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Ch.5 THERMOCHEMISTRY

Energy, E

work, w

1st Law Thermo

Calorimetry

Enthalphy, H, heat, q

heat of rxn;

enthalphyies of formation

Hess’ Law

∆HRXN

∆H, Enthalpy

Specific heat

relationship bet chem rxns & E es due to heat

capacity to do work or transfer heat

Energy, E

Work, w

E or force that causes a in direction or position of

an object w = F*d

Heat, q

E to cause increase of temp of an object

hotter -----> colder

sys ----> surr exothermic, sys losses q

surr ----> sys endothermic, sys gains q

ENERGY

PE: potential E

stored E, amt E sys has available

KE: kinetic E

E in motion, Ek = 0.5 mv2

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

∆Ealways => system surr

∆E = ∑Ef - ∑Ei

= ∑Epdt - ∑Ereact

∆Esys = -∆Esurr

Transfer of E results in work &/or heat

H2

O2

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

What kind

of Energies

would be

involved?

E UNITS

Joule, 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

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

∑PE(H2O+ CO2) < ∑PE(O2+ CH4)

system

2 mol O2

1 mol CH4

∆PE

E released to surroundings

as Heat

2 mol H2O

1 mol CO2

E

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

1. ice cube melts

2. 1 g butane gas burned to give CO2 & H2O

must predict if

heat absorbed

or released

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

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

1st Law of Thermodynamics: total E of universe is constant

- E is neither created nor destroyed but es form

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

Property 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 & w

2 types: electrical, PV

- movement of charged particles

- w of expanding gas

w = -P∆V

@ constant P

∆H = ∆E + P∆V

q + w

#1 no gas involved

s, l, ppt, aq phases have little or not V change; P∆V ≈ 0, then ∆H ≈ ∆E

#2 amt of gas no change

H2 (g) + I2 (g) -- 2 HI (g)

P∆V = 0, then ∆H = ∆E

#3 amt of gas does change

N2 (g) + 3 H2 (g) -- 2 NH3 (g)

P∆V ≠ 0, then ∆H ≈ ∆E

∆E mostly transfer as Heat

∆H Enthalpy Changes

∆Hcomb ∆Hf ∆Hfus ∆Hvap

combines w/ O2 cmpd formed subst melts subst vaporizes

s -- l l -- g

Calculate the work associated with the expansion

of a gas from 46 L to 64L @ 15 atm.

w = -P∆V

w = -(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 from

4.00*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 atm

1 L-atm = 101.3 J

∆V = 5.0*105 L

∆E = q + w

w = -P∆V

w = -(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 J

More E added by heating than gas expanding,

net increase in q, ∆E “+”

CALORIMETRY changes from

Heats of Reaction

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

Not as simple as: ∆Hfinal - ∆Hinitial

Solar-heated homes use rocks to store heat. An increase of 120C

in 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 release

of 450 kJ?

Heat Capacity, CSpecific Heat, Cs

T when object absorbs heat C of 1 g of subst

+q or -q?

gains loss

endo exo

q = Cs*m*T

How much Heat is transferred when 720 g of antifreeze cools 25.5 oC?

Cs = 2.42 J/g-K

THN IK!!!!

q = Cs * mass * ∆T

∆ T = -25.5 ?

K or oC?

K = oC

T?

q = (2.42 J/g-K) * (720 g) * (-25.5K)

= -44400 J or -44.4 kJ

HESS’S LAW 25.5

Heat Summation

Hess’ states:

overall H is sum of individual steps

Rxn are multi-step processes

Calculate H from tabulated values

REACTS ======> PDTS

Calculate 25.5 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) Hof = - 635.1 kJ

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

“o”??

Enthalpies of

Formation

Ho?

STANDARD STATES 25.5

Set 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

Individual ∆Hf0 values from book table, appendix C, pg1100

NOTE: look at state

Ca (s) + 0.5 O 25.5 2 (g) -----> CaO (s) H = - 635.5 kJ

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

Solar-heated homes use rocks to store heat. An increase of 120C

in 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 release

of 450 kJ?

What is the change in enthalpy for the reaction of sulfur dioxide

and oxygen to form sulfur trioxide. All in gas form.

Is this endo- or exo-thermic?

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

q = 12Cs*m*T

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

4.9 * 105J

T = q/[Cs*m]

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

11O decrease

Find 12Hof 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

H = Hf Pdts - Hf reacts

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

Exothermic, -H

What if rxn were reversed?????

Write balanced eqn for the formation of 1 mol of NO 122 gas from

nitrogen monoxide gas and oxygen gas. Calculate DHOrxn

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, CH 123OH(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

a. Does KE increase or decrease

b. As ball goes higher, want effect to PE

Decrease, KE converts to PE

Increases

Define

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

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

Explain

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

E not created nor destroyed, changes form

Total E of system, KE + PE

System absorbs heat or work done on system

Calculate 12E of system, is endo- or exo- thermic

a. 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) 122(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)

H = 109 kJ

Ca(OH)2(s)

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