Brown Ch 5-6 Chemistry Cottrell

Front 1

unit of energy
1 J =

Back 1

1 J = 1 kg(m/s)^2

Front 2

unit of energy
1 calorie =

Back 2

4.184 J

Front 3

(∆T) =
(∆H) =
(∆E) =

Back 3

Change in Temperature, Heat, Energy = (Final - Initial)

Front 4

Exothermic helpers

Back 4

- ; Gives off heat; releases heat; adding heat; hot; reacts

Front 5

Endothermic helpers

Back 5

+ ; Removing heat; absorbs heat; takes in heat; cold; decomposes; dissolving in water is always endo

Front 6

kinetic energy formula =

Back 6

Ek = 1/2 mv^2
m = mass = kg
v = velocity = m/s

Front 7

kinetic energy question example

Back 7

Calculate the kinetic energy in Joules of a 1200 kg automobilie moving at 18 m/s. Convert energy from Joules to calories.; use kinetic energy formula and calorie to Joule conversion

Front 8

internal energy formula

Back 8

answer in Joule; (∆E) = q + w
q = heat, w = work

Front 9

q helpers

Back 9

+ means system gains heat;
- means system loses heat

Front 10

w helpers

Back 10

+ means work done on system
- means work done by system

Front 11

- w quote

Back 11

"if system works on the surrounding..."

Front 12

+ work quote

Back 12

"if the surrounding works on the system..."

Front 13

∆E helpers

Back 13

+ means net gain of energy by system; decompose
- means net loss of energy by system; reacts

Front 14

internal energy questions example

Back 14

Calculate the ∆E of the system and determine whether the process is endothermic or exothermic. A balloon is cooled by removing .655 kJ of heat. It shrinks on cooling and the atmosphere (surrounding) does 382 J of work on the balloon (system).

Front 15

enthalpies of reaction question example

Back 15

4 step problem; endo or exo; calculate heat transferred when g is decomposed; given a reaction, ∆H during reaction is kJ. How many grams of product produced; how many J heat released when g reacts w/O2 at constant pressure

Front 16

Conversion clue

Back 16

if question as for a conversion, ∆ mol to mol; if no conversion is asked, use mols from equation

Front 17

when reactions reverses, you must...

Back 17

switch the sign given for ∆H

Front 18

first law of thermodynamics

Back 18

energy is conserved; not created or destroyed; any energy lost by a system is gained by surroundings, and vice versa

Front 19

Hess's Law

Back 19

states that if a reaction is carried out in a series of steps, ∆H for the overall reaction will equal the sum of the enthalpy changes for the individual steps

Front 20

enthalpy

Back 20

a thermodynamic function; means to warm; accounts for heat flow in processes occurring at constant pressure; the change equals the heat gained or lost at constant pressure

Front 21

basis of Hess's Law

Back 21

We can calculate for any process, as long as we find the route; used for more difficult processes to measure

Front 22

Hess's Law helpers

Back 22

must multiply/divide both process and ∆H before cancellations; same side add, opposite side cancels; if process reverses, must switch ∆H sign

Front 23

calorimetry formula

Back 23

Heat = mass x specific heat x ∆T
q = kg x c x ∆T

Front 24

specific heat = (rule)

Back 24

= heat/massx∆T = J / g℃

Front 25

calorimetry question example

Back 25

specific heat of ethylene glycol is 2.42 J/gK. How many J of heat are needed to raise the temp. of 62.0 g of ethylene glycol from 31.1 ℃ to 40.5 ℃.

Front 26

calorimetry rule

Back 26

never convert ℃ to K and vice versa; always equal

Front 27

specific heat of water (l)

Back 27

indicates the amount of heat that must be added to 1 gram of a substance to raise its temperature by 1 K (or 1 ℃); at 14.5-15.5℃ is 4.184 J/ g-K

Front 28

molar heat capacity of water (l)

Back 28

heat capacity of one mole of a substance

Front 29

specific heat

Back 29

= q / m x change T= J / g x degree C
-can rearrange to form new formuula

Front 30

relating heat, temp change and heat capacity

Back 30

change in H(f rxn) = change in H(f product) - change in H(f reactant) = in kJ

Hf = heat of formation, rxn = reaction, small o = standard enthalpy and not degree

use values for Hf in kJ/mol given by appendix C

Front 31

c = lambda x v

Back 31

Light as a wave:
c = speed of light = 3.0 x 10^8 m/s, lambda = wavelength in meters, v = frequency in s^-1 or 1/s; (can be rearranged)

Front 32

E = hv

Back 32

light as a particle (photon):
E = energy of photon in J, h = Plank's constant (6.626 x 10^-34 J-s, v = frequency in s^-1 or 1/s; (can be rearranged)

Front 33

lambda = h/mv

Back 33

matter as a wave:
lambda = wavelength, h = Plank's constant, m = mass of object in kg; v - speed of object in m/s

Front 34

change in x = change (mv) >= h/4pie

Back 34

Heisenberg's uncertainty principle; the uncertainty in position (change in x) and momentum (change in mv) of an object cannot be zero; the smallest value of their product is h/4pie

Front 35

Max Plank

Back 35

gave the name quantum (meaning 'fixed amount') to teh smallest quantity of energy that can be emitted or absorbed as whole numbers of electromagnetic radiation; proposed E=hv;

Front 36

Albert Einstein

Back 36

used Plank's theory to explain the photoelectric effect; E=hv; when radiant energy strikes a metal surface, it behaves as a stream of tiny energy packets called photons

Front 37

Bohr - 3 postulates

Back 37

proposed a model of the hydrogen atom that explains the line spectrum 1. only orbits of certain radii, are permitted for the electron in an H atom; 2, an electron in a permitted orbit has a specific energy and is in an "allowed' energy state; 3. energy is emitted or absorbed as a photon

Front 38

Johann Balmer

Back 38

showed wavelengths of four visible lines of hydrogen on spectra in a formula that related the wavelength of visible line spectrum to integers; extended to Rydberg's equation

Front 39

Rydberg's equation

Back 39

1/lambda = Rh(1/n squared,1 - 1/n squared,2); n2 is larger than n1
Rh is constant = 1.096776 x 10^7 m^-1

Front 40

Rutherford

Back 40

discovered the nuclear nature of the atom

Front 41

Bohr's formula

Back 41

E = (-hcRh)(1/n squared) = -2.18 x 10^-18 J(1/n squared); and he gave us orbits

Front 42

Louis de Broglie

Back 42

proposed wavelength of the electron depons on mass and its velocity; lambda = h/mv

Front 43

Pauli's excursion principle

Back 43

states no two electrons in an atom can have the same set of four quantum numbers n, l, m l, m s; and the orbitals are filled in order of increasing energy, with no more than two electrons per orbital

Front 44

Schrodinger

Back 44

wave equation; his work opened a new way of dealing with subatomic particles; quantum mechanics or wave mechanics; also gave us orbitals

Front 45

Hund's rule

Back 45

for degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximised