordered transfer of energy.

5.2. The first Law of Thermodynamics.

Energy is conserved

This means that energy cannot be created or destroyed, but only converted from one kind of energy to another.

Internal energy

The internal energy of a system (E) is the sum of all kinetic and potential energies

We don’t know the internal energy of the system, and can generally only calculate ΔE, the change in E that accompanies a change in the system.

ΔE = Efinal - Einitial

kinetic energy isenergy of moleculesrapidly moving about

moleculesalso havevibrationalenergy

system

Relating ΔE to heat and work.

ΔE = q + w

Where:

q is the heat transferred to the system, and

w is the work done on the system

Heat transferred into the system, and work done on the system, are positive.

Practice exercise:

Calculate the change in internal energy where the system absorbs 140 J of heat from the surroundings, and does 85 J of work on the surroundings:

q = + 140 J (absorbs heat = +ve)w = - 85 J (does work on______________ surroundings = -ve)

ΔE= + 55 J ______________________

Endothermic and Exothermic processes:

When a chemical process absorbs heat as it occurs, it is referred to as endothermic. When heat is given off, it is exothermic.

The burning of H2

in O2 is exothermic,because a large amount of heat isgiven off:

2 H2 + O2 2 H2O + heat

Exothermic / Endothermic Processes

H2O (l) H2O (g)

Endothermic:system gains heat

H2O (l) H2O (s)

Exothermic:system loses heat

Water Water vapor

Water ice

State functions.

The value of a state function depends only on the present conditions, not on how it got there. Examples of state functions are temperature and ΔE, the change in internal energy. Q and w are not state functions, because one can get to a particular value of ΔE by a variety of combinations of q and w. E itself is also a state function.

ΔE = q + w

The same valueof ΔE can beachieved e.g. with

large q and smallw, or small q and

large w

The height difference between Denver and Chicago is a state function.

The height difference between Denver and Chicago is astate function because it is independent of the route taken to travel from one to the other. The travel distance is not a state function because one can travel by different routes.

Route B

Route A

Denver

Chicago

4684 ft

Work done in a chemical reaction:

The work done in a chemical reaction at constant pressure is given by PΔV, where V is the change in volume during the reaction. For a reaction involving a gas, this can be a considerable contribution.

} increase in volume due to H2 gasgiven off

Bubbles of H2HClHCl

Znmetal

Zndissolving

air

piston

H2 formedplus air

Enthalpy (ΔH):

If a reaction is carried out at constant P, which is true for all reactions open to the atmosphere, e.g. in a beaker, then the work done by the system is equal to -P ΔV. However, the change in volume of a solution will generally be very small, and so this can be ignored. The heat given off or absorbed during a reaction at constant pressure is known as the enthalpy, and is given the symbol ΔH.

Enthalpy (ΔH):

It can be shown that the change at constant pressure is given by

ΔH = qp

where the subscript ‘p’ denotes constant pressure. When ΔH is positive, the system has gained heat from the surroundings, and is endothermic. When ΔH is negative, the process is exothermic.

5.4. Enthalpies of reaction.

In a thermochemical equation, the heat of reaction for the equation is written as:

2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ

This is the heat given off when 2 moles of H2 combine with one mole of O2 to give 2 moles of water, all in the gas phase. Note that a large negative value of ΔH such as we have here is associated with a very exothermic reaction.

1. Heat is an extensive property

With an extensive property such as heat, the amount of heat given off is proportional to the amount of substance reacted.

small log log twice as bigburning = twice as much heat

Enthalpy is an extensive property. If we burn one mol of H2 with ½ mol of O2, we will get 483.6/2 = -241.8 kJ, as shown below:

2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ2 moles 1 mole ΔH = -483.6 kJ1 mole ½ mole ΔH = -241.8 kJ

Factor = _____moles we have______ moles in balanced equation

= 1 mole = 0.52 moles

So multiply everything in the equation by the factor of 1/2 , including the enthalpy.

2. The enthalpy of a reaction is of opposite sign to its reverse reaction.

If we burn H2 a large amount of heat is given off:

If we break H2O up into H2 and O2, an equal amount of heat energy has to be put into this reverse reaction:

2 moles H2 1 mole O2 2 moles H2O

2 moles H2O 2 moles H2 1 mole O2

ΔH = - 483.6 kJHeat given off

ΔH = + 483.6 kJHeat put back in

The enthalpy of a reaction is equal in magnitude but opposite in sign for the reverse reaction.

2 H2(g) + O2(g) → 2 H2O(g) ΔH = - 483.6 kJ

but for the reverse reaction:

2 H2O(g) → 2 H2(g) + O2(g) ΔH = + 483.6 kJ

For the reverse reaction one simply changes

the sign of ΔH.

3. The enthalpy change for a reaction depends on the state of the reactants.

2 H2(g) + O2(g) →2 H2O(g) ΔH = -483.6 kJ

but

2 H2(g) + O2(g) →2 H2O(l) ΔH = -659.6 kJ

since

H2O(l) → H2O(g) ΔH = +88 kJor

H2O(g) → H2O(l) ΔH = -88 kJ

Note that -659.6 + (2 x 88) = -483.6 kJ (discussed later)

water vapor

liquid water

liquid water water vapor

water vapor liquid water

Example:

How much heat is given off by burning 3.4 g of H2 in excess O2?

2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ

2 moles 1 mole

‘Excess O2’ means that H2 is the limiting reagent, and so we don’t need to bother with the O2. So we know that 2 moles of H2 burns in O2 to give off -483.6 kJ, so we need to know how many moles of H2 we have in 3.4 g.

Problem (contd.)Molecular mass H2 = 1.0 + 1.0 = 2.0 g/mol

Moles H2 = 3.4 g x 1 mole = 1.7 moles 2.0 g

2 H2(g) + O2(g) → 2 H2O(g) ΔH = -483.6 kJ2 moles 1 mole1.7 moles ΔH = ?

ΔH = - 483.6 kJ x moles we have moles in balanced

equation

= - 483.6 kJ x 1.7 moles 2 moles

= - 411.1 kJ

Specific heat

Specific heat is the amount of heat in joules it takes to raise the temperature of a substance by 1 K. The units of specific heat are J/g.K. Some examples are:

Substance Specific heat (J/g.K)

H2O(l) 4.184

N2(g) 1.04

Al(s) 0.90

Fe(s) 0.45

Hg(l) 0.14

Calculating heat produced from rise in temperature and a knowledge of the specific

heat:

Example: 5 ml of H2SO4 (at 21.2 ºC) is added to 50 ml of water in a coffee-cup calorimeter. The temperature of the solution in the calorimeter rises from 21.2 to 27.8 ºC. How much heat was liberated by the dissolution of the H2SO4? (assume all 55 ml of solution has specific heat of water = 4.184 J/g.K, and density of water = 1g/ml).

add 5 ml H2SO4

50 ml H2O 55 ml H2Oplus H2SO4

Temperature = 21.2 ºC

Temperature = 27.8 ºC

Coffee-cup calorimeter

thermometer

Problem (contd.)

55 ml x 1g = 55 g of solution1 ml

4.184 = heat in J (q) weight (g) x temperature rise (K or

ºC)

= ________q (J)__________

55 g x (27.8 – 21.2) ºC

q = 4.184 J x 55 g x 6.6 ºC1 g x 1 ºC

= -1519 J

specific heat of water

rises in K orºC will be the same

(q is negative because heat is evolved)

Example on calculating heat evolved per mole:

When 9.55 g of NaOH dissolves in 100.0 g of water in a coffee-cup calorimeter, the temp. rises from 23.6 to 47.4 oC. Calculate ΔH for the process:(Assume specific heat is as for pure water = 4.18 J/g.K.)

NaOH (s) → Na+ (aq) + OH- (aq)

We assume that when presented with the balanced equation we need to calculate ΔH for the numbers of moles indicated by the coefficients, i.e. 1 mole NaOH

Problem (contd.)Wt. of solution = (100.0 g + 9.95 g) = 109.95 gchange in K = 47.4 - 23.6 = 23.8 K.

q = specifc heat x mass in g x temp. rise in K

q = 4.18 J x 109.95 g x 23.8 K = -10938 J g x K

= -10.9 kJ

F. Wt. NaOH = 23 + 16 + 1 = 40 g/molMoles NaOH = 9.95 g x 1 mole

40 g= 0.249 mol

Problem (contd.)

NaOH (s) → Na+ (aq) + OH- (aq)

1 mole 1 mole 1 mole

0.249 mole 0.249 mole 0.249 mol ΔH = -10.9 kJ

ΔH = -10.9 kJ x 1 mol/0.249 mol

= - 43.8 kJ/mol

Note: If the temperature rises in a process, then ΔHwill be negative.

5.3 Calorimetry (covered in labs):

The ‘system’ in a coffee cup calorimeter is usually the solution in the calorimeter. A thermometer is used to monitor the temperature rise due to the chemical reaction being studied. One assumes the heat capacity of the solution is that of water.Coffee-cup

calorimeter

thermometer stirrer

two nestedcoffee cupsto providebetter insulation

the solution = the ‘system’

lid

5.6 Hess’s Law.

Hess’s 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.

Hess’s Law provides a method for calculating ΔH values that are impossible to measure directly.

Hess’s law

IncreasingEnthalpy (H)

H2O (g)

H2O (l)

H2O (s)

-44kJ

-6 kJ

-50kJ

The enthalpyof going fromH2O (g) (watervapor) to H2O (s)(ice) in one step(-50 kJ) is thesum of the twosteps of goingfirst from H2O (g)to H2O (l) (-44kJ)and then fromH2O (l) to H2O (s)(-6 kJ)

Adding enthalpies following Hess’ Law:

We cannot measure directly the heat of burning graphite to give CO. However, we can calculate this by combining two equations:

C(s) + O2(g) → CO2(g) ΔH = -395.5 kJCO2(g) → CO(g) + ½O2(g) ΔH = 283.0 kJ____________________________________________________________________

C(s) + O2(g) + CO2(g) → CO2(g) + CO(g) + ½ O2(g) ΔH = -110.5 kJ

C(s) + ½O2(g) → CO(g) ΔH = -110.5 kJ

{

add theequations

}

add theΔH values

net equation

Cancel things that occur on both sides of equation

Graphite is the stable form of carbon: C(graphite) → C(diamond) ΔH = +1.9 kJ

This value of ΔH could not be measured directly, but could be obtained from the enthalpy of combustion of graphite and diamond using Hess’ Law:

C(graphite) + O2(g) → CO2(g) ΔH = -393.5 kJ

C(diamond) + O2(g) → CO2(g) ΔH = -395.4 kJ___________________________________________________________________

C(diamond) → C(graphite) ΔH = -1.9 kJor C(graphite) → C(diamond) ΔH = +1.9 kJ

The enthalpy of conversion of graphite to diamond from Hess’ Law

subtract

Using Hess’ Law to calculate the energy of formation of ethylene from C (graphite) and H2

gas:An impossible (so far) reaction to carry out would be:

2 C(s) + H2(g) = C2H2(g) (acetylene).

We can calculate the energy of the above by combining the heats of combustion of the components in the reaction:

C2H2(g) + 2½O2(g) → 2 CO2(g) + H2O(l) ΔH = -1299.6 kJC(s) + O2(g) → CO2(g) ΔH = -393.5

kJH2(g) + ½ O2(g) → H2O(l) ΔH = -285.8 kJ

We first want to get the products on the right hand side, so we reverse the first equation:

C2H2(g) + 2½ O2(g) → 2 CO2(g) + H2O(l) ΔH = -1299.6 kJ

2 CO2(g) + H2O(l) → C2H2(g) + 2½O2(g) ΔH = +1299.6 kJ

Then we double the second equation because there are two C-atoms in the desired reaction:

C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ

2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = -787.0 kJ

We now add them together in two steps (it’s easier that way):

2 CO2(g) + H2O(l) → C2H2(g) + 2½ O2(g) ΔH = +1299.6 kJ

2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = -787.0 kJ_______________________________________________________________________

2 C(s) + H2O(l) → C2H2(g) + ½ O2(g) ΔH = +512.6 kJ

2 C(s) + H2O(l) → C2H2(g) + ½ O2(g) ΔH = +512.6 kJ H2(g) + ½ O2(g) → H2O(l) ΔH = -285.8 kJ

_______________________________________________________________________

2 C(s) + H2(g) → C2H2(g) ΔH = +226.8 kJ

5.7. Standard enthalpies of formation

The standard enthalpy change ΔHo is defined as the enthalpy change when all the reactants and products are in their standard states. The standard state is 25 oC (298 K) and 1 atm pressure.

2 H2(g) + O2(g) → 2 H2O(l) ΔHº = -659.6 kJ

Standard states for H2 and O2 Standard state for H2O is a are gases at 25 oC and 1 atm liquid at 25 oC and 1 atm

Superscript ‘o’ indicates standard enthalpy change

Standard enthalpies of formation, ΔHof

The standard enthalpy of formation of a compound ΔHo

f is the enthalpy of formation of one mole of the substance from its constituent elements, all being in their standard states. For elements, the standard state is the most stable form of the element at 298 K and 1 atm, e.g. C is graphite, not diamond. For elements in their standard state (e.g. C(graphite) or O2(g)), ΔHo

f is zero.

(See Table of ΔHof values on p. 192)

ΔHof for some substances (kJ/mol):

______________________________________

C2H2(g) 226.7 HCl(g) -92.3

NH3(g) -46.19 HF(g) -268.6

C6H6(l) +49.0 CH4(g) -74.8

CO2(g) -393.5 AgCl(s) -127.0

Diamond +1.88 NaCl(s) -410.9

C2H5OH(l) -277.7 H2O(l) -285.8

C6H12O6(s) -1273 Na2CO3(s) -1130.9

______________________________________

Using Enthalpies of formation to calculate Enthalpies of reaction:

One can show from

Hess’s Law that:

ΔHorxn = ΣnΔHo

f(products) – ΣmΔHof(reactants)

Upper case Greek ‘sigma’means ‘sum of ’

Standard sum of standard sum of standard enthalpy heats of formation of heats of formation of reaction all products of all reactants

coefficients in thebalanced equation

Heat of reaction = sum of heats of formation of products minus sum of heats

of formation of reactants

What the equation on the previous slide is saying is that the standard enthalpy change for a chemical reaction (ΔHo

rxn) is given by the sum of the standard heats of formation of all the products minus the sum of the standard heats of formation of all the reactants. This is best illustrated by some examples.

Example – the standard enthalpy of formation of benzene:

Calculate the standard enthalpy change (ΔHo

rxn) of combustion of 1 mol of benzene from standard enthalpies of formation.

1)write out the balanced equation:

C6H6(l) + 7½ O2(g) → 6 CO2(g) + 3 H2O(l)

1 mole

Benzene, C6H6

H

C

Products: (ΔHof)

6 CO2(g) = 6 x (-393.5) = -2361.0 kJ.mol3 H2O(l) = 3 x (-285.8) = -857.4 kJ/mol

-3218.4 kJ/mol

Reactants:

C6H6(l) = (+49) = +49 kJ/mol7 ½ O2(g) = (7.5 x 0.0) = 0 kJ/mol

+ 49 kJ/mol

ΔHorxn = -3218.4 – (+49) = 3267 kJ/mol

(products) (reactants)