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Brown, LeMay Ch 5AP Chemistry

Monta Vista High School

From Greek therme (heat); study of energy changes in chemical reactions

Energy: capacity do work or transfer heat Joules (J) or calories (cal); 1 cal = 4.184

J

1. Kinetic: energy of motion; dependent on mass & velocity Applies to motion of large objects &

molecules Linked to thermal energy (object’s T

above 0 K)

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2

2

1mvKE

James Prescott Joule

(1818-1889)

2. Potential: stored in “fields” (gravitational and electrical/magnetic); dependant on position relative to another object

Applies to large objects where gravity is overriding force, but not significantly to molecules where gravity is negligible and electrostatic forces dominate

Associated with chemical energy; stored in arrangement of atoms or subatomic particles (electrostatic & nuclear forces, bonding between atoms)

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mghPE

Link for Kinetic and Potential Energy Conversion

System: “isolated” portion of study (typically just the chemicals in a reaction)

Surroundings: everything else (container, room, Earth, etc.)Closed system: easiest to study because

exchanges energy with surroundings but matter is not exchanged.

Force: a push or pull on an object Work: energy transferred to move an

object a certain distance against a force: W = (F)(d)

Heat: energy transferred from a hotter object to a colder one 4

“0th Law”: 2 systems are in thermal equilibrium when they are at the same T. Thermal equilibrium is achieved when the

random molecular motion of two substances has the same intensity (and therefore the same T.)

1st Law: Energy can be neither created nor destroyed, or, energy is conserved.

2nd and 3rd Laws: discussed in Ch. 195

Includes: Translational motion Rotational motion of particles through space Internal vibrations of particles.

It is difficult to measure all E, so the change in internal energy (E) is typically measured:

E = Efinal - Einitial

E > 0 Increase in energy of system (gained

from surroundings)E < 0Decrease in energy of system (lost to

surroundings)6

When a system undergoes a chemical or physical change, the change in internal energy (E) is equal to the heat (q) added or liberated from the system plus the work (w) done on or by the system:

E = q + w

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First Law of Thermodynamics

q > 0 Heat is added to system

q < 0 Heat is removed from system(into surroundings)

w > 0 Work done to system

w < 0 System does work onsurroundings

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Octane and oxygen gases combust within a closed cylinder in an engine. The cylinder gives off 1150 J of heat and a piston is pushed down by 480 J during the reaction. What is the change in internal energy of the system? q is (-) since heat leaves system; w is (-) since

work is done by system. Therefore,E = q + w = (-1150 J) + (-480 J) = - 1630 J

1630 J has been liberated from the system (C8H18 and O2) added to the surroundings (engine, atmosphere, etc.)

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http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/activa2.swf

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E

E

Property of a system that is determined by specifying its condition or its “state” The value of a state function depends only on its

present state and not on the history of the sample.

T & E are state functions.Consider 50 g of water at 25°C: EH2O does not depend on how the water got to be 25°C (whether it was ice that melted or steam that condensed or…)

Work (w) and heat (q) are not state functions because the ratio of q and w are dependent on the scenario.Consider the combustion of gasoline in a car engine vs. burning in the open.

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Measurement of heat flow Heat capacity, C: amount of heat required

to raise T of an object by 1 K

q = C T Specific heat (or specific heat capacity,

c): heat capacity of 1 g of a substance

q = m c T

Ex: How much energy is required to heat 40.0 g of iron (c = 0.45 J/(g K) from 0.0ºC to 100.0ºC?

q = m c T = (40.0 g)(0.45 J/(g K))(100.0 – 0.0 ºC)

= 1800 J12

  Calorimetry is an experimental technique used

to measure the heat transferred in a physical or chemical process.

The apparatus used in this procedure is called as a “Calorimeter”. There are two types of calorimeters- Constant pressure (coffee cup) and constant volume (bomb calorimeter). We will discuss constant pressure calorimeter in detail.

Constant Pressure Calorimeter: The coffee cup calorimeter is an example of this type of calorimeter. The system in this case is the “contents” of the calorimeter and the surroundings are cup and the immediate surroundings.

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During the rxn:

qrxn + q solution = 0

where qrxn is the heat gained or lost in the chemical reaction and qsolutionis the heat gained or lost by solution. Heat exchange in this system (qrxn), is equal to enthalpy change. Assuming no heat transfer takes place between the system and surroundings, qrxn + q solution = 0

chemlab.truman.edu

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chemlab.truman.edu

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Sample Problem #1:1. 0.500g of magnesium chips are placed in a coffee-cup calorimeter and 100.0 ml of 1.00 M HCl is added to it. The reaction that occurs is:

Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq)

The temperature of the solution increases from 22.2oC (295.4 K) to 44.8 oC (318.0 K). What’s the enthalpy change for the reaction, per mole of Mg? (Assume specific heat capacity of solution is 4.20 J/(g * K) and the density of the HCl solution is 1.00 g/ml.)Ans. -4.64 x 105 J/mol Mg

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Sample Problem #2:

200. ml of 0.400 M HCl is mixed with the same amount and molarity of NaOH solution, inside a coffee-cup calorimeter. The temperature of the solutions before mixing was 25.10 oC, and 27.78

oC after mixing and letting the reaction occur. Find the molar enthalpy of the neutralization of the acid, assuming the densities of all solutions are 1.00 g/ml and their specific heat capacities are 4.20 J/(g * K).

Since most reactions occur in containers open to the air, w is often negligible. If a reaction produces a gas, the gas must do work to expand against the atmosphere. This mechanical work of expansion is called PV (pressure-volume) work.

Enthalpy (H): change in the heat content (qp) of a reaction at constant pressure

H = E + PVH = E + PV (at constant P)H = (qp + w) + (-w)H = qp

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Sign conventionsH > 0 Heat is gained from surroundings

+ H in endothermic reaction

H < 0 Heat is released to surroundings- H in exothermic reaction

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Also called heat of reaction:1. Enthalpy is an extensive property (depends on

amounts of reactants involved).

Ex: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

Hrxn = - 890. kJ Combustion of 1 mol CH4 produces 890. kJ

… of 2 mol CH4 → (2)(-890. kJ) = -1780 kJ

What is the H of the combustion of 100. g CH4?

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kJ 5550CH mol 1

kJ 890. -

CH g 6216.04

CH mol 1

1

CH g 100.

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2. Hreaction = - Hreverse reaction

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

H = - 890. kJ

CO2 (g) + 2 H2O (l) CH4 (g) + 2 O2 (g)

H = +890. kJ

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88kJ

802 kJ CO2 (g) + 2 H2O

(g)

3. Hrxn depends on states of reactants and products.

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

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

So:CO2 (g) + 2 H2O (l) CH4 (g) + 2 O2 (g) H =

890. kJCH4 (g) + 2 O2 (g)

2 H2O (l)

2 H2O (g)

CO2(g) + 2 H2O(l)

CH4(g) + 2 O2(g) 890.

kJ

If a rxn is carried out in a series of steps,Hrxn = (Hsteps) = H1 + H2 + H3 + …

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Ex. What is Hrxn of the combustion of propane?C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

3 C (s) + 4 H2 (g) C3 H8 (g) H1 = -103.85 kJ

C (s) + O2 (g) CO2 (g) H2 = -393.5 kJ

H2 (g) + ½ O2 (g) H2O (l) H3 = -285.8 kJHrxn = 103.85 + 3(- 393.5) + 4(- 285.8) = - 2219.8 kJ

3[ ] 3( )4[ ] 4( )

Germain Hess(1802-1850)

C3H8 (g) 3 C (s) + 4 H2 (g) H1 = +103.85 kJ

Formation: a reaction that describes a substance formed from its elements

NH4NO3 (s)

Standard enthalpy of formation (Hf): forms 1 mole of compound from its elements in their standard state (at 298 K)

C2H5OH (l)Hf = - 277.7 kJ

Hf of the most stable form of any element equals zero.H2, N2 , O2 , F2 , Cl2 (g)Br2 (l), Hg (l)C (graphite), P4 (s, white), S8 (s), I2 (s) 24

Ex: 2 N2 (g) + 4 H2 (g) + 3 O2 (g) 2

2 C (graphite) + 3 H2 (g) + ½ O2 (g)

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The complete (long) way to calculate the

ENTHALPY of any reaction is to take apart the reactant molecules piece

by piece to elements,

and then build the product molecules

from all of the pieces, bond by

bond.

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A GREAT SHORTCUT exists however:

To estimate the enthalpy of a

reaction ΔRH ° we need only to know the enthalpies of formation of the

reactants and the products.

(we calculate the large RED arrow by

the difference between the known

small GREEN arrows)

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An example: The combustion of liquid benzene : 

Firstly, we need to define the overall reaction, and balance it to 1 mole of the molecule we’re interested in : 1 C6H6 (l) + 15/2 O2 (g) 6 CO2 (g) + 3 H2O (l)

ΔRH ° = ??? ΔRH ° = ΔfH ° { products } - ΔfH ° { reactants } Secondly, we need to look up the ΔfH ° values corresponding to both the reactants and the products, and multiply by the moles:

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 ΔfH ° { products } = 6 ( - 394 kJ/mol ) + 3 ( - 286 kJ/mol ) = - 3222 kJ/mol to form the products  ΔfH ° { reactants } = ( + 49 kJ/mol ) + 15/2 ( zero ) = + 49 kJ/mol to form the reactants -----------------------------------------------Overall reaction ΔRH ° = - 3268 kJ/mol

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Enthalpies of combustion ΔCH ° are also commonly tabulated :

( converted into enthalpies per gram, for convenience )

 methane ΔCH ° = - 55 kJ/goctane ΔCH ° = - 48 kJ/gdodecane ΔCH ° = - 51 kJ/gmethanol ΔCH ° = - 23 kJ/g

 glucose ΔCH ° = - 16 kJ/gcarbohydrates ΔCH ° ~ - 18 kJ/gtristearin ΔCH ° = - 38 kJ/g

 human beings at age 20 need around 10,000 kJ per day of energy from combustions (~12,000 kJ

for men, and ~9000 kJ for women).

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)(reactantsΔHn - (products)ΔHnΔH 0f

0f

0rxn

Ex. Combustion of propane:C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Given: Compound Hrxn (kJ/mol)C3H8 (g) -103.85CO2 (g) -393.5H2O (l) -285.8H2O (g) -241.82

Hrxn = [3(- 393.5) + 4(- 285.8)] – [1(-103.85) + 5(0)]

= - 2219.8 kJ

mount of energy required to break a particular bond between two elements in gaseous state. Given in kJ/mol. Remember, breaking a bond always requires energy!

Bond enthalpy indicates the “strength” of a bond. Bond enthalpies can be used to figure out Hrxn .

Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) Hrxn = ?1 C-H & 1 Cl-Cl bond are broken (per mole)1 C-Cl & 1 H-Cl bond are formed (per mole)

Hrxn ≈ (Hbonds broken) - (Hbonds formed)

Note: this is the “opposite” of Hess’ Law whereHrxn = Hproducts – hreactants

Bond Enthalpy link31

Ex: CH4 (g) + Cl2 (g) → CH3Cl (g) + HCl (g) Hrxn = ?Bond Ave H/mol Bond Ave H/molC-H 413 Cl-Cl 242H-Cl 431 C-Cl 328

C-C 348 C=C 614

Hrxn ≈ (Hbonds broken) - (Hbonds formed)

Hrxn ≈ [(1(413) + 1(242)] – [1(328) + 1(431)]

Hrxn ≈ -104 kJ/mol

Hrxn = -99.8 kJ/mol (actual)

Note: 2 C-C ≠ 1 C=C2(348) = 696 kJ ≠ 614 kJ 32

H

Absorb E, break 1 C-H and 1 Cl-Cl

bond

Release E, form 1 C-

Cl and 1 H-Cl bondCH4(g) + Cl2(g)

CH3Cl (g) + HCl (g)Hrxn

*CH3(g) + H(g) + 2 Cl(g)

Hrxn = (Hbonds broken) + (- Hbonds formed)Hrxn = (Hbonds broken) - (Hbonds

formed)

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