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CH 6: Thermochemistry Renee Y. Becker Valencia Community College CHM 1045 1

CH 6: Thermochemistry

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CH 6: Thermochemistry. Renee Y. Becker Valencia Community College CHM 1045. Energy. Energy : is the capacity to do work, or supply heat. Energy = Work + Heat Kinetic Energy : is the energy of motion. E K = 1 / 2 mv 2 (1 Joule = 1 kg m 2 /s 2 ) (1 calorie = 4.184 J) - PowerPoint PPT Presentation

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Page 1: CH 6:  Thermochemistry

CH 6: Thermochemistry

Renee Y. BeckerValencia Community College

CHM 1045

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Page 2: CH 6:  Thermochemistry

Energy

• Energy: is the capacity to do work, or supply heat.

Energy = Work + Heat

• Kinetic Energy: is the energy of motion.

EK = 1/2 mv2 (1 Joule = 1 kgm2/s2)

(1 calorie = 4.184 J)

• Potential Energy: is stored energy.2

Page 3: CH 6:  Thermochemistry

Ek & Ep

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Page 4: CH 6:  Thermochemistry

Example 1: KE

Which of the following has the greatest kinetic energy?

1. A 12 kg toy car moving at 5 mph?

2. A 12 kg toy car standing at the top of a large hill?

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Page 5: CH 6:  Thermochemistry

Energy

• Thermal Energy is the kinetic energy of molecular motion

• Thermal energy is proportional to the

temperature in degrees Kelvin. Ethermal T(K)

• Heat is the amount of thermal energy transferred between two objects at different temperatures.

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Page 6: CH 6:  Thermochemistry

• In an experiment: Reactants and products are the system; everything else is the surroundings.

• Energy flow from the system to the surroundings has a negative sign (loss of energy). (-E or - H)

• Energy flow from the surroundings to the system has a positive sign (gain of energy). (+E or +H)

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Page 7: CH 6:  Thermochemistry

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Page 8: CH 6:  Thermochemistry

• The law of the conservation of energy: Energy cannot be created or destroyed.

• The energy of an isolated system must be constant.

• The energy change in a system equals the work done on the system + the heat added.

E = Efinal – Einitial = E2 – E1 = q + w

q = heat, w = work8

Page 9: CH 6:  Thermochemistry

• Pressure is the force per unit area.(1 N/m2 = 1 Pa)

(1 atm = 101,325 Pa)

• Work is a force (F) that produces an object’s movement, times the distance moved (d):

Work = Force x Distance

AF

AreaForce=Pressure

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Page 10: CH 6:  Thermochemistry

The expansion in volume that occurs during a reaction forces the piston outward against atmospheric pressure, P.

Work = -atmospheric pressure * area of piston * distance piston moves

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Page 11: CH 6:  Thermochemistry

Example 2: Work

How much work is done (in kilojoules), and in which direction, as a result of the following reaction?

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Page 12: CH 6:  Thermochemistry

• The amount of heat exchanged between the system and the surroundings is given the symbol q.

q = E + PV

At constant volume (V = 0): qv = E

At constant pressure: qp = E + PV = H

Enthalpy change: H = Hproducts – Hreactants

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Page 13: CH 6:  Thermochemistry

Example 3: Work

The explosion of 2.00 mol of solid TNT with a volume of approximately 0.274 L produces gases with a volume of 489 L at room temperature. How much PV (in kilojoules) work is done during the explosion? Assume P = 1 atm, T = 25°C.

2 C7H5N3O6(s) 12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s)

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Page 14: CH 6:  Thermochemistry

• Enthalpies of Physical Change:

Enthalpy is a state function, the enthalpy change from solid to vapor does not depend on the path taken between the two states.

Hsubl = Hfusion + Hvap 14

Page 15: CH 6:  Thermochemistry

• Enthalpies of Chemical Change: Often called heats of reaction (Hreaction).

Endothermic: Heat flows into the system from the surroundings and H has a positive sign.

Exothermic: Heat flows out of the system into the surroundings and H has a negative sign.

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Page 16: CH 6:  Thermochemistry

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Bromination vs. Chlorination

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Page 17: CH 6:  Thermochemistry

• Reversing a reaction changes the sign of H for a reaction.

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) H = –2219 kJ

3 CO2(g) + 4 H2O(l) C3H8(g) + 5 O2(g) H = +2219 kJ

• Multiplying a reaction increases H by the same factor.

3 [C3H8(g) + 15 O2(g) 9 CO2(g) + 12 H2O(l)] H = 3(-2219) kJ

H = -6657 kJ17

Page 18: CH 6:  Thermochemistry

Example 4: Heat

• How much heat (in kilojoules) is evolved or absorbed in each of the following reactions?

a) Burning of 15.5 g of propane:

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

H = –2219 kJ/mole

b) Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride:

Ba(OH)2·8 H2O(s) + 2 NH4Cl(s) BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)

H = +80.3 kJ/mole

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Page 19: CH 6:  Thermochemistry

• Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and 25°C; 1 M concentration for all substances in solution.

• These are indicated by a superscript ° to the symbol of the quantity reported.

• Standard enthalpy change is indicated by the symbol H°.

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Page 20: CH 6:  Thermochemistry

Example 5:

Is an endothermic reaction a favorable process thermodynamically speaking?

1) Yes2) No

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Page 21: CH 6:  Thermochemistry

Hess’s Law

• Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.(not a physical change, chemical change)

3 H2(g) + N2(g) 2 NH3(g) H° = –92.2 kJ

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Page 22: CH 6:  Thermochemistry

• Reactants and products in individual steps can be added and subtracted to determine the overall equation.

(1) 2 H2(g) + N2(g) N2H4(g) H°1 = ?

(2) N2H4(g) + H2(g) 2 NH3(g) H°2 = –187.6 kJ

(3) 3 H2(g) + N2(g) 2 NH3(g) H°3 = –92.2 kJ

H°1 + H°2 = H°reaction

Then H°1 = H°reaction - H°2

H°1 = H°3 – H°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ

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Page 23: CH 6:  Thermochemistry

Example 6: Hess’s Law

• The industrial degreasing solvent methylene chloride (CH2Cl2, dichloromethane) is prepared from methane by reaction with chlorine:

CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g)

Use the following data to calculate H° (in kilojoules) for the above reaction:

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) H° = –98.3 kJ

CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g) H° = –104 kJ

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Page 24: CH 6:  Thermochemistry

• Standard Heats of Formation (H°f): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states.

• The standard heat of formation for any element in its standard state is defined as being ZERO.

H°f = 0 for an element in its standard state

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Page 25: CH 6:  Thermochemistry

Standard Heats of Formation

• Calculating H° for a reaction:

H° = H°f (Products) – H°f (Reactants)

• For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient.

aA + bB cC + dD

H° = [cH°f (C) + dH°f (D)] – [aH°f (A) + bH°f (B)]

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Page 26: CH 6:  Thermochemistry

-1131Na2CO3(s)49C6H6(l)-92HCl(g)

-127AgCl(s)-235C2H5OH(g)95.4N2H4(g)

-167Cl-(aq)-201CH3OH(g)-46NH3(g)

-207NO3-(aq)-85C2H6(g)-286H2O(l)

-240Na+(aq)52C2H4(g)-394CO2(g)

106Ag+(aq)227C2H2(g)-111CO(g)

Some Heats of Formation, Some Heats of Formation, HHff° ° (kJ/mol)(kJ/mol)

Standard Heats of Formation

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Page 27: CH 6:  Thermochemistry

Example 7: Standard heat of formation

Calculate H° (in kilojoules) for the reaction of ammonia with O2 to yield

nitric oxide (NO) and H2O(g), a step in the Ostwald process for the commercial production of nitric acid.

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Page 28: CH 6:  Thermochemistry

Example 8: Standard heat of formation

Calculate H° (in kilojoules) for the photosynthesis of glucose and O2 from

CO2 and liquid water, a reaction carried

out by all green plants.

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Page 29: CH 6:  Thermochemistry

Example 9:

Which of the following would indicate an endothermic reaction? Why?

1. -H

2. + H

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Page 30: CH 6:  Thermochemistry

Heat of Phase Transitions from Hf

Calculate the heat of vaporization, Hvap of water, using standard enthalpies of formation

HfH2O(g) -241.8 kJ/mol

H2O(l) -285.8 kJ/mol

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Page 31: CH 6:  Thermochemistry

Calorimetry and Heat Capacity

• Calorimetry is the science of measuring heat changes (q) for chemical reactions. There are two types of calorimeters:

• Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q = E.

• Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q = H.

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Page 32: CH 6:  Thermochemistry

Constant PressureBomb

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Page 33: CH 6:  Thermochemistry

Calorimetry and Heat Capacity

• Heat capacity (C) is the amount of heat required to raise the temperature of an object or substance a given amount.

Specific Heat: The amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C.

q = s x m x tq = heat required (energy)

s = specific heat

m = mass in grams

t = Tf - Ti

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Page 34: CH 6:  Thermochemistry

Calorimetry and Heat Capacity

• Molar Heat: The amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C.

q = MH x n x t

q = heat required (energy)

MH = molar heat

n = moles

t = Tf - Ti

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Page 35: CH 6:  Thermochemistry

Example 10: Specific Heat

What is the specific heat of lead if it takes 96 J to raise the temperature of a 75 g block by 10.0°C?

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Page 36: CH 6:  Thermochemistry

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Page 37: CH 6:  Thermochemistry

Example 11: Specific Heat

How much energy (in J) does it take to increase the temperature of 12.8 g of Gold from 56C to 85C?

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Page 38: CH 6:  Thermochemistry

Example 12: Molar Heat

• How much energy (in J) does it take to increase the temperature of 1.45 x104

moles of water from 69C to 94C?

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