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Chapter 6

EnergyThermodynamics

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Energy is... Conserved

Made of heat and work.– Work is a force acting over a distance– Heat is energy transferred between objects

because of temperature difference.

A state function which means that the result is independent of the path, or how you get from point A to B.

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The universe Is divided into two halves: the

system and the surroundings. The system is the part you are

concerned with. The surroundings are the rest.

Unfortunately, it is easier to measure the effect on the surroundings than the system directly

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The universe Exothermic reactions release

energy to the surroundings.

Endothermic reactions absorb energy from the surroundings.

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CH + 2O CO + 2H O + Heat4 2 2 2

CH + 2O 4 2

CO + 2 H O 2 2

Pote

nti

al energ

y

Heat

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N + O2 2

Pote

nti

al energ

y

Heat

2NO

N + O 2NO2 2 + heat

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Direction Every energy measurement has

three parts.1. A unit ( Joules or calories).2. A number how many.3. A sign to tell direction. negative – exothermic

– System to the surrounding positive- endothermic

– Surroundings to the system

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System

Surroundings

Energy

DE <0

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System

Surroundings

Energy

DE >0

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Same rules for heat and work Heat given off is negative. Heat absorbed is positive. Work done by system on

surroundings is negative. Work done on system by

surroundings is positive. Thermodynamics- The study of

energy and the changes it undergoes.

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First Law of Thermodynamics The energy of the universe is

constant. Law of conservation of energy. q = heat w = work DE = q + w Take the systems point of view to

decide signs. Punch Line: DE = q (At constant P)

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What is work? Work is a force acting over a

distance. w= F x Dd P = F/ area d = V/area w= (P x area) x D (V/area)= PDV Work can be calculated by

multiplying pressure by the change in volume at constant pressure.

units of liter - atm L-atm

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Work needs a sign If the volume of a gas increases,

the system has done work on the surroundings.

work is negative w = - PDV Expanding work is negative. Contracting, surroundings do work

on the system w is positive. 1 L atm = 101.3 J

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Example #1 What amount of work is done

when 15 L of gas is expanded to 25 L at 2.4 atm pressure? – volume increase – work is being done by the

system on the surroundings (-)

w = - PDVw = -(2.4atm)(25L-15L)w = -24L atm-24 L atm x (101.3J/Latm) = -2431 J = -2.4kJ

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Example #2If 2.36 J of heat are absorbed by the gas above. what is the change in energy?DE = q + wDE = 2.36 J + -2431 J DE = -2429 J

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Enthalpy (H) H = E + PV (that’s the

definition)– at constant pressure.

DH = DE + PDV

– the heat at constant pressure qp can be

calculated from

DE = qp + w = qp – PDV (w=-

PDV)

qp = DE + P DV = DH– PUNCH LINE qp = DH

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Calorimetry Measuring heat using a

calorimeter. Two kinds – the first kind being a

– Constant pressure calorimeter (called a coffee cup calorimeter)

heat capacity for a material, C is calculated (heat required to change a substances temperature)

C= heat absorbed/ DT = DH/ DT

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Calorimetry Specific heat capacity = C/mass Molar heat capacity = C/moles

heat = specific heat x mass x DT heat = molar heat x moles x DT

Make the units work and you’ve done the problem right.

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Calorimetry A coffee cup calorimeter measures

DH.

The specific heat of water is 1 cal/gºC or 4.184 J/gC

Heat of reaction= DH = SH x mass x DT

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Examples The specific heat of graphite is

0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

DH = SH x mass x DT DH = 0.71J/gC (75000g)(54C) DH = 2876 kJ

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Examples A 46.2 g sample of copper is heated to

95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

DHsurr = -DHsys 75.0g (4.184J/g C)(2.2C) = -[46.2g(SH)(-73.6C)]690J = 3400g C (SH)

.203 J/g C = SH

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Calorimetry Constant volume calorimeter is

called a bomb calorimeter. Material is put in a container with

pure oxygen. Wires are used to start the combustion. The container is put into a container of water.

The heat capacity of the calorimeter is known and tested.

Since DV = 0, PDV = 0, DE = q

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Bomb Calorimeter

thermomet

er

stirrer

full of water

ignition wire

Steel bomb

sample

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Bomb Calorimeter A bomb calorimeter works in a

similar manner as the coffee cup calorimeter, but there is one significant difference.

In a coffee cup calorimeter, the reaction takes place in the water. In a bomb calorimeter, the reaction takes place in a sealed metal can, which is then placed in the water (contained in an insulated container).

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Bomb Calorimeter Analysis of the heat flow is a bit more

complex than it was for the coffee cup calorimeter, because the heat flow absorbed by the metal parts of the calorimeter (the “bomb” part) must be taken into account:

qrxn = - (qwater + qbomb)– where qwater = 4.18 J/(g·°C) x mwater x ΔT– The heat flow of the bomb is:

– qbomb = Ccal x ΔT

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One More Twist – Phase ChangesMolar Heat of Fusion – Energy required to change 1 mol from a solid to a liquid (visa versa)

6kJ/mol for water

Molar Heat of Vaporization – Energy required to change 1 mol from a liquid to a gas (vise versa)

40.7kJ/mol for water

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Phase ChangesIf 20g of ice at 0C is placed in a calorimeter containing 100ml of water at 60C, what is the final temperature of the system?

qphasechange + qheating = -qwater

20g (1mol/18.02g) = 1.10mol

1.10mol(6000J/mol) + 20g(4.184J/gC)(Tf-0) = -100g(4.184J/gC)(Tf-60) 6600 + 83.68Tf = -418.4Tf + 25104

502.88 Tf = 18504

Tf = 36.8 C

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Problem Solving Strategies The 1st step is to ascertain whether the

process is constant pressure (open to the atmosphere) or constant volume. – If it’s constant pressure, use ΔH = −ΔEsurr; – for constant volume it’s ΔErxn = −ΔEsurr.

In many problems – ΔEsurr = (mass)(specific heat)ΔT. – To use this equation, you must determine the part of the

overall system that is changing temperature.

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Problem Solving Strategies For example, if we add 1.0 g of Mg to

100.0 mL of 1.0 M HCl, it is not the Mg that is changing temperature, but rather the 100.0 mL of acidic solution in which the Mg is reacting.

In this example, then, – ΔEsurr = (100.0 g)(4.184 J/[g oC])ΔT, – where we have used the usual assumptions stated above. We

would not use the mass of Mg (1.0 g) and the specific heat of Mg (1.02 J/[g oC]) in the ΔEsurr equation because it’s not the Mg that

is changing temperature.

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Problem Solving Strategies Always ask yourself this question:

“What is actually changing temperature in this process?”

Wherever the thermometer goes, that’s what is changing temperature.

*Note: in some problems – the heat capacity is given, so you’ll use ΔEsurr = CΔT, where C is heat capacity)

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Problem Solving Strategies In the equations ΔH = −ΔEsurr or ΔErxn =

−ΔEsurr, the units on both sides are joules (J)!

Therefore, if the problem asks you to find an answer in J or kJ per mole (or g), you need to find J first and then divide by the appropriate moles (or g)

Likewise, if the problem gives you ΔH (or ΔErxn) and asks you to find one of the terms in ΔEsurr, you first need to make sure that the units on ΔH (or ΔErxn) are J and not J per mole or J per gram

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Hess’s Law Enthalpy is a state function.

We can add equations to to come up with the desired final product, and add the DH

Two rules– If the reaction is reversed the sign of DH is

changed– If the reaction is multiplied, so is DH

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N2 2O2

O2 NO2

68 kJ

NO2180 kJ

-112 kJ

H (

kJ)

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Standard Enthalpy The enthalpy change for a reaction

at standard conditions (25ºC, 1 atm , 1 M solutions)

Symbol DHº When using Hess’s Law, work by

adding the equations up to make it look like the answer.

The other parts will cancel out.

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H (g) + 1

2O (g) H (l) 2 2 2O

C(s) + O (g) CO (g) 2 2DHº= -394 kJ

DHº= -286 kJ

C H (g) + 5

2O (g) 2CO (g) + H O( ) 2 2 2 2 2 l

Example Given

calculate DHº for this reaction

DHº= -1300. kJ

2C(s) + H (g) C H (g) 2 2 2

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Example

O (g) + H (g) 2OH(g) 2 2 O (g) 2O(g)2 H (g) 2H(g)2

O(g) + H(g) OH(g)

Given

Calculate DHº for this reaction

DHº= +77.9kJDHº= +495 kJ

DHº= +435.9kJ

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Standard Enthalpies of Formation Hess’s Law is much more useful if

you know lots of reactions. Made a table of standard heats of

formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states.

Standard states are 1 atm, 1M and 25ºC

For an element it is 0

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Standard Enthalpies of Formation Need to be able to write the

equations.

What is the equation for the formation of NO2 ?

½N2 (g) + O2 (g) ® NO2 (g)

Have to make one mole to meet the definition.

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Since we can manipulate the equations

We can use heats of formation to figure out the heat of reaction.

Lets do it with this equation:

ΔH = ∑ nΔHf (products) – ∑ nΔHf (reactants)  

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Example Use the table of standard enthalpies of

formation at 25°C to calculate ΔH for the reaction

4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g)

ΔH = ∑ nΔHf (products) – ∑ nΔHf (reactants)   

   = [6 ΔHf (H2O) + 4 ΔHf (NO)] – [4 ΔHf (NH3) + 5 ΔHf (O2)]

       = 6(–241.8) kJ mol–1 + 4(90.3) kJ mol–1 – 4(–46.1 kJ mol–1) – 5 × 0       = –1450.8 kJ mol–1 + 361.2 kJ mol–1 + 184.4 kJ mol–1

       = –905.2 kJ mol–1