Calorimetry

How is heat measured?

It isn’t

Temperature measured

T (K)

C = heat capacity

(J/K) = q (J)C

= heat to raise T 1o

x

heat capacity

specific heat capacity

q (J) =

qsystem =

C (J / K g) x T (K) x mass (g)

-

molar heat capacity molg

qsurroundings

J / KJ / K

(thermometer)(reaction)

heat capacity

K.E.translational

heat

K.E.rotational

compare: water Helium

He

- raise T of system

+ K.E.vibrational + P.E.

heat capacities

substance specific m.w. molarheat capacity (g/mol) heat capacity

(J / K g) (J / K mol)

Al (s)Fe (s) 0.44

0.89 26.98 24.055.85 24.8

H2O(s) 36.5

H2O(l) CCl4(l)

75.2133.0

Calorimetry

1. Measure T (Tfinal - Tinitial)

2. Convert to q

T (K) x mass (g) =x C (J/K g) q (J)

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)

Calorimetry

1. Measure T (Tfinal - Tinitial)

2. Convert to q

T (K) x mass (g) =x C (J/K g) q (J)

q is a path function E = q + w

q = EE- w = + Pext V

q = E + PextV H

q = E + PV

At constant P,

H = E + PV

qq

Enthalpy H

H = E + PV +VP

P = 0

H =E +PV

E + PV

pp= = HH

Enthalpy

H E + PV

Extensive State function Hrxn = qrxn = qsystem = - qsurroundings

(thermometer)(reaction)

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)

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)

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)