chemistry 101 : chap. 5 thermochemistry (1) the nature of energy (2) the first law of thermodynamics...

Chemistry 101 : Chap. 5Chemistry 101 : Chap. 5

Thermochemistry

(1) The Nature of Energy

(2) The First Law of Thermodynamics

(3) Enthalpy

(4) Enthalpy of Reaction

(5) Calorimetry

(6) Hess’s Law

(7) Enthalpy of Formation

NOTE: We will discuss chapter 19, “Chemical Thermodynamics”, after this chapter

Nature of EnergyNature of Energy

Thermodynamics : The study of energy and its transformation

Energies can be in many different forms. These energy changes canbe as simple as those associated with a falling object to complexesprocesses like metabolism

Thermochemistry : Thermodynamics associated with chemical processes.

The study of energy changes accompanying chemical reactions


Energy : The capacity to do work or transfer heat

Work : Energy used to move an object with mass

work = force distance

Heat : Energy used to increase the temperature of an object

An object can do work against the environment or transfer heat to theenvironment. In such case, the energy of an object needs to be converted to work or heat.

An object can possess two fundamentally different kinds of energy

: Kinetic energy and Potential energy


Kinetic Energy : The energy an object has due to its motion

2

2

1mvEk

massvelocity

Ex. (1) A moving car (macroscopic object)

(2) Atoms and molecules are constantly moving.

This is the source of thermal energy

(temperature increase or decrease)


Potential Energy : The energy an object has due to its position

relative to other objects

All objects in the universe are interacting with each other. In other words,they push or pull surrounding objects (exerting force). Potential energy arises when there is any kind of force acting on an object.

Ex. (1) Gravitation Energy : r

mmGEg

21

221

r

mmGFg Gravitational Force :

For the gravitational energy between an object (m1 = m) on earth and

earth (m2 = mearth) itself, r ~ radius of earth = rearth.

Fg, earth = m (Gmearth/rearth2) = mg This is the formal

definition of “weight”

Gravitation potential energy (on earth) = mgh

(or gravitational potential energy)


(2) Electrostatic Energy :d

QQEel

21

Electrostatic interaction is either attractive (charges with opposite

signs) or repulsive (charges with same signs).

(3) Chemical Energy : Potential energy stored in the arrangement of

the atoms in the substance of interest (mostly electrostatic)

One of the important goal in chemistry is to relate the energy changethat we see in macroscopic world (i.e. in laboratory experiment) to thekinetic energy and potential energy of substances at the atomic andmolecular level.

electrical charge (in C)

The Units of EnergyThe Units of Energy

joule (J) :1 J = 1 kgm2/s2

The kinetic energy of a 2kg object with

the speed of 1 m/s is equal to 1J:

Ek = (1/2)mv2 = (1/2)2 kg (1 m/s)2

= 1 kgm2/s2 = 1 J James Joule (1818 ~ 1889)

NOTE: kg, m and s are the SI units of mass, distance and time. Therefore, joule is the SI unit of energy.

calorie (cal) : This is an older unit of energy. But, it is still

frequently used in chemistry.

1 cal = 4.184 J

The Units of EnergyThe Units of Energy

Example : If a hot dog is burned, it produces 180 dietary Calories (1 Cal = 1000 cal). How many joules is this?

System and SurroundingSystem and Surrounding

We divide the universe into two parts: system and surrounding

System = The portion of the universe that we are interested in

Surrounding = Everything else

system

surrounding

* No exchange of particles between system and surrounding (closed system)

* Energy (heat and work) exchange is allowed between system and surrounding

Transferring EnergyTransferring Energy

Energy can be transferred between a system and its surrounding

in the form of heat (q) and/or work (w)

Work = Force Distance ( w = f d )

Energy (work) is transferred from surrounding to an object (system) when the object is moved by external force.

Heat :

Energy (heat) is transferred from an object at higher temperature to one at a lower temperature


Heat Transfer: Whenever two objects at different temperature are brought into contact, heat (thermal energy) will flow from the hot object to the cold object.

hot cold heat flow

Heat is transferred from system to surrounding

Heat is transferred from surrounding to system


Work Transfer: Work can be done on a system by surrounding or workcan be done by the system to surrounding.

Work done on the system Work done by the system

The First Law of ThermodynamicsThe First Law of Thermodynamics

Energy is conserved

Energy can neither be created nor destroyed, only converted from one form to another (Law of conservation of energy)

Energy that is lost by the system must be gained by thesurrounding, and visa versa.

Internal Energy of a system (E): The sum of all the kinetic and

potential energy of all its components.

e.g, translation, vibrational and rotational motions of atoms,

molecules


Internal Energy change (E) : The change in E that accompanies a change in the system

E = Efinal - Einitial

Efinal < Einitial

Heat (q) transferred from system to surroundings:

E < 0 Efinal > Einitial

Heat (q) transferred from surroundings to system:

E > 0

q q


First Law of Thermodynamics and internal energy change

E = q + w

heat (thermal energy)transferred in +q endothermic

heat (thermal energy)transferred out -q exothermic

work performed by surrounding on system +w

work performed by system on surrounding -w


A(g) + B(g) C (s)

E

q = -1200J

heat

work

w = +500 J

E = q + w = -1200 J + 500 J = -700 J

The system lost 700 J of energy to surrounding


Most of the work involved in chemical or physical changesare associated with volume change.

V > 0 w < 0w done by the system

V < 0 w > 0w done on the system

State FunctionsState Functions

State Function : A property of a system that is determined by the

system’s present condition, or state.

The value of a state function depends only on the present state

of the system, not on the path the system took to reach that state.

Example of state function : Internal Energy (E).

State FunctionsState Functions

Because E is a state function, E is also a state function

NOTE: Although E is a state function, q and w are NOT state functions

EnthalpyEnthalpy

Most of chemical reactions we study take place in open container

under constant pressure condition (atmospheric pressure)

Enthalpy Change (H) : A measure of the amount of heat

exchanged when a reaction takes place under constant pressure

H = qp

NOTE: Enthalpy is also a state function (H = Hfinal – Hinitial) and

it

is defined mathematically as H = E + PV where P

is

the pressure.

EnthalpyEnthalpy

Properties of enthalpy change :

H2O (l) H2O (g) H = 44kJ (endothermic)

H2O (g) H2O (l) H = -44kJ (exothermic)

2H2O (l) 2H2O (g) H = 88kJ (endothermic)

Enthalpy changes, like all energy changes, are extensive properties.

The amount of heat (or enthalpy change) liberated or absorbed

depends on the amount of material undergoing reaction

EnthalpyEnthalpy

Example: What is the enthalpy change (in kJ) for the sublimation

of 15 g of iodine

I2 (s) I2(g) Hsub = 62.4kJ

MW=254

Enthalpy of ReactionsEnthalpy of Reactions

Example : How much heat is given off by burning 3.4 g of H2 gas?

2 H2 (g) + O2 (g) 2 H2O (l) H = - 483.6 kJ


Example : What is the enthalpy change for the formation 12 g

of CO2(g) from the combustion of CH4 (g)?

CH4(g) + 2O2 (g) CO2(g) + 2H2O(g) H = - 806 kJ

MW=44

H

2 H2 (g) + O2 (g)

2 H2O (l)q given off

H < 0

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

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

q absorbed

H > 0


CalorimetryCalorimetry

Calorimetry : The measure of heat flow during physical and

chemical changes

Calorimeter : A device for measuring heat flow during physical

and chemical processes

+ NaOH

dissolves

heat (q)How can wemeasure q ?


We need to carry out the reaction in a container that is thermally insulated

heat(q)

NaOH dissolves

simple calorimeter

This is under constant pressure condition


When NaOH is dissolved in water, the temperature of the solution

will increase. But, how much?

Heat Capacity : The amount of energy required to raise the

temperature of an object by 1 oC

Heat capacity is an extensive property

Specific Heat Capacity : The amount of energy required to raise

the temperature of 1 g of a substance by 1 oC

Specific heat capacity is an unique property of each substance


Unit of specific heat capacity

1 g of H2O

11oC

1 g of H2O

12oC

+ 4.184 J of heat

Tm

qC

KgJ

Cm

qT

heat

mass temperature change

unit of specificheat capacity

CmTq


Example :Which of the following substances requires the smallest

amount of energy to increase the temperature of 50.0 g

of that substance by 10K?

CH4(g) Hg(l) H2O(l)

specific heat 2.20 0.14 4.18capacity (J/gK)


Example :Which of the following substances requires the smallest

amount of energy to increase the temperature by 1K?

CH4(g) Hg(l) H2O(l)

C (J/gK) 2.20 0.14 4.18

mass (g) 10 100 2


Example : Determine the specific heat of Al if it takes 16 J of

thermal energy to raise the temperature 6.0g of Al

by 3.0 oC


Example : What is the change in thermal energy (in kJ) for

300 g iron rod when it is cooled from 250oC to 50oC ?

(C for iron is 0.451 J/gK)


Molar Heat Capacity (Cm) : The amount of energy required to raise

the temperature of 1 mol of substance by 1 oC

T

qCm

mol Kmol

J

mC

qT

mol

heat

mol temperature change

unit of molarheat capacity

mCTq mol


Example : The specific heat capacity of iron is 0.451 J/goC.

Determine the molar heat capacity of iron.


Example : A 5.00 g pellet of copper at 75.0oC is placed in a beaker containing 35.0g of water at 25.0oC. What is the final temperature

of the water and copper? (Cu = 0.385 J/goC, H2O = 4.18 J/goC)


Example : A 2.00 g of sodium is placed in 150 g of water at 22.0oC in a calorimeter. When the reaction is complete, the temperature of the solution is found to be 41.0oC. Calculate H for the reaction

between Na and water: 2Na(s) + 2H2O(l) H2(g) + 2NaOH(aq)

Strategy to solve the problem:

(1) compute the total amount of heat gained by water

(2) That is the -H value for 2.00 g of sodium

(3) We need H value for 2 mol of sodium

(conversion from 2.00 g to 2 mol of sodium for H )

Hess’s LawHess’s Law

If a reaction is carried out in a series of steps, H for the overall

reaction will equal the sum of the enthalpy changes for the

individual steps

HH2O (g)

H2O (l)

H2O (s)

-44kJ

-6 kJ

-50kJ


Consider the following imaginary reactions.

(A) A + B → C H = -125 kJ

(B) D3 → C + B H = -41 kJ

How can we calculate the H for the following reaction?

(C) A + 2B → D3

To use Hess’s law, we have to rearrange reaction (A) and (B)

so that the overall reaction becomes (C) A+2B D3. Be sure

that you make appropriate changes for H’s when you rearrange

chemical equations.


This is how we can rearrange the reactions.

(1) Compare (A) and (C):

(A) A + B → C

(C) A + 2B → D3

This is what we have.

This is what we want.

We need one more B on the left and one D3 on the right and removeC on the right

(2) Consider (B) : D3 → C + B H = -41 kJ

If we reverse (B), it pretty much satisfies our need: one B on the left and one D3 on the right

(B’) C + B D3 H = +41 kJ


(3) Add (A) and (B’)

A + B C H = -125kJ

C + B D3 H = + 41kJ

A + C + 2B C + D3 H = -84 kJ

A + 2B D3 H = -84 kJ


Example : Given the enthalpy changes of following two reactions,

calculate the enthalpy change for the combustion of

methane to form liquid water and CO2

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

H2O(g) H2O(l) H = -44 kJ



calculate the enthalpy change for making diamond from

graphite : C(graphite) C(diamond)

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

C(diamond) + O2(g) CO2(g) H = -395.4 kJ

graphite diamond



calculate the enthalpy change for the reaction between

hydrogen and ozone : 3 H2 (g) + O3 (g) → 3 H2O (g)

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

3 O2 (g) → 2 O3 (g) H = +284.6 kJ

Standard Enthalpy of FormationStandard Enthalpy of Formation

H for the formation of one mole of compound from elements,

with all substances in their standard states [25oC, 1atm]

H2 (g) + ½ O2 (g) → H2O (l) Hof = -285.8 kJ

½ N2 (g) + ½ O2 (g) → NO (g) Hof = 90.4 kJ

2 C (s) + 3 H2 (g) + ½ O2 (g) → C2H5OH (g) Hof = -277.7 kJ

graphite

NOTE : Hof of the most stable form of any element is

zero

i.e. Hof for O2 (g) , N2 (g) , H2 (g), Br2 (l) etc. = 0


Example : For which of the following reactions would H represent

a standard enthalpy of formation?

2 Na (s) + ½ O2 (g) → Na2O (s)

K (l) + ½ Cl2 (g) → KCl (s)

CO (g) + ½ O2 (g) → CO2 (g)

2 Na (s) + Cl2 (g) → 2 NaCl (s)


Enthalpies of formations can be used to calculate enthalpies

of reactions (under standard conditions)

Horxn = sum of all Ho

f(products) - sum of all Hof(reactants)

Horxn = Σ n Ho

f(products) – Σ m Hof(reactants)

summoles of products

moles of reactants


Example : Determine the standard enthalpy change for the

following reaction

C2H4(g) + H2O(g) C2H5OH(l)

Hof [C2H4(g)] = 52.30kJ, Ho

f [H2O(g)] = -241.82 kJ, Hof[C2H6O(l)] =-277.7kJ


Example : What is the enthalpy change of the following reaction

at standard condition?

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

Hof [CO2 (g)] = -394kJ, Ho

f [C3H8(g)]= -104kJ, Hof [H2O(l)] = -286kJ

chemistry 101 : chap. 5 thermochemistry (1) the nature of energy (2) the first law of thermodynamics...

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