Announcements If you havent made it to a talk yet for your seminar homework

assignment, then you might be like the following Public Talk. (A public talk is usually meant to be very accessible.) Astronomy Public Talk by Dr. Jerey BenneI PAA A-102, Wednesday April 15, 7:30pm

The exam seaTng chart is posted online. Check out where your seat is for the Mock Exam (Mon.) and Midterm 1 (Wed.)

Your TAs will be in charge of the Mock Exam and Midterm 1. AYer class on Friday, Ill rush o to the airport. Ill be giving seminars at Stanford and Princeton about my own labs research. Ill be back late Wednesday night. The TAs (and I) will be checking the Canvas discussion board in the mean Tme.

Lecture 8 Topics

Calorimetry (9.4) Hesss Law (9.5)

QuesTons well answer How are temperature and heat related? How do we measure E and H? Ho do we determine E or H for

reacTons that we cant study directly?

Constant Pressure

calorimeter

Constant Volume

calorimeter

Changes in Enthalpy If a process is conducted at constant pressure,

all the heat transferred is the enthalpy. Most chemical and physical processes have an

associated enthalpy change.

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) Hrxn = -802 kJ/mol of rxn

2 CO2(g) 2 CO(g) + O2(g) Hrxn = +556 kJ/mol of rxn What if we double it? 4CO2(g) 4CO(g) + 2O2(g) Hrxn = ?? H2O(s) H2O(l) Hrxn = +6.02 kJ/mol of rxn What if we reverse the reacTon? H2O(l) H2O(s) Hrxn = ??

Enth

alpy

Enth

alpy

Enth

alpy

Hinitial

Hfinal

H > 0

qin

Enth

alpy Hinitial

Hfinal

H < 0

qout

pq H= Exothermic

Endothermic

6.02 kJ/mol of rxn The magnitude and sign of the enthalpy in a thermochemical equa6on is for the process as wri'en!

+1112 kJ/mol of rxn

Gejng to Know q Add 100 J of energy to

A swimming pool full of water A beaker full of water

Which system will experience the largest increase in temperature?

Beaker has fewer molecules to divide the 100 J of energy between.

greater increase in average molecular kine6c energy

greater increase in T

Gejng to Know q Add enough heat to the pool and beaker of water to raise both systems to 40oC.

Which system has the largest amount of thermal energy?

Pool has more molecules, so more energy is required to increase its T to 40oC than the beaker.

Extensive vs. Intensive ProperTes Extensive physical property: value is proporTonal to the

amount of substance present in the system it describes. Heat content depends on the number of parTcles available to store

the heattherefore q is an extensive property. Other extensive proper4es: mass, volume, length, internal energy,

enthalpy

Intensive physical property: value does not depend on the system size or the amount of substance in the system. Temperature depends only on the average molecular kineTc energy.

Since dierent-sized collecTons of molecules can have the same average molecular kineTc energy, T is an intensive property.

Other intensive proper4es: density, pressure, concentraTon, color, phase (gas, liquid, solid), viscosity, melTng point, boiling point

We can directly measure the enthalpy of a chemical process if we conduct that process at constant pressure.

By monitoring the change in temperature of the soluTon during the process, we can determine the magnitude and sign of the enthalpy.

The heat gained/lost by the reacTon is exactly equal to

that lost/gained by the soluTon.

Constant-Pressure Calorimetry

q s m T=

specic heat capacity of soluTon, J/g.oC

mass of the soluTon

2NaOH( ) + HCl( ) H O( ) + NaCl( )aq aq l aq

temp change experienced by

soluTon

??rxnH =

solution rxnq q+ =

= 58 kJ/(mol of reaction)

solution rxnor, q q = +

TiniTal = 25.00oC

solution rxnq q = +

q

solution solution solution solutionq s m T= NH4NO3(s)

H2O(l)

H2O(l) q

Heat balance relaTonship:

qrxn = nreaction H rxn

Tnal = 23.79oC

This is what we want to nd.

5.00 g of ammonium nitrate is dissolved in 500. mL of water at 25.00 oC. The nal temperature of the soluTon is 23.79 oC. What is the enthalpy of the dissoluTon process, expressed per mole of ammonium nitrate? (ssoln = 3.20 J/g.oC, dH2O = 1.00 g/mL)

NH4NO3(s) NH4+(aq) + NO3(aq) Hrxn = ?? J/(mol of reacTon)

For each mole of reacTon, one mole of NH4NO3 is consumed.

4 3rxn NH NO rxnq n H=

1955 J=

3 3

1955 J1 mol5.00 g NH NO

80.05 g

+=

31.3 kJ/mol= +

solution rxnq q = +Recall: rxn 1955 Jq = + 4 3NH NO rxnn H=

4 3

rxnNH NO

rxnqHn

=

1 g500. mL + 5.00 g1 mL

( )o23.79 C 25.00 C ( )J3.20 g C= solution solution solution solutionq s m T=

5.00 g of ammonium nitrate is dissolved in 500. mL of water at 25.00 oC. The nal temperature of the soluTon is 23.79 oC. What is the enthalpy of the dissoluTon process, expressed per mole of ammonium nitrate? (ssoln = 3.20 J/g.oC, dH2O = 1.00 g/mL)

NH4NO3(s) NH4+(aq) + NO3(aq) Hrxn = ?? J/(mol of reacTon)

Constant-P calorimetry is appropriate for soluTon-phase reacTons, because the volume of the system doesnt change.

But what about something like this:

Number of moles of gas change, so volume would change at constant-P, making it very dicult to measure the heat transferred.

What if we hold volume constant?

We can nd H by considering the ngas:

0 since = PV

Fixed volume: w = PV = 0

Constant-Volume Calorimetry

3 8 2 2 2C H ( ) + 5 O ( ) 3 CO ( ) 4 H O( )g g g l +

VE q w = +

( )H E PV = + ( )E nRT= + gasE RT n= +

What is the enthalpy change associated with changing graphite into diamond?

Cgraphite Cdiamond Hrxn = ?? This process is dicult to study directly.

But we can readily determine the heats of combusTon of these two forms of carbon using constant volume calorimetry:

2 2C( ) + O ( ) CO ( )s g g

Hesss Law In going from a par4cular set of reactants to a par4cular set of products, the overall change in enthalpy is the same whether the reac4on takes place in one step or in a series of steps.

Cgraphite Cdiamond

C(s, diamond) + O2(g) CO2(g); Hrxn = -396 kJ/mol

C(s, graphite) + O2(g) CO2(g); Hrxn = -394 kJ/mol

CO2(g) C(s, diamond) + O2(g) Hrxn = +396 kJ/mol

C(s, graphite) C(s, diamond) Hrxn = +2 kJ/mol > 0endo

+

C(s, diamond) C(s, graphite) Hrxn = -2 kJ/mol < 0exo

A Quick Note about Units Throughout this lecture, Ive tried to be quite careful about heats

of reacTons by pujng a subscript rxn and by showing values in units of kJ/(mole of reacTon) as in:

N2 (g) + 3H2 (g) 2NH3 (g) Hrxn = -92 kJ/mol of reacTon However, when you look up heats of reacTons in a table, values

will usually be denoted in simply Joules or kiloJoules, as in:

N2 (g) + 3H2 (g) 2NH3 (g) H = -92 kJ

In other words, when doing problems and using real data out in the world, you may need to remind yourself that a heat of reacTon is for one mole of that reacTon.

A Second Hesss Law Example Calculate Hrxn for the following reacTon: H2(g) + Cl2(g) 2HCl(g) Given the following:

NH3 (g) + HCl (g) NH4Cl(s) Hrxn = -176 kJ/mol of reacTon N2 (g) + 3H2 (g) 2NH3 (g) Hrxn = -92 kJ/mol of reacTon N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) Hrxn = -629 kJ/mol of reacTon

We need more room. Lets conTnue this on the next slide.

Step 1: Only the rst reacTon contains the product of interest (HCl). Therefore, reverse the reacTon and mulTply by 2 to get the stoichiometry correct.

NH3 (g) + HCl (g) NH4Cl(s) Hrxn = -176 kJ/mol

2NH4Cl(s) 2NH3 (g) + 2HCl (g) Hrxn = 352 kJ/mol

Find Hrxn for: H2(g) + Cl2(g) 2HCl(g) Given the following: NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ/mol N2 (g) + 3H2 (g) 2NH3 (g) H = -92 kJ/mol N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ/mol

Step 2. Need Cl2 as a reactant, therefore, add reacTon 3 to result from step 1 and see what is leY.

2NH4Cl(s) 2NH3 (g) + 2HCl (g) Hrxn = 352 kJ/mol

N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) Hrxn = -629 kJ/mol

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)

Hrxn = -277 kJ/mol

Find Hrxn for: H2(g) + Cl2(g) 2HCl(g) Given the following: NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ/mol N2 (g) + 3H2 (g) 2NH3 (g) H = -92 kJ/mol N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ/mol

Step 3. Use remaining known reacTon in combinaTon with the result from Step 2 to get nal reacTon.

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) Hrxn = -277 kJ/mol

N2 (g) + 3H2(g) 2NH3(g) Hrxn = -92 kJ/mol

2NH3(g) 3H2 (g) + N2 (g) Hrxn = +92 kJ/mol

H2(g) + Cl2(g) 2HCl(g) H = -185 kJ/mol

Need to take the second reaction and reverse it

Find Hrxn for: H2(g) + Cl2(g) 2HCl(g) Given the following: NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ/mol N2 (g) + 3H2 (g) 2NH3 (g) H = -92 kJ/mol N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ/mol

1

Use the informaTon below to determine Hrxn for the reacTon: 3 C(graphite) + 4 H2(g) C3H8(g)

C3H8(g) + 5 O2(g) 3 CO2(g) + 4H2O(l) H1 = -2219.9 kJ/mol C (graphite) + O2(g) CO2(g) H2 = -393.5 kJ/mol H2(g) + 1/2 O2(g) H2O(l) H3 = -285.8 kJ/mol 3 CO2(g) + 4H2O(l) C3H8(g) + 5 O2(g) -H1 = +2219.9 kJ 3 C (graphite) + 3 O2(g) 3 CO2(g) 3(H2) = -1180.5 kJ 4 H2(g) + 2 O2(g) 4 H2O(l) 4(H3) = -1143.2 kJ

3 C(graphite) + 4 H2(g) C3H8(g) -H1 + 3(H2) + 4(H3) Hrxn = -103.8 kJ/mol

+

Another Hess Law Example