Transcript Review - University of Illinois at Urbana–Champaign

Calorimetry
How is heat measured?
It isn’t
Temperature measured
 T (K) x C (J/K) = q (J)
C = heat capacity
= heat to raise T 1o
heat capacity
specific heat capacity J / K g
J / K mol
molar heat capacity
q (J) = C (J / K g) x T (K) x mass (g)
qsystem = - qsurroundings
(reaction)
(thermometer)
heat capacity
K.E.translational - raise T of system
heat
K.E.rotational + K.E.vibrational + P.E.
compare: water
Helium
He
heat capacities
substance
Al (s)
Fe (s)
H2O(s)
H2O(l)
CCl4(l)
specific
m.w.
heat capacity (g/mol)
(J / K g)
0.89
0.44
26.98
55.85
molar
heat capacity
(J / K mol)
24.0
24.8
36.5
75.2
133.0
Calorimetry
1. Measure T (Tfinal - Tinitial)
2. Convert to q
T (K) x C (J/K g) x mass (g) = q (J)
Heat transfer experiments
150 mL
Tf
50 mL
100o
100 mL
25o
q = C x T x mass
q1 = (4.184 J/oC g) x (Tf - 100) x (50 g)
q2 = (4.184 J/oC g) x (Tf - 25) x (100 g)
q1 = - q2
(Tf - 100) x (50) = - (Tf - 25) x (100)
Tf = 50o C
Calorimetry
1. Measure T (Tfinal - Tinitial)
2. Convert to q
T (K) x C (J/K g) x mass (g) = q (J)
q is a path function
E=q+w
q = E- w = E + Pext V
Enthalpy
H
H  E + PV
H =E +PV
H = E + PV +VP
At constant P, P = 0
q = E + PextV
q = E + PV
q p = H
H = E + PV
Enthalpy
H  E + PV
Extensive State function
Hrxn = qrxn = qsystem = - qsurroundings
(reaction)
(thermometer)
Enthalpy of reaction
Hrxn = qrxn
coffee cup calorimeter
10.5 g KBr in 125g water at 24o
KBr(s)  K+ (aq) + Br-(aq)
Tf = 21o
Calculate Hrxn
qsystem = - qsurroundings = Hrxn
Hrxn
qsystem = - qsurroundings = Hrxn
qsurroundings = C x T x mass
qsurroundings = (4.184 J/goC) (21- 24oC) (10.5 g + 125 g)
= -1756 J
qsystem = - qsurroundings = +1756 J = Hrxn
H is extensive
= 167 J/g = 19873 J/mol
Hrxn = 1756 J
10.5 g KBr
E = q + w
E = q - PV
At constant V,
E = qv
Bomb calorimeter
qrxn = qsystem = -qcalorimeter
qcalorimeter = C (J / oC) x T (oC)
Constant Volume calorimetry
2Fe (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) (31.9oC)
Erxn = - 82.2 kJ / 0.1 mol Fe2O3
= - 822 kJ/mol Fe2O3
qv v.s. qp
qv = E
qp = H
H = E + PV
H = E + PV = E + nRT
if n = 0
H = E
2Fe (s) + 3/2 O2 (g)  Fe2O3 (s)
n = (0 - 3/2) = - 3/2
H = - 822 kJ/mol + (- 3/2)(8.314 x 10-3 kJ)(298)
H = -826 kJ/mol