thermochemistry: chemical energy chapter 8. energy is the capacity to do work thermal energy is the...

Thermochemistry: Chemical Energy

Chapter 8

Energy is the capacity to do work

• Thermal energy is the energy associated with the random motion of atoms and molecules

• Chemical energy is the energy stored within the bonds of chemical substances

• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom

• Electrical energy is the energy associated with the flow of electrons

• Potential energy is the energy available by virtue of an object’s position, or stored energy

• Kinetic energy is moving energy.

E = qrxn + w

Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

Temperature is a measure of the thermal energy.

Temperature = Thermal Energy

900C400C

greater thermal energy

Thermochemistry is the study of heat change in chemical reactions.

The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing

SYSTEMSURROUNDINGS

Calorimetry and Heat CapacityMeasure the heat flow at constant pressure (H).

Calorimetry and Heat CapacityMeasure the heat flow at constant volume (E).

First Law of Thermodynamics

• Energy is neither created nor destroyed.• In other words, the total energy of the universe is a

constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Use Fig. 5.5

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

H < 0Hproducts > Hreactants

H > 0

Enthalpy Changes 01

• Enthalpies of Physical Change:

Enthalpies of Physical and Chemical Change

Enthalpy of Fusion (Hfusion): The amount of heat necessary to melt a substance without changing its temperature.

Enthalpy of Vaporization (Hvap): The amount of heat required to vaporize a substance without changing its temperature.

Enthalpy of Sublimation (Hsubl): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase.

Thermochemical Equations

H2O (s) H2O (l) H = 6.01 kJ

Is H negative or positive?

System absorbs heat

Endothermic

H > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890.4 kJ

Is H negative or positive?

System gives off heat

Exothermic

H < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

H2O (s) H2O (l) H = 6.01 kJ

• The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations

• If you reverse a reaction, the sign of H changes

H2O (l) H2O (s) H = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O (s) 2H2O (l) H = 2 x 6.01 = 12.0 kJ

H2O (s) H2O (l) H = 6.01 kJ

• The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical Equations

H2O (l) H2O (g) H = 44.0 kJ

How much heat is transfered when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x-3013 kJ1 mol P4

x = -6470 kJ

The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = ms

Heat (q) absorbed or released:

q = mst

q = Ct

t = tfinal - tinitial

Calorimetry and Heat CapacityAssuming that a can of soda has the same specific heat as water, calculate the amount of heat (in kilojoules) transferred when one can (about 350 g) is cooled from 25 °C to 3 °C.

Specific heat (Water) = 4.18g °C

J

Temperature change = 3 C-25 °C = - 22 °C

Mass = 350 g

q = (specific heat) x (mass of substance) x T

Calorimetry and Heat CapacityCalculate the amount of heat transferred.

= -32 000 J -22 °C

1000 J

1 kJ

x 350 gxHeat evolved =

= -32 kJ

g °C

4.18 J

x-32 000 J

Calorimetry and Heat Capacity

Bomb Calorimetry

• Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.

E = qrxn + w

• W = VΔP

• For most reactions, the difference is very small.

• ΔE almost = qrxn

Constant-Volume Calorimetry

Constant-Volume Calorimetry

No heat enters or leaves!

qsys = qwater + qCal + qrxn

qsys = 0

qrxn = - (qwater + qCal)

qwater = mst

qCal = CCalt

Constant-Pressure Calorimetry

No heat enters or leaves!

qsys = qwater + qcal + qrxn

qsys = 0

qrxn = - (qwater + qcal)

qwater = mst

qcal = Ccalt

Reaction at Constant PH = qrxn

Hess’s Law

Haber Process:

Multiple-Step Process

N2H4(g)2H2(g) + N2(g)

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.

2NH3(g)3H2(g) + N2(g) H° = -92.2 kJ

H°1 = ?

2NH3(g)N2H4(g) + H2(g)

H°1+2 = -92.2 kJ2NH3(g)3H2(g) + N2(g)

H°2 = -187.6 kJ

H°1 = H°1+2 - H°2

Hess’s Law

= -92.2 kJ - (-187.6 kJ) = 95.4 kJ

H°1 + H°2 = H°1+2

Hess’s Law 01

• 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.

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

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

Standard Heats of Formation

H2(g) + 1/2 O2(g) H2O(l) H°f = –286 kJ/mol

3/2 H2(g) + 1/2 N2(g) NH3(g) H°f = –46 kJ/mol

2 C(s) + H2(g) C2H2(g) H°f = +227 kJ/mol

*Standard state (°):

P = 1atm, t = 25°C,

Concentration = 1 mol/Lit

Standard Heats 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

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)]

Standard Heats of Formation

-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)

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x(–393.5) + 6x( -286 ) – [ 2x(49.04) ] = -6542.0

kJ-6542.0

2 mol= - 3271.0 kJ/mol C6H6

See Page 292 for Standard Heat of Formation

Standard Heats of FormationUsing standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O.

C6H12O6(s) + 6O2(g)6CO2(g) + 6H2O(l) H° = ?

H° = [H°f (C6H12O6(s))] - [6 H°f (CO2(g)) + 6 H°f (H2O(l))]

= 2816 kJ

[(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]

H° = [(1 mol)(-1260 kJ/mol)] -

Bond Dissociation Energy• Bond Dissociation Energy: Can be used to determine an

approximate value for H°f .

H°f = D (Bonds Broken) – D (Bonds Formed)

ΔHorxn = D (Reactant bonds) - D (Product bonds)

• For the reaction: H-H (g) + Cl-Cl (g) 2H-Cl (g)

H°rxn = (D(H–H) + D(Cl-Cl))-(2 X D(H–Cl) ) H°rxn = -185 kJ

H°f = -92.5 KJ/mol

Bond Dissociation Energy 02

Calculate an approximate H° (in kilojoules) for the synthesis of

ethyl alcohol from ethylene:C2H4(g) + H2O(g) C2H5OH(g)

HH

HH HH

HH

CC CC

O-HO-H

CC HH

HH

+ H-O-H+ H-O-H

HH

CC HH

HH

ΔHorxn = D (Reactant bonds) - D (Product bonds)

ΔHorxn = (DC=C + 4 DC H + 2 DO H) - (DC C + DC O + 5 DC H + DO H)

ΔHorxn = [(1 mol)(611 kJ/mol) + (4 mol)(410 kJ/mol)

+(2 mol)(460 kJ/mol)] - [(1 mol)(350 kJ/mol) + (1 mol)( 350 kJ/mol) + (5 mol)(410 kJ/mol) + (1 mol)(460 kJ/mol)]

ΔHorxn = 39 kJ

ΔHorxn = (DC=C + 4 DC H + 2 DO H) - (DC C + DC O + 5 DC H + DO H)

Thermodynamics

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

energy, pressure, volume, temperature

Predicting spontaneity of a chemical Reaction

• Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings.

• A spontaneous process is one that proceeds on its own without any continuous external influence.

• A nonspontaneous process takes place only in the presence of a continuous external influence.

Introduction to Entropy 02

• The measure of molecular disorder in a system is called the system’s entropy; this is denoted S.

• Entropy has units of J/K (Joules per Kelvin).

S = Sfinal – Sinitial

– Positive value of S indicates increased disorder.

– Negative value of S indicates decreased disorder.

Introduction to Entropy 03

Introduction to Entropy 05

• Predict whether S° is likely to be positive or negative for each of the following reactions.

• a. 2 CO(g) + O2(g) 2 CO2(g)

b. 2 NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)

c. C2H4(g) + Br2(g) CH2BrCH2Br(l)

d. 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)

Introduction to Entropy 04

• To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered:

• Spontaneous process:

Decrease in enthalpy (–H), Increase in entropy (+S).

• Nonspontaneous process:

Increase in enthalpy(+H),Decrease in entropy (–S).

Introduction to Free Energy 01

• Gibbs Free Energy Change (G): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process.

G = H – TS

G < 0 Process is spontaneous

G = 0 Process is at equilibrium

G > 0 Process is nonspontaneous

Which of the following reactions are spontaneous under

standard conditions at 25°C?

•AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) G° = –55.7 kJ

2 C(s) + 2 H2(g) C2H4(g)

G° = 68.1 kJ

N2(g) + 3 H2(g) 2 NH3(g)

H° =-92.2 kJ S° = -199 J/K

ΔGo = ΔHo -TΔSo = ( -92.2 kJ) - (298 K)(- 0.199 kJ/K) = -32.9 kJBecause ΔGo is negative, the reaction is spontaneous.

Introduction to Free Energy 04

• Equilibrium (G° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature?

– N2(g) + 3 H2(g) 2 NH3(g)

H° = –92.0 kJ S° = –199 J/K

– Equilibrium is the point where G° = H° – TS° = 0

– T = 462 K