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Heat transfer experiments. 150 mL. T f. q = C x T x mass. 50 mL 100 o. 100 mL 25 o. q 1 =. (4.184 J/ o C g). x (T f - 100). x (50 g). q 2 = (4.184 J/ o C g) x. (T f - 25) x. (100 g). q 1 = - q 2. (T f - 100) x (50) =. -. (T f - 25) x (100). T f = 50 o C. - PowerPoint PPT Presentation
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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) xq1 = - 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 reactionHrxn = 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 J
qsystem = - 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 + wE = 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/oCTcalorimeter = + 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’ Lawextensive, State function
Al2O3(s) + 2Fe(l)
2Al(s) +Fe2O3(s) Al2O3(s) + 2Fe(l)
2Al(s) H = -1676 kJ/molH = - 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 statesfor 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 Energieschemical reactions = bond breakage and
bond 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 + PVif n = 0
2Fe (s) + 3/2 O2 (g) Fe2O3 (s)
n =H =
H = -826 kJ/mol
= E + nRTH = E
(0 - 3/2) = - 3/2- 822 kJ/mol + (- 3/2)(8.314 x 10-3 kJ)(298)