50 mL 100o

100 mL 25o

150 mL

(4.184 J/oC g)

q = C x T x mass

q2 = (4.184 J/oC g) x

q1 = - q2

(Tf - 100) x (50) =

Tf = 50o C

q1 = x (Tf - 100) x (50 g)

Tf

Heat transfer experiments

(Tf - 25) x (100 g)

- (Tf - 25) x (100)

Enthalpy of reaction

Hrxn = qrxn

coffee cup calorimeter

10.5 g KBr

Tf = 21o

Calculate Hrxn

KBr(s) K+ (aq) + Br-(aq)

qsystem = = Hrxn

in 125g water at 24o

- qsurroundings

Hrxn

qsystem = - qsurroundings = Hrxn

qsurroundings = C x

qsurroundings = (4.184 J/goC)(21 (10.5 g

= -1756 Jqsystem = - qsurroundings = +1756 J = Hrxn

H is extensive

Hrxn = 1756 J = 167 J/g = 19873 J/mol

T x mass

- 24oC) + 125 g)

a) endothermic b) exothermic

10.5 g KBr

E = q + w

E = q - PV

At constant V,

Bomb calorimeter

qrxn = qsystem = -qcalorimeter

qcalorimeter = C (J / oC) x

E = qv

T (oC)

Constant Volume calorimetry2Fe (s) + 3/2 O2 (g) Fe2O3 (s)

11.2 g Fe(s), 1 atm O2

Ccalorimeter = 2.58 kJ/oC

Tcalorimeter = + 31.9 oC qrxn = -qcalorimeter = Erxn

= - (2.58 kJ/oC)

= - 82.2 kJErxn

= - 822 kJ/mol Fe2O3

/ 0.1 mol Fe2O3

(31.9oC)

Thermite reaction

H is an

2Al(s) +Fe2O3(s)

Hess’ Law

extensive, State function

Al2O3(s) + 2Fe(l)

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

2Al(s) H = -1676 kJ/mol

H = - 822 kJ/mol

2Al(s) -854 kJ/mol

Fe(s) Fe(l) +15 kJ/mol_______________________________ __________

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

Hrxn = -824 kJ/mol

2 ( )2 2

2 Fe(s) + 3/2 O2(g) Fe2O3(s)Fe2O3(s) 2Fe(s) + 3/2 O2(g)_______________________________ __________+

+ 3/2 O2(g)

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

Al2O3(s)

Hess’ Law

• Always end up with exactly the same reactants and products

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

• If you change the stoichiometry, change H

Heats of formation, Hof

H = heat lost or gained by a reaction

“o” = standard conditions:

all solutes 1M

all gases 1 atm

“f” = formation reaction:

1mol product

from elements

in standard states

for elements in standard states,

Hof = 0

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

reactants productselements

2 Al(s) 2 Al(s) Al2O3(s)

Fe2O3

2 Fe(s)

3/2 O2(g)2 Fe (l)

Hof

Hof Al2O3(s)

+ 2 Hof Fe (l) Fe2O3 Al(s)

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

reactants productselements

2 Al(s) 2 Al(s) Al2O3(s)

Fe2O3

2 Fe(s)

3/2 O2(g)2 Fe (l)

Fe2O3 Hof Al2O3(s)

+ 2 Hof Fe (l)

Hrxn = nHof products -

- Hof

nHof reactants

Hof

- Hof Al(s)

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

Hrxn =

Hrxn =

Hrxn =

= -824 kJ

nHof products - nHo

f reactants

[Hof Al2O3(s) +

-

[(-1676) -

Hof Fe(l)]2

[Hof Fe2O3(s) + Ho

f Al(s)]2

+ (15)]2 [(-822)+ 0]kJ

Bond Energies

chemical reactions = bond breakage andbond formation

bond energiesenergy required to break bondbond breakage a) endothermic

b) exothermic(raise P.E.)

bond formation (lower P.E.)exothermic

positive

Bond energies

CH4 (g)

Hrxn=

C-H 413 kJO=O 495 kJC=O 799 kJO-H 467 kJ

Hrxn =

Hrxn =

bonds broken - bonds formed

[ (C-H)+ (O=O)] [ (C=O)+ (O-H)] = -824 kJ

Hof products- Ho

f reactants =- 802 kJ

-4 22 4

+ 2O2 (g) CO2 (g) + 2H2O (g)

qv v.s. qp

qv = E qp = H H = E + PV

H = E + PV

if n = 0

2Fe (s) + 3/2 O2 (g) Fe2O3 (s)

n =H =

H = -826 kJ/mol

= E + nRT

H = E

(0 - 3/2) = - 3/2- 822 kJ/mol + (- 3/2)(8.314 x 10-3 kJ)(298)