Chapter 6: Thermochemistry Thermochemistry: energy considerations associated with chemical and physical change

Energy: the capacity of a system to do work or produce heat

potential energy – energy of position; stored energy

kinetic energy – energy of motion

kinetic energy = ! mυ2

kinetic energy = 3⁄2 RT

In a system, energy can be transferred in the form of heat or work:

heat, q – energy transfer results in temperature change (ΔT); ΔT = Tfinal – Tinitial

work, w – energy transfer during the act of moving an object against an opposing force

overall: ΔU = q + w

An Example of a System Configured for Energy Transferred as Work:

Units of EnergySI unit of energy: Joule, J

1 Joule is the amount of energy required to raise the temperature of 0.2390 g H2O by 1ºC.

1 kJ = 103 J1 MJ = 106 J

thermochemical calorie, cal

1 calorie is the amount of energy required to raise the temperature of 1 g of H2O by 1ºC.

1 cal = 4.184 J1 kcal = 103 cal

1 nutritional calorie, 1 Cal = 1000 cal

The System, The Surroundings, The Universe & The First Law of Thermodynamics

The System:

◆ what we’re interested in

The Surroundings:

◆ everything else

The Universe:

◆ System + Surroundings = the Universe

The 1st Law of Thermodynamics:

The energy of the Universe is constant.

Endothermic Change:

◆ energy is absorbed by the system

◆ energy flow is from the surroundings into the system

◆ Esys increases Esurr decreases

◆ for the system: Efinal > Einitial

◆ if energy transferred in the form of heat, Tsys increases and Tsurr decreases

◆ E lost by surroundings = E gained by the system

Exothermic Change

◆ energy is released by the system

◆ energy flow is from the system into the surroundings

◆ Esys decreases Esurr increases

◆ for the system: Efinal < Einitial

◆ if energy transferred in the form of heat, Tsys decreases and Tsurr increases

◆ E lost by system = E gained by the surroundings

Changes in Thermchemical Quantities:importance of sign and magnitude

ΔX = Xfinal – Xinitial

if ΔX is + : Xfinal > Xinitial

if ΔX is ! : Xfinal < Xinitial

if ∆E is +Efinal > Einitial

energy is absorbed by the system

endothermic change

if ∆E is !Efinal < Einitial

energy is released by the system

exothermic change

if w is +

surroundings do work on the system

endothermic

if q is +qfinal > qinitial

heat is absorbed by the system

endothermic

if w is !

system does work on the surroundings

exothermic

if q is !qfinal < qinitial

heat is released by the system

exothermic

example:

Calculate the change in internal energy of a system that releases 26.8 kJ of heat as it does 68.7 kJ of work on the surroundings.

PV Workexpansion or compression of an ideal gas

w = !P∆V

notes:

◆ w and ∆V are always opposite in sign; why?

◆ 1 L•atm = 101.3 J

example:

Calculate the quantity of work done (in kJ) by an ideal gas as it expands from an initial volume of 2.50 L to a final volume of 27.50 L against a constant external pressure of 0.980 atm.

Enthalpy, HH = E + PV

◆ We will be focussing on changes in enthalpy that accompany processes that occur at constant P.

at constant P: ∆H = qP

∆H = Hfinal ! Hinitial

if ∆H is + : heat is absorbed by the systemendothermic process

if ∆H is ! : heat is released by the systemexothermic process

Enthalpy Changes for Chemical Reactions:Thermochemical Equations

∆Hrxn = Hfinal ! Hinitial

OR ∆Hrxn = Hproducts ! Hreactants

Thermochemical Equation: balanced chemical equation + thermochemical data

example: 2 Na (s) + 2 H2O (l) ! 2 NaOH (aq) + H2 (g); ∆H = –367.5 kJ

notes:◆ reaction is exothermic as written◆ specifically, 367.5 kJ of energy is released when:

2 mol Na & 2 mol H2O are consumed 2 mol NaOH & 1 mol H2 are formed

An Enthalpy Diagram for This Reaction:interpretation of energy changes associated with a reaction must take into consideration the following:

◆ phases of reactants and productsex. 2 H2 (g) + O2 (g) ! 2 H2O (g); ∆H = –483.7 kJ

vs.2 H2 (g) + O2 (g) ! 2 H2O (l); ∆H = –571.5 kJ

◆ direction of the reaction:a reaction is endothermic in one direction and exothermic in the reverse direction

ex. CH4 (g) + 4 Cl2 (g) ! CCl4 (l) + 4 HCl (g); ∆H = –433 kJ vs.

CCl4 (l) + 4 HCl (g) ! CH4 (g) + 4 Cl2 (g); ∆H = +433 kJ

Enthalpy Changes for Chemical Reactions:Thermochemical Equations

Enthalpy Changes for Chemical Reactions:Thermochemical Equations

◆ stoichiometry of balanced equation

energy released or absorbed is an extensive property (depends on size of sample)

ex. combustion of propaneC3H8 (g) + 5 O2 (g) ! 3 CO2 (g) + 4 H2O (g); ∆Hrxn = –2200 kJ

combustion of 100 g (2.27 mol) C3H8 generates ~5000 kJvs.

combustion of 2.5 kg (56.7 mol) C3H8 generates ~125,000 kJ

example:

Determine the quantity of heat released (in kJ) in the synthesis of 907 kg NH3.

example:

12.0 L N2 (g) at STP and 5.00 L H2 (g) at 30ºC and 3.02 atm are combined and allowed to react. What is the maximum amount of heat (in kJ) that can be generated?

Stoichiometric Calculations:Using ∆Hrxn as a Conversion Factor

N2 (g) + 3 H2 (g) ! 2 NH3 (g); ∆H = –91.8 kJ

Calorimetry

A laboratory technique in which a reaction proceeds in an insulated container (i.e. calorimeter) at constant P or constant V.

◆ adiabatic conditions for the system; no heat is lost to or gained from the surroundings

◆ record an observed ∆T

◆ calculate ∆H or ∆E

at constant P: ∆H = qP

at constant V: ∆E = qV

Heat Capacity (C) and Specific Heat (s)

How does a substance respond to heating?

Compare the observed temperature changes when 50.0 J energy are added to 10.0 g samples of diamond and tungsten:

diamond tungsten m = 10.0 g m = 10.0 g q = 50.0 J q = 50.0 J

∆T = 9.8ºC ∆T = 37.3ºC

Heat Capacity of a substance, C – heat required to raise the temperature of a sample by 1º (C or K).

Heat Capacity = –––––––––––– ; units J∕ºC or K

OR C = q∕∆T

Specific Heat, s – heat required to raise the temperature of 1g of substance by 1º (C or K)

Specific Heat = –––––––––––– ; units J/g•ºC or K

OR s = q/(m•∆T)

heat supplied∆T produced

heat supplied(∆T)(mass) note:

use molar mass to convert between specific heat and molar heat capacity of a substance

example:

Calculate the amount of heat required (in J) to raise the temperature of a 25.0 g block of nickel from 22ºC to 104ºC. For Ni, s = 0.44 J/g•ºC.

example:

30.0 g H2O at 280 K is mixed with 50.0 g H2O at 330 K. What will be the final temperature of the mixture? For H2O, s = 4.184 J/g•ºC.

Constant Pressure Calorimetry to Determine ∆H

adiabatic conditions: ◆ no heat exchanged between system and surroundings

◆ ∆Hsys = 0

if: ∆Hsys = ∆H1 + ∆H2

then: ∆H1 = –∆H2

◆ for an exothermic reaction:

∆H1 = heat released by chemical reaction occurring in calorimeter

∆H2 = heat absorbed by solution & calorimeter resulting in increase in T

What do you measure in lab?◆ amounts of reactants (mass, volume, mol)◆ observed ∆T

order of determination (for the following examples):

∆H2 + , units kJ

∆H1 ! , units kJ

molar enthalpy of reaction ! , units kJ/mol

example:

Consider the neutralization reaction of hydrochloric acid and sodium hydroxide:

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

Determine the molar enthalpy change (in kJ/mol) for this reaction if, in a constant pressure calorimetry experiment 33. 0 mL of 1.20 M HCl (aq) is combined with 42.0 mL of 1.20 M NaOH (aq) resulting in an increase in temperature from an initial 25.00ºC to a final 31.80ºC.

You can assume that, for the solutions, volume is additive; d = 1.00 g/mL; s = 4.184 J/gºC.

example:

When 23.6 g CaCl2 is dissolved in water in a constant pressure calorimeter, the temperature rose from 25.0ºC to 38.7ºC.

If the heat capacity of the calorimeter and solution is 1258 J/ºC (i.e. Ccal = 1258 J/ºC), determine the heat released by the dissolution of 1.20 mol CaCl2.

CaCl2 (s) ! Ca2+ (aq) + 2 Cl– (aq)

Constant Volume Calorimetry to Determine ∆Eadiabatic conditions: ◆ no heat exchanged between system and surroundings

◆ ∆Esys = 0

if: ∆Esys = ∆E1 + ∆E2

then: ∆E1 = –∆E2

◆ for an exothermic reaction:

∆E1 = heat released by chemical reaction occurring in calorimeter

∆E2 = heat absorbed by solution & calorimeter resulting in increase in T

example:

The combustion of ethanol is studied in a constant volume (bomb) calorimeter with a calorimeter constant of 9.63 kJ/ºC (i.e Ccal = 9.63 kJ/ºC).

The combustion of 2.84 g C2H5OH results in an increase in temperature from Ti = 25.00ºC to Tf = 33.73ºC.

Determine the energy released in the combustion of 1 mol of C2H5OH.

Determine the heat of combustion per gram of ethanol (in kJ/g)

Hess’s Law of Heat Summation

◆ We can determine of ∆H for one chemical reaction based on the ∆H values for related reactions using Hess’s Law:

For a chemical reaction that can be written as the sum of 2 or more steps, ∆H for the overall reaction is equal to the sum of ∆H’s for the individual steps.

so: for a reaction that is a result of the sum of 3 individual steps:

∆Hrxn = ∆H1 + ∆H2 + ∆H3

Hess’s Law of Heat Summation

Why does Hess’s Law of Heat Summation work?

◆ Enthalpy is a state property:

dependent only on initial and final states

independent of path

Hess’s Law of Heat Summation

Determine ∆H for the following reaction using equations 1 & 2 given below:

2 C (s) + O2 (g) ! 2 CO (g) ∆H = ????

use: (1) 2 C (s) + 2 O2 (g) ! 2 CO2 (g) ∆H1 = –787 kJ

(2) 2 CO2 (g) ! 2 CO (g) + O2 (g) ∆H2 = +566 kJ

overall: 2 C (s) + O2 (g) ! 2 CO (g) ∆H = –221 kJ

An Enthalpy Diagram for This Process: To solve a Hess’s Law of Heat Summation problem:

◆ consider the equations you have to work with

◆ identify how to manipulate them so that when you add them together, you end up with your target, overall equation

◆ your options are:

multiply all stoichiometric coefficients in the equation by some factor, n " multiply ∆H by the same factor, n

write an equation in reverse (i.e. change direction) " change the sign on ∆H

example:

Determine ∆H for: W (s) + C (s) ! WC (s); ∆H = ????

using:

(1) 2 W (s) + 3 O2 (g) ! 2 WO3 (s) ∆H = –1680.6 kJ

(2) C (s) + O2 (g) ! CO2 (g) ∆H = –393.5 kJ

(3) 2 WC (s) + 5 O2 (g) ! 2 WO3 (s) + 2 CO2 (g) ∆H = –2391.6 kJ

solution:

Determine ∆H for: W (s) + C (s) ! WC (s); ∆H = ????

using:

(1) 2 W (s) + 3 O2 (g) ! 2 WO3 (s) ∆H = –1680.6 kJ

(2) C (s) + O2 (g) ! CO2 (g) ∆H = –393.5 kJ

(3) 2 WC (s) + 5 O2 (g) ! 2 WO3 (s) + 2 CO2 (g) ∆H = –2391.6 kJ

actions to take:reverse equation (3)multiply equation (1) by factor of "multiply equation (3) by factor of "do nothing to equation (2)

solution:

Determine ∆H for: W (s) + C (s) ! WC (s); ∆H = ????

using:

(1) W (s) + 3/2 O2 (g) ! WO3 (s) ∆H = –840.3 kJ

(2) C (s) + O2 (g) ! CO2 (g) ∆H = –393.5 kJ

(3) WO3 (s) + CO2 (g) ! WC (s) + 5/2 O2 (g) ∆H = +1195.8 kJ–––––––––––––––––––––––––––––––––––––––––––––––––––––––––

overall: W (s) + C (s) ! WC (s) ∆H = –38.0 kJ

Standard Reaction Enthalpies

Standard Reaction Enthalpies, ∆Hº: reaction enthalpy corresponding to reactants in their standard states forming products in their standard states

standard state: the physical state of a pure substance at P = 1 atm and a defined temperature

Standard Enthalpies of Formation, ∆Hºf

◆ ∆Hºf corresponds to the formation of 1 mol of a substance in its standard state from its elements in their standard states.

ex: Na (s) + " Cl2 (g) ! NaCl (s)6 C (s) + 6 H2 (g) + 3 O2 (g) ! C6H12O6 (s)

◆ ∆Hºf of an element in its most stable form = 0

◆ units of ∆Hºf are kJ/mol

◆ tabulated in Table 6.2 and Appendix C of text

◆ use ∆Hºf’s to calculate ∆Hºrxn

Using Standard Enthalpies of Formation, ∆Hºf

∆Hºrxn = ∑n•∆Hºf products – ∑n•∆Hºf reactants

example:

Calcualte ∆Hº (in kJ) for the following reaction using the given ∆Hºf values:

3 NO2 (g) + H2O (l) ! 2 HNO3 (aq) + NO (g)∆Hºf: 33.2 –285.8 –206.6 90.3(in kJ/mol)

example:

Determine ∆Hºf (in kJ/mol) for HCl (g) using the following data:

CH4 (g) + 4 Cl2 (g) ! CCl4 (l) + 4 HCl (g); ∆Hº = –433.3 kJ∆Hºf : –74.9 0 –139 ???(in kJ/mol)

solution:let x = ∆Hºf of HCl (g)

–433.3 kJ = [(1 mol CCl4)(–139 kJ/mol) + (4 mol HCl)(x)] ! [(1 mol CH4)(–74.9 kJ/mol)]

–508.2 kJ = –139 kJ + (4 mol)(x)–369.2 kJ = 4 mol(x)

so x = ∆Hºf for HCl (g) = –92.3 kJ/mol

example:

Determine ∆Hºf (in kJ/mol) for HCl (g) using the following data:

CH4 (g) + 4 Cl2 (g) ! CCl4 (l) + 4 HCl (g); ∆Hº = –433.3 kJ∆Hºf : –74.9 0 –139 ???(in kJ/mol)