Chemistry Study Guide

Only available on StudyMode
  • Topic: Atom, Ionization energy, Energy
  • Pages : 5 (612 words )
  • Download(s) : 260
  • Published : April 14, 2002
Open Document
Text Preview
Chemistry Study Guide
Oct 2nd 1 hour Exam

Chapter 9- Thermodynamics

KE= ½ mv2

w= F∆x
w= force × distance

∙ A state function refers to a property of the system that depends only on its present state. ∙Internal Energy = heat + work
∆E = q + w
∙Pressure = Force/Area = P = F/A
∙Work= - external pressure × change in volume
w = - P∆V

Enthalpy

H = E + PV
qp = ∆E + P∆V
∆H = qp

∆H = H products – H reactants

Ideal Gas Law

PV = nRT
Energy "heat" = 3/2 R∆T
Cv = 3/2 R = "heat" required to change the temp of 1 mol of gas by 1K at constant volume
Energy required = "heat" – energy needed -energy needed to do to change the translationalthe PV work
energy
Cp = 3/2 + R = 5/2 R
= Cv + R = Cp

E = 3/2 RT (per mole)
∆E = Cv∆T (per mole)
∆E = nCv∆T

"Heat" required = qp = nCp∆T
= n (Cv + R) ∆T
= nCv∆T + nR∆T
(∆E) (PV)

∆H = ∆E + ∆(PV)
∆H = ∆E + ∆(nRT) = ∆E + nR∆T
∆H = nCp∆T

Energy released = energy absorbed
= m × Cp × ∆T
(mass)(specific heat)(change in Temperature)
∆E = q + w = q= qv

Bomb calorimeter =
∆H = ∆TCp

∆Hºreaction= ∑∆Hºf (products) - ∑∆Hºf (reactants)

Chapter 12

c = λν
speed of light (3.0E8 m ) = (wavelength)(frequency)

∆E = nhv or
∆E = hv if n=1
Planck's constant = h= 6.626 × 10-34 J s

Ephoton = hv = hc/λ
KEelectron = ½ mv2 = hv – hvo

E=mc2

De Broglie

λ = h/mv

PE = -Z e2/r
Z= atomic number
R= distance between nucleus

Ηψ = Εψ
Ψ2 = Probability Density

n = energy level
l = shape
ml = orientation
ms = spin
Effective nuclear charge = Zeff = Zactual – effect of electron-electron repulsion 1 eV = 1.602 × 10-19 J

Ionization Energy

X(g) → X+(g) + e-

∙ First Ionization energy increases as we go across a period from left to right due to shielding effects

∙ Shielding occurs because electrons repel each other

∙ First ionization energy decreases as we go down a group.

∙ As n increases, the size of the orbital increases, and the electron is easier to remove.

∙ Exceptions :
Be to B – decrease in IE shows that the electrons in 2s orbital effectively shield the 2p electron.

N to O- drop in IE because of addition of electron in first p orbital that results in a pair that repel each other and make either of them easier to remove.

Electron Affinity

X(g) + e-→ X- (g)

∙ Down a group – more positive since electron is added at increasing distances from the nucleus.

∙ Increase across period because effective nuclear charge is increasing.

Electronegativity = (Electron Affinity + Ionization Potential)/2

Lattice Energy

Li+ (g)+ Cl- (g) → LiCl(s)
Lattice Energy = k(Q1Q2)/r

∆H = ∑ D(Bonds broken) - ∑D(Bonds formed)

Formal charge = (# of valence electrons on a free atom) –
(# of valence electrons assigned to the atom in molecule)

(Valence electron)assigned = (# of lone pair electrons) + ½ (# of shared electrons)

1.Lone pair electrons belong entirely to the atom in question 2.2. Shared electrons are divided equally between the 2 sharing atoms

Chapter 14

Bond Order = (# of bonding electrons - # of antibonding electrons)/2

Chap 20

CO2 > CN- > NO2- > en > NH3 > H2O > OH- > F- > Cl- > Br- > I- (strong field) (weak field) (diamagnetic) (paramagnetic)
(Low spin) (High spin)
Magnitude of ∆ for ligand increases as charge on metal ion increases.
tracking img