Chapter 5 Thermochemistry. Energy The capacity to do work (or produce heat) We cannot see or directly measure energy Energy is a substance-like quantity.

Chapter 5

Thermochemistry

Energy• The capacity to do work (or produce

heat)• We cannot see or directly measure

energy• Energy is a substance-like quantity that

is always conserved• Energy is responsible for any change• Without energy, there would be no

chemistry

Types of Energy• potential energy- (PE) due to position or

composition–ex. attractive or repulsive forces (+ and )

• kinetic energy- (KE) due to motion of the object (thermal/translational)–KE = ½mv2 : depends on mass and speed

SI Units of Energy

• calorie - amount of energy to raise 1 gram of water by 1 degree C

• Joule 1 cal = 4.184 Joules

• In nutrition, 1 Calorie = 1000 cal = 1kcal

Transfer of Energy: Work & Heat

• Heat- (q) transfer of energy between two objects because of a temperature difference

Heat always flows from Hot to Cold

• Work- (w) force acting over a distance (w = F x d)

• Temperature- measure of the average kinetic energy of the particles

Chemical Energy

Exothermic Reactions:

• Gives off energy as it progresses

• PE stored in chemical bonds is converted to thermal energy (random KE) through heat.

• Products are generally more stable (stronger bonds) than reactants.

Endothermic Reactions:

• Energy is absorbed from the surroundings

• Energy flows into the system to increase the potential energy of the system

• Products are generally less stable (weaker bonds) than the reactants.

Transfer of Energy

Combustion of Methane Gas is exothermic

CH4 + 2O2 ---> 2H2O + CO2 + energy

Transfer of Energy

Reaction between nitrogen and oxygen is endothermic

N2 + O2 + energy ---> 2 NO

State Functions• Dependent only on the current condition.

• Independent of the past or the future.

• Independent of the pathway taken to get to that state

Ex: A liter of water behind a dam has the same potential energy for work regardless of whether it flowed downhill to the dam, or was taken uphill to the dam in a bucket. The PE is a state function dependent only on the current position of the water, not on how the water got there.

Thermodynamics

Thermodynamics – the study of energy and its transformations.

First Law of Thermodynamics –

• Energy can be neither created nor destroyed… energy is conserved.

• Energy that is lost by the system must be gained by the surroundings, and vice versa. ∆E = q + w

Internal Energy

• Eint = PE + KE

• ∆Eint = q + w

• can be changed by work, heat, or both

Signs

• signs will always reflect the system’s point of view unless otherwise stated

• q = heat

(+) for endothermic reactions

(-) for exothermic reactions

• w = work

(+) work done on the system

(-) system does work

E = change in energy

1. +q and +w, then + E

2. -q and –w, then – E

3. If the signs of q and w are opposite of one another, then the sign for E will depend on the magnitude of q and w.

Signs

Work done by gases

• w = -P V

• Through Expansion:

V (+) and w is (-)

• Through Compression:

V (-) and w is (+)

Example 1

• Find the ∆E for endothermic process where 15.6 kJ of heat flows and 1.4 kJ of work is done on system

• Since it is endothermic, what does that mean about the sign of q and w?

• q is (+) and w is (+)

kJ

kJkJ

wqE

0.17

4.16.15

Example 2

• Calculate the work of expansion of a gas from 46 L to 64 L at a constant pressure of 15 atm.

• Since it is an expansion, ∆V is + and w is -

atmL

LLatm

VPw

270

)4664)(15(

Example 3• A balloon was inflated from 4.00 x 106 L

to 4.50 x 106 L by the addition of 1.3 x 108 J of heat. Assuming the pressure is 1.0 atm, find the ∆E in Joules. (1 L∙atm=101.3 J)

Since it is an expansion, ∆V is + and w is -

VPqwqE

)1000.4105.4)(0.1(103.1 668 LLatmJE

JJJE 778 100.8)101.5(103.1

Enthalpy (H)• H = Eint + PV

• Since E, P, and V are all state functions… so is H

• If system at constant P, then qp = H

• H is the amount of energy exchanged between a system and its surroundings at constant P.

• heat of reaction and change in enthalpy are used interchangeably for a reaction at constant P∆H = Hproducts - Hreactants

endo: + ∆H exo: - ∆H

Calorimetry

• science of measuring heat

• calorimeter- device used to experimentally find the heat associated with a chemical reaction

• substances respond differently when heated

Heat Capacity• (C) how much heat it takes to raise a

substance’s T by one °C or K

• the amount of energy depends on the amount of substance

K)or C(in Tin increase

J)(in absorbedheat Ccapacityheat

Heat Capacity

• specific heat capacity (c) •heat capacity per gram

–in J/°C•g or J/K•g

• molar heat capacity

– heat capacity per mole

–in J/°C•mol or J/K•mol

Constant-Pressure Calorimetry

• uses simplest calorimeter (like coffee-cup calorimeter) since it is open to air

• used to find changes in enthalpy (heats of reaction) for reactions occurring in a solution since qP = ∆H

• heat of reaction is an extensive property (dependent on amount of substance), so we usually write them per mole so they are easier to use

Constant-Pressure Calorimetry

• when 2 reactants are mixed and T increases, the chemical reaction must be releasing heat so is exothermic

• the released energy from the reaction increases the motion of molecules, which in turn increases the T

Constant-Pressure Calorimetry

• If we assume that the calorimeter did not leak energy or absorb any itself (that all the energy was used to increase the T), we can find the energy released by the reaction:

E released by rxn = E absorbed by soln

∆H = qP = m c ∆T

Constant-Volume Calorimetry

• uses a bomb calorimeter

• weighed reactants are placed inside the rigid, steel container and ignited

• water surrounds the reactant container so the T of it and other parts are measured before and after reaction

Constant-Volume Calorimetry

E released by rxn = ∆T x Ccalorimeter

Example 1

• When 1 mol of CH4 is burned at constant P, 890 kJ of heat is released. Find ∆H for burning of 5.8 g of CH4 at constant P.

• 890 kJ is released per mole of CH4

CH4 + 2O2 ---> CO2 + 2H2O ∆H = 890 kJ

5.8gCH4 1moleCH4

16.0gCH4

890kJ

1molCH4

320kJ

Example 2

• When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0°C in a coffee-cup calorimeter, solid BaSO4 forms and the T increases to 28.1°C. The specific heat of the solution is 4.18 J/g•°C and the density is 1.0 g/mL. Find the enthalpy change per mole of BaSO4 formed.

Example 2• Write the net ionic equation for the

reaction:

Ba2+ (aq) + SO42- (aq) BaSO4(s)

• Is the energy released or absorbed? What does that mean about ∆H and q?

exothermic: ∆H and –qP

• How can we calculate ∆H or heat?heat = q = m c ∆T

• How can we find the m?use density and volume

Example 2

• Find the mass:

• Find the change in T:

• Calculate the heat created:

gmLmLg

VDmV

mD

33 100.2)100.2)(/0.1(

CT 1.30.251.28

q(2.0 103g)(4.18J

Cg)(3.1C) 2.6 104 J

Example 2

• since it is a one-to-one ratio and the moles of reactants are the same, there is no limiting reactant

• 1.0 mol of solid BaSO4 is made so

∆H= -2.6x104 J/mol = -26 kJ/mol

Example 3

• Compare the energy released in the combustion of H2 and CH4 carried out in a bomb calorimeter with a heat capacity of 11.3 kJ/°C. The combustion of 1.50 g of methane produced a T change of 7.3°C while the combustion of 1.15 g of hydrogen produced a T change of 14.3°C. Find the energy of combustion per gram for each.

Example 3

• methane: CH4

• hydrogen: H2

gkJgkJH

kJCC

kJH

TCH rcalorimete

/555.1/83

83)3.7()3.11(

gkJgkJH

kJCC

kJH

TCH rcalorimete

/14115.1/162

162)3.14()3.11(

The energy released by HThe energy released by H22 is about is about 2.5 times the energy released by CH2.5 times the energy released by CH44

Hess’s Law

• If a reaction is carried out in a series of steps, ∆H for the reaction will equal the sum of the enthalpy changes for the individual steps.

Examples

• One Step:

N2(g) + 2O2 (g) 2NO2 (g) H1 = 68kJ

• Two Step:

N2(g) + O2 (g) 2NO (g) H2 = 180kJ

2NO(g) + O2 (g) 2NO2 (g) H3 = -112kJ

N2(g) + 2O2 (g) 2NO2 (g) H2 + H3 = 68kJ

Hess’s Law Rules

• If the reaction is reversed, then the sign on H is reversed

• The magnitude of H is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of H is multiplied by the same integer

Using Hess’s Law

1. Work backward from the final reaction

2. Reverse reactions as needed, being sure to also reverse H

3. Remember that identical substances found on both sides of the summed equation cancel each other

Example 2

• Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to diamond.

Cgraphite (s) Cdiamond (s) ∆H=?

Example 2 Cgraphite (s) Cdiamond (s) ∆H=?

(1) Cgraphite(s) + O2(g) CO2(g) ∆H=-394kJ/mol

(2) Cdiamond(s) + O2(g) CO2(g) ∆H=-396kJ/mol

• to get the desired equation, we must reverse 2nd equation:

(1) Cgraphite(s) + O2(g) CO2(g) ∆H=-394kJ/mol

+(2) CO2(g) Cdiamond(s) + O2(g) ∆H= 396kJ/mol

Cgraphite (s) Cdiamond (s)∆H=-394 + 396

∆H=2 kJ/mol

Example 3

• Find ∆H for the synthesis of B2H6, diborane:

2B(s) + 3H2(g) B2H6(g) ∆H =?

Given: 2B(s) + 3/2O2(g) B2O3(s) ∆H1=-1273kJ

B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g) ∆H2=-2035kJ

H2(g) + ½O2(g) H2O(l) ∆H3=-286kJ

H2O(l) H2O(g) ∆H4=44 kJ

Example 3 2B(s) + 3H2(g) B2H6(g) ∆H =?• Start by paying attention to what needs to be

on reactants and products side

(1) 2B(s) + 3/2O2(g) B2O3(s) ∆H1=-1273kJ

-(2) B2O3(s) + 3H2O(g) B2H6(g) + 3O2(g) -∆H2=2035kJ

(3) H2(g) + ½O2(g) H2O(l) ∆H3=-286kJ

(4) H2O(l) H2O(g) ∆H4=44 kJ

Example 3

• Underline what you want to keep- that will help you figure out how to cancel everything else:

(1) 2B(s) + 3/2O2(g) B2O3(s) ∆H1=-1273kJ

-(2) B2O3(s) + 3H2O(g) B2H6(g) + 3O2(g) -∆H2=2035kJ

(3) H2(g) + ½O2(g) H2O(l) ∆H3=-286kJ

(4) H2O(l) H2O(g) ∆H4=44 kJ

Example 3• Need 3 H2 (g) so 3 x (3)• Need 3 H2O to cancel so 3 x (4)

(1) 2B(s) + 3/2O2(g) B2O3(s) ∆H1=-1273kJ-(2) B2O3(s) + 3H2O(g) B2H6(g) + 3O2(g)

-∆H2=-(-2035kJ)3x(3) 3H2(g) + 3/2O2(g) 3H2O(l) 3∆H3=3(-286kJ)3x(4) 3H2O(l) 3H2O(g) 3∆H4=3(44 kJ)

2B(s) + 3H2(g) B2H6(g)∆H = -1273 + -(-2035) + 3(-286) + 3(44) = 36kJ

Standard Enthalpy of Formation ∆Hf°

• change in enthalpy that accompanies the formation of one mole of a compound from its elements in standard states

• ° means that the process happened under standard conditions so we can compare more easily

Standard States

• For a COMPOUND:– for gas: P = 1 atm– For (s) or (l): pure liquid or solid– in solution: concentration is 1 M

• For an ELEMENT:– form that exists in at 1 atm and 25°C

O: O2(g) K: K(s) Br: Br2(l)

Writing Formation Equations

• always write equation where 1 mole of compound is formed (even if you must use non-integer coefficients)

NO2(g):

½N2(g) + O2(g) NO2(g)

∆Hf°= 34 kJ/mol

CH3OH(l):C(s) + 2H2(g) + ½ O2(g) CH3OH(l)

∆Hf°= -239 kJ/mol

Using Standard Enthalpies of Formation

where

– n = number of moles of products/reactants

– ∑ means “sum of”

– ∆Hf° is the standard enthalpy of formation for reactants or products

• ∆Hf° for any element in standard state is zero so elements are not included in the summation

∆Hrxn = ∑n ∆H∆Hff° ° (products) ∑n ∆H∆Hff° ° (reactants)

Example 1• Calculate the standard enthalpy change for

the reaction that occurs when ammonia is burned in air to make nitrogen dioxide and water

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(l)

∆Hfo values (in appendix in back of book (pg.

1041):Ammonia = -46.19 kJ/molOxygen = 0 (because it is in its standard state)Nitrogen Dioxide = 33.84 kJ/molWater (l) = - 285.8 kJ/mol

[(4 x 33.84 kJ/mol) + (6 x -285.8)] – [(4 x -46.19) + (7 x 0)] =

• -1394.68 kJ/mol

• -1395 kJ/mol

Example 1

Example 2

• Calculate the standard enthalpy change for the following reaction (H Iron (III) Oxide = -1676 kJ and luminum Oxide = -826 kJ)

2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s)

s)f(reactantr)f(productspreaction ˘Hn•˘Hn•˘H

kJ.850˘H

)]1676(1)0(2[)]0(2)826(1[˘H

reaction

reaction

Example 3

• Compare the standard enthalpy of combustion per gram of methanol with per gram of gasoline (it is C8H18).

• Write equations:

2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(l)

2C8H18(l) + 25O2(g)16CO2(g) + 18H2O(l)

Example 3• Calculate the enthalpy of combustion per mole:

2CH3OH(l) + 3O2(g) 2CO2(g) + 4H2O(l)

]2[]42[ )()()(3 322

OHCHfOHfCOfOHCH HHHH

²H CH3OH 1454 kJ for 2 moles CH3OH

]2[]1816[ )()()( 18822188

HCfOHfCOfHC HHHH

²H C8H18

1.09 104 kJ for 2 moles C8H18

2C2C88HH1818(l) + 25O(l) + 25O22(g)(g)16CO16CO22(g) + 18H(g) + 18H22O(l)O(l)

Example 3

• Convert to per gram using molar mass:

• so octane is about 2x more effective

g

kJ22.7

g 32.0

OHCH mol 1

OHCH mol 2

kJ 1454 3

3

g

kJ.874

g 114.2

HC mol 1

HC mol 2

kJ 101.09 188

188

4