1 vanessa prasad-permaul valencia college chm 1045

1

Vanessa Prasad-PermaulValencia CollegeCHM 1045

2

•INTRODUCTION TO THERMODYNAMICS•KEY DEFINITIONS•FIRST LAW OF THERMODYNAMICS•HEAT, WORK & INTERNAL ENERGY•ENTHALPY•THERMODYNAMIC SYSTEM•HESS’ LAW•STANDARD ENTHALPIES OF FORMATION•ENERGY AND THE ENVIRONMENT

THERMOCHEMISTRYENERGY

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

Energy = Work + Heat

Kinetic Energy is the energy of motion.Ek = 1/2 mv2 Ek = kinetic energy

m = mass (kg) v = velocity (m/s)

(1 Joule = 1 kgm2/s2) (1 calorie = 4.184 J)

Potential Energy is stored energy.

3


4

EXAMPLE 6.1 A regulation baseball weighing 143 grams travels 75 miles per hour. What is the kinetic energy of this baseball in Joules? Convert to calories.

Ek = ½ mv2

m = 143g x 1kg = 0.143kg 1000g

v = 75miles x 1hr x 1min x 1609.3m =

33.527 m/s 1 hour 60min 60sec 1 mile

Ek = ½ x 0.143kg x (33.527m/s)2 = 80.3kg.m2/s2 = 80.3J

80.3J x 1 cal = 19.2cal 4.184J


5

EXERCISE 6.1 An electron whose mass is 9.11 x 10-31 kg is accelerated by a positive charge to a speed of 5.0 x 106 m/s. What is the kinetic energy of the electron in Joules? In calories?

Ek = ½ mv2

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 = work

6

THE FIRST LAW OF THERMODYNAMICS:

Thermal Energy is the kinetic

energy of molecular motion

Thermal energy is proportional to the

absolute temperature. Ethermal T(K)

Heat is the amount of thermal

energy transferred between two

objects at different temperatures.

7

Enthalpies of Physical Change:

8

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

9

ILLUSTRATION OF A THERMODYNAMIC SYSTEM

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)

10

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

Endothermic: Heat flows into the system from the surroundings, heat is absorbed.

H has a positive sign. Energy added q is (+)

Exothermic: Heat flows out of the system into the surroundings, heat is evolved.

H has a negative sign. Energy subtracted q is (-)

11

12

q (heat) +System gains

thermal energy

-System loses

thermal energy

w (work) +Work done

on a system

-Work done

by the system

E (change in internal energy)

+Energy flows

into the system

-Energy flows

out of the system

SIGN CONVENTIONS FOR HEAT, WORK AND INTERNAL ENERGY

1313

EXAMPLE 6.2Barium hydroxide octahydrate reacts with ammonium nitrate:

Ba(OH)2.8H2O(s) + 2NH4NO3(s) 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)

When 1mol of barium hydroxide octahydrate reacts with 2 mol of ammonium nitrate, the reaction mixture absorbs 170.8kJ of heat. Is this reaction exothermic or endothermic? What is the heat of reaction (q)?

Energy is absorbed so this reaction is endothermic.

q is +170.8kJ

14

EXERCISE 6.2 Ammonia burns in the presence of a platinum catalyst to give nitric oxide

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l)

4 mol of ammonia is burned and 1170kJ of heat is evolved. Is the reaction endothermic or exothermic? What is the value of q?

Pt

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 of Reaction: H = Hproducts – Hreactants

15

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

w = -PV 16

PRESSURE-VOLUME WORK

1717

EXAMPLE 6.2AIf a balloon is inflated from 0.100L to 1.85L against an external pressure of 1.00atm, how much work is done (J)?

w = -PV = -1.00atm (1.85L – 0.100L)

= -1.75 L.atm

Conversion 1L.atm = 101.3J

-1.75 L.atm x 101.3J 1L.atm-177J

18

EXERCISE 6.2AA cylinder equipped with a piston expands against an external pressure of 1.58 atm. If the initial volume is 0.485L and the final volume is 1.245L, how much work is done (in J)?

19

EXAMPLE 6.3Aqueous sodium hydrogen carbonate reacts with hydrochloric acid to produce aqueous sodium chloride, water and carbon dioxide gas. The reaction absorbs 12.7kJ of heat at constant pressure for each mole of NaHCO3. Write the thermochemical equation.

NaHCO3(aq) + HCl NaCl(aq) + H2O(l) + CO2(g)

H = +12.7kJ

2020

EXERCISE 6.3A propellant for rockets is obtained by mixing the liquids hydrazine (N2H4) and dinitrogen tetroxide. These compounds react to give gaseous nitrogen and water vapor and 1049kJ of heat is evolved (at constant pressure when 1 mol reacts). Write the thermochemical equation for this reaction

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 kJ21

22

EXAMPLE 6.4When 2 mol H2(g) and 1mol O2 react to give liquid water, 572kJ of heat evolves:

2H2(g) + O2(g) 2H2O(l) ; H = -572kJWrite this equation for 1 mol of liquid water. Give the reverse equation, in which 1 mol of liquid water dissociates into hydrogen and oxygen.

2H2(g) + O2(g) 2H2O(l) ; H = -572kJ2

1H2(g) + ½ O2(g) 1H2O(l) ; H = -286kJ

Reversing the equation:1H2O(l) 1H2(g) + ½ O2(g) ; H = +286kJ

23

EXERCISE 6.4Write the thermochemical equation for the reaction described in Exercise 6.3 for the case involving 1 mol N2H4. Write the reverse reaction.

24

Applying Stoichiometry to Heats of Reaction

Grams of A

(reactant or

product)

Conversion Factor: grams of A to mols of A (using molar

mass)

Conversion Factor: mols of

A to kJ (H)

kiloJoules of Heat

x x

=

25

EXAMPLE 6.5How much heat is involved when 9.007x105g of ammonia is produced according to the following reaction (assuming the reaction is at constant pressure)?

N2(g) + 3H2(g) 2NH3(g); H = -91.8kJ

9.07 x 105g x 1 mol NH3 x -91.8kJ = -2.45 x 106 kJ

17.0g NH3 2 mol NH3

26

EXERCISE 6.5How much heat evolves when 10.0g of hydrazine reacts according to the reaction described in EXERCISE 6.3?

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

qcal = Ccal x T

Specific Heat: The amount of heat required to raisethe temperature of 1.00 g of substance by 1.00°C atconstant pressure.

q = s x m x t

q = heat required (energy)

s = specific heat

m = mass in grams

t = Tf - Ti

27

28

SPECIFIC HEATS AND MOLAR CAPACITIES FOR SOME COMMON SUBSTANCES @ 25oC

• 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

29

30

EXAMPLE 6.6Calculate the heat absorbed by 15.0g of water to raise the temperature from 20.0oC to 50.0oC (at constant pressure). The specific heat of water is 4.18 J/g.oC

t = 50.0oC – 20.0oC = 30.0oC

q = s x m x t

q = 4.18J x 15.0g x 30.0oC = 1.88 x 103 J g.oC

31

EXERCISE 6.6Iron metal has a specific heat of 0.449 J/(g.oC). How much heat is transferred to a 5.00 g piece of iron, initially at 20.0oC when placed in a pot of boiling water? (Assume that the temperature of the water is 100.0oC and the water remains at this temperature, which is the final temperature of the iron).

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.

32

33

Constant PressureBomb

34

EXAMPLE 6.7Suppose 0.562g of graphite is placed in a calorimeter with an excess of oxygen at 25.00oC and 1atm pressure. Excess O2 ensures that all carbon burns to form CO2. The graphite is ignited, and it burns according to the following equation

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

On reaction, the calorimeter temperature rises from 25.00oC to 25.89oC. The heat capacity of the calorimeter and it’s contents was determined in a separate experiment to be 20.7 kJ/oC. What is the heat of reaction at 25.00oC and 1 atm pressure? Express in a thermochemical equation.

35

EXAMPLE 6.7 continued…

qrxn = -Ccalt = -20.7kJ/oC x (25.89oC – 25.00oC) = -20.7kJ/oC x 0.89oC = -18.4kJ

0.562g x 1 mol = 0.0468 mol 12 g C

-18.4kJ = -3.9 x 102 kJ/mol 0.0468mol

C(graphite) + O2(g) CO2(g) ; H = -3.9 x 102 kJ

36

EXERCISE 6.7Suppose 33mL of 0.120M HCl is added to 42mL of a solution containing excess sodium hydroxide in a coffee cup calorimeter. The solution temperature, originally at 25oC, rises to 31.8oC. Give the enthalpy change H for the reaction:

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Assume that the heat capacity and the density of the final solution in the cup are those of water.s = 4.184kJ/g.oCd = 1.000g/mLExpress the answer as a thermochemical equation.

THERMOCHEMISTRYHESS’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

37


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 kJ38


39

EXAMPLE 6.8What is the enthalpy of reaction, H, for the formation of tungsten carbide (WC) from the elements?

W(s) + C(graphite) WC(s)

2W(s) + 3O2(g) 2WO3(s) ; H

= -1685.8 kJ

C(graphite) + O2(g) CO2(g) ; H =

-393.5 kJ

2WC(s) + 5O2(g) 2WO3(s) + 2CO2(g) ; H =

-2391.8kJ

LAST EQUATION NEEDS TO BE REVERSED2CO2(g) + 2WO3(s) 2WC(s) + 5O2(g) ; H

= 2391.8kJ


40

2W(s) + 3O2(g) 2WO3(s) ; H = -

1685.8 kJ 2 2

C(graphite) + O2(g) CO2(g) ; H = -

393.5 kJ

2CO2(g) + 2WO3(s) 2WC(s) + 5O2(g) ; H =

2391.8kJ

2

2


EXAMPLE 6.8 cont…

41

W(s) + 3/2 O2(g) WO3(s) ; H =

-842.9 kJ

C(graphite) + O2(g) CO2(g) ; H =

-393.5 kJ

CO2(g) + WO3(s) WC(s) + 5/2 O2(g) ;

H = 1195.9kJ


W(s) + C(graphite) WC(s) H = -40.5kJ

EXAMPLE 6.8 cont…

42

EXERCISE 6.8Manganese metal can be obtained by the reaction of manganese dioxide and aluminum

4Al(s) + 3MnO2(s) 2Al2O3(s) + 3Mn(s)

What is the H for this reaction? Use the following data:

2Al(s) + 3/2O2(g) Al2O3(s) ; H = −1676 kJ

Mn(s) + O2(g) MnO2(s) ; H = −520 kJ

THERMOCHEMISTRYSTANDARD ENTHALPIES OF FORMATION

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

43

-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))


44


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°.

45


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)]

46


47

EXAMPLE 6.9Use the values of Ho

f to calculate the heat of vaporization Ho

vap of carbon disulfide at 25oC. The vaporization process is:

CS2(l) CS2(g)

Hof = 89.7kJ/mol

Hof = 116.9kJ/mol

Hovap = nHo

f(products) – mHof (reactants)

Hof[CS2(g)] - Ho

f [CS2(l)]116.9kJ - 89.7kJ = 27.2kJ


48

EXERCISE 6.9Calculate the heat of vaporization, Ho

vap of water using standard enthalpies of formation


49

EXAMPLE 6.10Large quantities of ammonia are used to prepared nitric acid. The first consists of the catalytic oxidation of ammonia to nitric oxide

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) 45.9kJ 0kJ 90.3 6kJ -241.8kJ

What is the standard enthalpy change for this reaction?

Hovap = nHo

f(products) – mHof (reactants)

= [4(90.3) + 6(-241.8)]kJ - [4(-45.9) +

5(0)]kJ = -906kJ


50

EXERCISE 6.10Calculate the enthalpy change for the following reaction:

3NO2(g) + H2O(l) 2HNO3(aq) + NO(g)


51

EXERCISE 6.11Calculate the standard enthalpy change for the reaction of an aqueous solution of barium hydroxide with an aqueous solution of ammonium nitrate at 25oC.

THERMOCHEMISTRYFUELS

52

FUELS is any substance that is burned or react to provide heat and other forms of energy.

FOODS AS FUELS

FOSSIL FUELS

COAL GASIFICATION AND LIQUEFICATION

ROCKET FUELS

THERMOCHEMISTRYFUELS

53

Sources of energy consumed in the United States.

Example 2: Work

How much work is done (in kilojoules) and in

which direction, as a result of the following

reaction?

54

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)

55

Example 5:

Is an endothermic reaction a favorable process thermodynamically speaking?

1) Yes2) No

56

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

57

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.

58

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.

59

Example 9:

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

1. -H

2. + H

60

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?

61

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?

62

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?

63

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

64

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

65

41

Bromination vs. Chlorination

66

1 vanessa prasad-permaul valencia college chm 1045

Documents

energy change

internal energy energy

thermal energysystem

stored energy

energy of motion

jpotential energy

energy subtractedq

conservation of energy