Change in Energy

Energy

Different types of energy exist (heat, potential, kinetic, chemical, nuclear etc.)

Heat - the energy transferred between objects that are at different temperatures. Unit of heat is joules J.

According to the Law of Conservation of Energy, energy cannot be created or destroyed. It can only by transferred or transformed. The total amount of energy in a system remains constant.

Potential Energy

Diagramhttp://www.kentchemistry.com/links/Kinetics/PEDiagrams.htm

Potential Energy Diagram

• Potential Energy Diagrams show changes in energy during chemical reactions.

• CHEMICAL EQUATIONS

A + B ---> C + D

Reactant Products

• In chemical reactions, reactants react to form products with an associated release or absorption of energy.

Potential Energy Diagram

Reaction Coordinate

Ene

rgy

CO + NO2

CO

NOO

CO

NOO

REACTANTS

PRODUCTS

CO

NO

O

No unit, fancy of saying “start to

finish”

Potential Energy in KJ

Potential Energy Diagram

Reaction Coordinate

Ene

rgy

CO + NO2

REACTANTS

PRODUCTS

Heat of Reactants

Heat of Activated Complex

Heat of Products

A

D

C

Potential Energy Diagram

Reaction Coordinate

Ene

rgy

CO + NO2

CO

NOO

CO

NOO

REACTANTS

PRODUCTS

CO

NO

O

Activation EnergyAmount of energy needed to start

the reaction

ΔH: ENTHALPYNet energy of the reaction

B

ΔH

Potential Energy Diagrams• A catalyst is a substance that changes the speed of reactions by

changing the activation energy.

• Catalysts speed up chemical reaction by lowering the activation energy

• Catalysts are not consumed during the reaction and can be recovered

Reaction Coordinate

Pote

ntial E

nerg

y→

Activation

Energy with

no catalyst

Enthalpy of

Reaction

Activation

Energy with

a catalyst

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

DH = H (products) – H (reactants)

DH = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

DH < 0Hproducts > Hreactants

DH > 0

Change in Heat Energy

1. Exothermic Reaction

– The change in heat (∆H) is NEGATIVE.

– Heat is released into the surroundings

2. Endothermic Reaction

– The change in heat (∆H) is POSITIVE.

– Heat is absorbed by the system

Click Below for the Video Lectures

Activation Energy

Reaction Coordinate

Catalyst

Exothermic and

Endothermic

Exothermic Reaction

A + B C + D + ENERGY

a) Heat is released as a product.b) This heat is released into the environment surrounding

the reaction, causing the temperature of the surroundings to increase.

c) The products have less energy than the reactants after release.

d) Since the products have less energy stored in their chemical bonds, they are more stable than the reactants were. Burning paper is exothermic. The ash formed by the burning is not flammable and will not burn.

General Exothermic Reaction

A + B C

ΔHA = 60 kJ ΔHB = 40 kJ ΔHc = 30 kJ

100 KJ

+ 70 kJ

Law of conservation of energy states that the total energy of an isolated system cannot change.

Energy Released as

product Exothermic

Exothermic Reaction

C (g) + O2 (g) CO2 (g) ΔH = -393.5 kJ

• The minus sign in front of the enthalpy

393.5 kJ indicates that this reaction is exothermic, that energy was released.

• This energy can be placed on the products side, as follows:

C (g) + O2 (g) CO2 (g) + 393.5 kJ

C (g) + O2 (g) CO2 (g) ΔH = -393.5 kJ

• If we were to synthesize 2.3 moles of CO2, how many kJ would be released?

1 mol of CO2

2.3 mol CO2

1 mol CO2

393.5 kJ= - 910 kJ

Synthesis

Decomposition

If ΔHsynthesis = - 393.5 kJ/mole, then Δ Hdecomposition = +393.5 kJ/mole

Potential Energy Diagram - Exothermic

Reaction Coordinate

Ene

rgy

A + B

REACTANTS

PRODUCTS

Activation Energy = 150 – 70

80 KJ

ΔH = 70 – 30

- 40 kJ, released

C + D30

70

150

A + B C + D + 40 kJ

Endothermic Reaction

A + B + ENERGY C + D

a) Heat is absorbed by the reactants.b) This heat is absorbed from the environment

surrounding the reaction, and the temperature of the surroundings decreases.

c) The products have more energy than the reactants after absorption.

d) This energy is stored in the chemical bonds of the products

General Endothermic Reaction

A + B C

ΔHA = 40 kJ ΔHB = 20 kJ ΔHc = 110 kJ

60 KJ

+ 50 kJ

Law of conservation of energy states that the total energy of an isolated system cannot change.

Energy absorbed as

product Endothermic

Potential Energy Diagram - Endothermic

Reaction Coordinate

Ene

rgy

A + BREACTANTS

PRODUCTS

Activation Energy = 150 – 30

80 KJ

C + D

30

70

150

A + B C + D+ 40 kJ

ΔH = 70 – 30

+ 40 kJ, Absorbed

Click Below for the Video Lectures

Exothermic and Endothermic

Calorimetry

Calorimetry

q = m C ΔTunits

What Each Variable Means

Joules (J) Grams (g) J/goC oC

q is the quantity of heat that is absorbed or released by a

physical or chemical change

m is the mass of water in the calorimeter cup that absorbs heat from the change or releases heat to the change.

C is the specific heat of water, the rate at which water gains or loses heat if energy is absorbed or removed from it.

∆T is the Temperature change of the water in the calorimeter cup as a result of the physical or chemical change.

CSpecific HeatThe amount of

heat required to raise the

temperature of one gram of

substance by one degree Celsius.

cal/g°C J/g°C (or J/gK)

water 1.00 4.18

aluminum 0.22 0.90copper 0.093 0.39silver 0.057 0.24gold 0.031 0.13

Use this equation on the back of periodic

table

Specific Heat of water. Use the one

for correct unit! They are NOT equations!!!

• How many joules are absorbed by 100.0 grams of water if the temperature is increased from 35.0oC to 50.0oC?

1. Circle the numbers, underline what you are looking for.

2. Make a list of number you circled using variables.

3. Write down the formula4. Derive the formula to

isolate the variable you are looking for.

5. Plug in the numbers

m =

T i=

T f=

q=

C=

q = mcΔT

q = (100.0 g)(4.18 j/goC)(50.0oC- 35.0oC)

100.0 g35.0oC50.0oC?4.18 j/goC

q = 6270 J

• 300. J of energy is absorbed by a 50. g sample of water in a calorimeter. How much will the temperature change by?

1. Circle the numbers, underline what you are looking for.

2. Make a list of number you circled using variables.

3. Write down the formula4. Derive the formula to

isolate the variable you are looking for.

5. Plug in the numbers

q =

m =

ΔT =

C=

q = mcΔT

ΔT = 300 J/ (50g)(4.18 j/goC)

300 J 50g?4.18 j/goC ΔT = = 1.4354066 oC

ΔT = q / mc

ΔT = = 1.4oC

Checking for understanding

1. How many joules are required to raise the temperature of 100. grams of water from 20.0oC to 30.0oC ?

2. How many grams of water can be heated from 20.0oC to 75.0oC using 2500. J?

Calorimetry

• We use an insulated device called a calorimeter to measure this heattransfer.

• A typical device is a “coffee cupcalorimeter.”

CalorimetryTo measure ΔH for a reaction

1.dissolve the reacting chemicals in known volumes of water

2.measure the initial temperatures of the solutions

3.mix the solutions

4.measure the final temperature of the mixed solution

Calorimetry

• The heat generated by the reactants is absorbed by the water.

• We know the mass of the water, mwater.

• We know the change in temperature, ∆Twater.

• We also know that water has a specific heat of Cwater = 4.18 J/°C-g.

• We can calculate the heat of reaction by:

• qsys = ∆H = −qsurr = -mwater × Cwater × ∆Twater

When 25.0 mL of water containing 0.025 mol of HCl at 25.0°C is added to 25.0 mL of water containing 0.025 mol of NaOH at 25.0°C in a coffee cup calorimeter, a reaction occurs. Calculate ∆H (in kJ) during this reaction if the highest temperature observed is 32.0°C. Assume the densities of the solutions are 1.00 g/mL.

Knowns: Vfinal = VHCl + VNaOH = (25.0 + 25.0) mL = 50.0 mL

Dwater = 1.00 g/mL

∆Twater = Tfinal − Tinitial = 32.0°C − 25.0°C = +7.0°C

Cwater = 4.18 J/°C-g

Calculation: mwater = 50.0 g

∆H = −m × C × ∆T

= −(50.0 g)(4.18 J/°C-g)(7.0°C)

= −1463 J

= −1.5×103 J

Phase Change Heating and

Cooling Curve

34 4/3/2017

Solid Liquid Gas

Melt Evaporate

CondenseFreeze

Sublime

Deposit

Phase Change

• During a phase change, there is a change in heat energy but no change in temperature.

At the melting point, heat energy isbeing used to break down the crystalline lattice.

At the boiling point, heat energy is being used to convert the liquid to a gas.

During a phase change, heat is being used to change phase, not increase the kinetic energy of the particles. It is for this reason that Temperature remains constant in a

phase change while heat energy changes.

Energy

Tem

per

atu

re

A

BC

D E

SOLID

LIQUID

GAS

SOLID to LIQUID

LIQUID TO GAS

BP

MPMELTING

EVAPORATING

Phase Change

• During phase change, temperature stays constant but energy increases.

Phase Change

• Energy needed to melt is called Latent Heat of Fusion ( ΔHfus)

• Energy need to vaporize is called Latent Heat of Vaporization (ΔHvaporization)

Triple point – the point on a phase diagram at which thethree states of matter: gas, liquid, and solid coexistCritical point – the point on a phase diagram at which thesubstance is indistinguishable between liquid and gaseousstates

q = mCΔT

q = mCΔT

q = mCΔT

q = m Δ Hfus

q = m Δ Hvap

How many joules does it take to melt 100. grams of water at its melting point?

q = m Δ Hfus

= 100. grams X 334 J/gram

= 33400 Joules

How many joules does it take to condense 50.0 grams of water at its boiling point?

q = m Δ Hvap

= 50.0 grams X 2260 J/gram

= 113 000 Joules

Calculating Enthalpy Change

The enthalpy change for a reaction is the enthalpy of the products minus the enthalpy of the reactants

DH = H (products) – H (reactants)

Hess’s Law

The overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process

Ex: Elemental carbon can be found as graphite and diamond at 25oC

C(diamond) C(graphite)– Takes millions of years to go from diamond to

graphite

– Reaction is too slow to measure the heat change

Hess’s Law

• Hess’s Law of Heat Summation:– If you add two or more thermochemical equations to

give a final equation, then you can also add the heats of reaction ΔH to give the final heat of reaction

• Example: Conversion of diamond to graphite

C(diamond) C(graphite)C(s, diamond) + O2 (g) CO2 (g) ΔH = -395.4kJ

C(s, graphite) + O2 (g) CO2 (g) ΔH = -393.5kJ• REVERSE the above equation, so that we can show diamond

being converted into graphite

CO2 (g) C(s, graphite) + O2 (g) ΔH = 393.5kJ

•Add the equations together!

C(s, diamond) C(s, graphite) ΔH = -1.9 kJ

•The overall ΔH is negative, so you can see this is an exothermic process!

kJ 393.5 H (g) O graphite) C(s, (g) CO

kJ 395.4- H (g) CO (g)O diamond) C(s,

22

22

D

D

Standard Enthalpy (Heat) of Formation

• Free elements at 298 K & 1 atm have an enthalpy of zero. Examples, H O N Cl Br I F

• Other ΔHºf values have been calculated

• ΔHºreaction can be calculated as follows:

o

reactantsf

o

productsf

o

reaction HHH DDD

o

fHD - standard heat of formation – the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states at 298 K (25oC)

Let’s do an example!

• What is the standard heat of reaction (ΔHº) for the following equation:O2 (g) + 2CO (g) 2CO2

• From your text:– ΔHºf O2 (g) = 0 (free element)– ΔHºf CO (g) = -110.5 kJ/mol– ΔHºf CO2 (g) = -393.5 kJ/mol

• Calculate the ΔHºf of the reactants:– 2 mol CO, 1 mol O

• 2 mol of CO 2 mol x -110.5 kJ/mol = -221.0 kJ• 1 mol O2 = 0 KJ (free element)• TOTAL: 0 kJ + -221.0 kJ = -221 kJ (reactants)

• Calculate the ΔHºf of the products:– 2 mol CO2

• 2 mol CO2 2 x -393.5 kJ/mol = -787.0 kJ

• Use to find the total!o

reactantsf

o

productsf

o

reaction HHH DDD

ΔHº= -787.0 kJ – (-221.0) kJ ΔHº = -566.0 kJ

Try a Practice Problem

2NO (g) + O2 (g) 2NO2 (g)

ΔHºf NO = 90.37

ΔHºf O2 = 0.0

ΔHºf NO2 = 33.85

2(33.85) – [2(90.37) + 0] = -113.04 kJ

Spontaneous ReactionsChanges in enthalpy and entropy determine

whether a process is spontaneous

Spontaneous Process: any physical or chemical change that once begun, occurs without any

outside intervention.

Examples:

Rusting of iron

Melting of ice

Vaporization of water

Entropy

• Entropy = a measure of the disorder or randomness of the particles that make-up a system

• The Second Law of Thermodynamics

– Spontaneous processes always proceed in a way that the entropy of the universe increases

EntropyWater

Melting

Ssolid < Sliquid

Evaporation

Sliquid < Svapor

Dissolving

Ssolute + Ssolvent < Ssolution

Energy Calculations

• Standard entropy (So) is measured at 25oC.Units on So = J/(K*mol)

– Predicting if ∆S is positive or negative:1. State changes – which would cause ∆S to increase?

Decrease?2. Dissolving a gas in a solvent?

Ex O2 (g) O2 (aq)

3. Number of gaseous particles present?Ex 2SO3 (g) 2SO2 (g) + O2 (g)

4. Dissolving?5. Temperature Changes?

DS0rxn = S0

products - S0reactants

Energy CalculationsWhich is more important if they are working against

each other: Enthalpy or Entropy?

Gibbs Free Energy (ΔG): the enthalpy of the

system minus the product of the temperature times the entropy of the system.

Kelvinsin TST-HG ooo DDD

In a spontaneous reaction, ΔG is always negative!

Reaction Spontaneity and the Signs of DH0, DS0, and DG0

DH0 DS0 -TDS0 DG0 Description

- + - -

+ - + +

+ + - + or -

- - + + or -

Spontaneous at all T

Nonspontaneous at all T

Spontaneous at higher T;

nonspontaneous at lower T

Spontaneous at lower T;

nonspontaneous at higher T

Practice Problems

• Using the values for ΔH and Sº, determine whether the reaction is spontaneous at 25ºC:

C(s, graphite) + O2 (g) CO2 (g)

Substance ΔH (kJ/mol) Sº (J/Kxmol)

C (s, graphite) 0.0 5.69

O2 0.0 205

CO2 -393.5 214