Ch.5 THERMOCHEMISTRY

Energy, E work, w1st Law ThermoCalorimetry

Enthalphy, H, heat, q heat of rxn; enthalphyies of formationHess’ Law

∆H, Enthalpy∆HRXN

Specific heat

Thermochemistry

Energy, E capacity to do work or transfer heat

Work, w E or force that causes a in direction or position of an object w = F*d

relationship bet chem rxns & E es due to heat

Heat, q E to cause increase of temp of an objecthotter -----> colder

sys ----> surr exothermic, sys losses qsurr ----> sys endothermic, sys gains q

ENERGY

PE: potential E stored E, amt E sys has available

E in motion, Ek = 0.5 mv2KE: kinetic E

2 objects mass1 > mass2 @ same speed which more Ek?1 object v1 < v2 @ same mass which more Ek?

Internal E total KE + PE of system

∆E = ∑Ef - ∑Ei = ∑Epdt - ∑Ereact

∆Esys = -∆Esurr

∆E always => system surr

Transfer of E results in work &/or heat

H2

O2

NOW, think of atoms & molecules in random motion colliding!!!!!!

What kindof Energieswould beinvolved?

E UNITSJoule, J EK = 0.5(2 Kg)(1 m/s)2 = 1 Kg-m2/s2 = 1 J

calorie, cal E needed to raise 1 g H2O by 1oC 1 cal = 4.184 J

1 Cal (food) = 1000 cal= 1 kcal

Transfer of E results in work &/or heat

System – Surroundings Defined as ………….? What???

System: a defined region

Surroundings: everything that will ∆ by influences of the system

OPEN: matter & E ex w/ surr

CLOSED: ex E, not matter w/ surr

E

2 mol O2

1 mol CH4

∆PE

2 mol H2O1 mol CO2

E released to surroundings as Heat

∑PE(H2O + CO2) < ∑PE(O2 + CH4)

system

∆PE

2 mol NO

E1 mol N2

1 mol O2

Heat absord from surroundings

∑PE(NO) > ∑PE(O2 + N2)

system

Determine the sign of DH in each process under 1 atm; eno or exo?

1. ice cube melts2. 1 g butane gas burned to give CO2 & H2O

1. ice is the sys, ice absorbs heat to melt, DH “+”, ENDO

2. butane + O2 is the sys, combustion gives off heat, DH “-”, EXO

must predict ifheat absorbedor released

1st Law of Thermodynamics: total E of universe is constant - E is neither created nor destroyed but es form

Conservation of E

q: Heat, Internal H E transfer bet sys & surr w/ T diff

w: work, other form E transfer mechanical, electrical,

∆E = q + w sum of E transfer as heat &/or work

∆E = q + w + + + - - - + - + : sys gain E; w > q - + - : sys lost E |w| > |q|

What is ∆E when a process in which 15.6 kJ of heat and 1.4 kJ of work is done on the system?

∆E = q + w 15.6 + 1.4 kJ = 17.0 kJ

State FunctionsProperty of variable depends on current state; not how that state was obtained

T, H, E, V, P use CAPITAL letters to indicate state fcts

∆ : state fcts depend on initial & final states

∆H Enthalpy

Must measure q & w2 types: electrical, PV - movement of charged particles - w of expanding gas

w = -P∆V@ constant P∆H = ∆E + P∆V q + w

3 Chemical Systems

#1 no gas involveds, l, ppt, aq phases have little or not V change; P∆V ≈ 0, then ∆H ≈ ∆E

#2 amt of gas no changeH2 (g) + I2 (g) -- 2 HI (g)

P∆V ≠ 0, then ∆H ≈ ∆E∆E mostly transfer as Heat

P∆V = 0, then ∆H = ∆E

#3 amt of gas does changeN2 (g) + 3 H2 (g) -- 2 NH3 (g)

∆H Enthalpy Changes∆Hcomb ∆Hf ∆Hfus ∆Hvap

combines w/ O2 cmpd formed subst melts subst vaporizes s -- l l -- g

PV WorkCalculate the work associated with the expansionof a gas from 46 L to 64L @ 15 atm.

w = -P∆Vw = -(15 atm)(18 L) = -270 L-atm

NOTE: “PV” work - P in P∆V always refers to external P - P that causes compression or resists expansion

A balloon is inflated by heating the air inside. The vol changes from4.00*106 L to 4.5*106 L by the addition of 1.3*108 J of heat. Find ∆E,assuming const P = 1.0 atm

Heat added, q = + P = 1.0 atm1 L-atm = 101.3 J∆V = 5.0*105 L

∆E = q + ww = -P∆Vw = -(1.0 atm)(5.0*105 L) = -5.0*105 L-atm

(-5.0*105 L-atm)(101.3 J / 1 L-atm) = -5.1*107 J

∆E = q + w = (1.3*108 J) + (-5.1*107 J) = 8*107 JMore E added by heating than gas expanding,net increase in q, ∆E “+”

REVIEW

PE - KE J - cal 1st LawEnthalpy sys - surr PV STATE Fcts

Measure of Heat flow, released or absorded, @ const P & V

CALORIMETRYHeats of Reaction

Heat Capacity, C Specific Heat, Cs

T when object absorbs heat C of 1 g of subst

Solar-heated homes use rocks to store heat. An increase of 120Cin temp of 50.0 Kg of rocks, will absorb what quantity of heat?Assume Cs = 0.82 J/Kg-K. What T would result in a releaseof 450 kJ?

Not as simple as: ∆Hfinal - ∆Hinitial

q = Cs*m*T

+q or -q?gains lossendo exo

KgJ

T*m

q

)T - (T*g

heat trans C

if

s

How much Heat is transferred when 720 g of antifreeze cools 25.5 oC?Cs = 2.42 J/g-K

q = Cs * mass * ∆T ∆ T = -25.5 ?

T?

K or oC? K = oC

THN IK!!!!

q = (2.42 J/g-K) * (720 g) * (-25.5K) = -44400 J or -44.4 kJ

HESS’S LAWHeat Summation

Rxn are multi-step processesCalculate H from tabulated values

Hess’ states:overall H is sum of individual steps

REACTS ======> PDTS

Figure 05.22

- reverse rxn?

- had 3X many moles?

What effect H if

Figure 05.22

THN IK!!!!

Calculate HRXN for Ca(s) + 0.5 O2(g) + CO2 (g) --------> CaCO3(s) HRXN = ? given the following steps: Ca (s) + 0.5 O2 (g) -----> CaO (s) Ho

f = - 635.1 kJ CaCO3 (s) -----> CaO (s) + CO2 (g) Ho

f = + 178.3 kJ

Note: to obtain overall rxn ==> (1st rxn) + (-2nd rxn)

Ca (s) + 0.5 O2 (g) -----> CaO (s) Hof = - 635.1 kJ

CaO (s) + CO2 (g) -----> CaCO3 (s) Hof = - 178.3 kJ

Ca(s) + 0.5 O2(g) + CO2 (g) ------> CaCO3(s) Horxn = - 813.4 kJ

Ho?“o”??

Enthalpies of

Formation

STANDARD STATESSet of specific conditions

- gas: 1 atm, ideal behavior - aq solution: 1 M (mol/L) - pure subst: most stable form @ 1 atm & Temp T usually 25oC - forms 1 mole cmpd; kJ/mol

Use superscript “o” indicates Std States

parts elemental its

fromsubst of formation

0rxn

0f H H

Individual ∆Hf 0 values from book table, appendix C, pg 1100

NOTE: look at state

Ca (s) + 0.5 O2 (g) -----> CaO (s) H = - 635.5 kJ

0.0 H Ca 0f(s)

0.0 H O 0.5 0f(g)2

635.5- H CaO 0f(s)

Orxn

Of

Of H (Reacts) H - (Pdts) H

-635.5 kJ - (0.0 + 0.0)kJ = -635.5 kJ

What is the change in enthalpy for the reaction of sulfur dioxideand oxygen to form sulfur trioxide. All in gas form.Is this endo- or exo-thermic?

2 SO2 (g) + O2 (g) -----> 2 SO3 (g)

Solar-heated homes use rocks to store heat. An increase of 120Cin temp of 50.0 Kg of rocks, will absorb what quantity of heat?Assume Cs = 0.82 J/Kg-K. What T would result in a releaseof 450 kJ?

q = (0.82 J/g-K)*(5.0*104 g)*(12.0 K) =4.9 * 105 J

q = Cs*m*T

T = q/[Cs*m]

T = (4.5*105 J)/[(0.82 J/g-K)*(5.0*104 g)] = 11O decrease

H = Hf Pdts - Hf reacts

H = (-792.0 kJ) - (-593.6 kJ) = -198.4 kJ

Exothermic, -H What if rxn were reversed?????

Find Hof per mole in tables (kJ/mol)

SO2 = -296.8 SO3 = -396.0 O2 = 0.0 free element

Sum Hf reactants using stoich coeff & also pdts

kJ 792.0- mol 1

kJ 0.396SO mol 2 3

kJ 593.6- mol 1

kJ 0.0O mol 1 2

mol 1

kJ 8.296SO mol 2 2

Write balanced eqn for the formation of 1 mol of NO2 gas fromnitrogen monoxide gas and oxygen gas. Calculate DHO

rxn

1 NO(g) + .5 O2(g) ---> NO2(g)

DHOf : 90.3 kJ + .5(0) kJ ---> 33.2 kJ

(33.2 kJ) - (90.3 + 0)kJ = -57.1 kJ

Find the overall rxn, CH3OH(l) + H2O(l) ---> CO2(g) + 3 H2(g),

from the given steps: H2(g) + CO(g) ---> CH3OH(l)

CO(g) + H2O(l) ---> CO2(g) + H2(g)

Calculate DHrxn for each step and find the overall DHrxn

1 CH3OH(l) ---> 2 H2(g) + 1 CO(g)

1 CO(g) + 1 H2O(l) ---> 1 CO2(g) + 1 H2(g)

DHf: -238.6 0 -110.5 DHrxn = 128.1 kJ DHf: -110.5 -285.8 -393.5 0 DHrxn = 2.8 kJ

1 CH3OH(l) + H2O(l) ---> CO2(g) + 3 H2(g) DHrxn = 130.9 kJ

Toss a ball upward. a. Does KE increase or decrease

b. As ball goes higher, want effect to PE

Definea. System b. Closed system c. Not part of system

Explaina. 1st Law b. Internal E c. How internal E of closed system increase

Decrease, KE converts to PE

Increases

Region of study w/ E changes exchange E not mass surroundings

E not created nor destroyed, changes formTotal E of system, KE + PESystem absorbs heat or work done on system

Calculate E of system, is endo- or exo- thermica. Balloon cooled, remove 0.655 kJ heat, shrinks, & atmosphere does 382 J work on

b. 100 g metal bar gains 25oC, absorbs 322 J of heat. Vol is constant

c. Surroundings do 1.44 kJ work compressing gas in perfectly insulated container

q “-” w “+” E =-0.655 kJ + 0.382 kJ = -0.273 kJ EXO

q “+” w = 0 E = +322 J ENDO

q = 0 (perfectly insulated) w “+” E = +1.44 kJ ENDO

Ca(OH)2(s) ----- CaO(s) + H2O(g)

Requires addition of 109 kJ of heat per mol of Ca(OH)2

a. Write balanced thermochemistry equation

b. Draw enthalpy diagram

Ca(OH)2(s) ----- CaO(s) + H2O(g) H = 109 kJ

CaO(s) + H2O(g)

Ca(OH)2(s)

H = 109 kJ