Chem 2 Exam 2

Factors that effect reaction rates

1. T↑, R↑

2. Catalyst

3. Collision ↑ (and orientation)

4. s«l«g

What does a catalyst do? how?

• lowers/decrease activation energy

• breaks down reaction into steps

Reaction mechanisms: Intermediate

• appears first as a product

• then as a reactant

• not apart of final

Reaction mechanisms: catalyst

• first appears as a reactant

• then later as a product

Reaction mechanisms: which step = rate law overall?

the slower step

Equilibrium condition

• concentration = constant

• rate law: rate (forward)= rate (reverse)

equilibrium constant expression

kc = [products]^cc / [reactants]^cc

kp expression

• kp=kc(RT)^Δn

• Δn = (product cc)-(reactant cc)

• T must be in K

Kelvin to Celsius conversion

K=^oC+273.15

Find Kc from steps

K_overall =K₁K₂

which side of reaction will be favored based on Kc

Kc » 1 → favor products

Kc « 1 → favor reactants

Kc = 1 → [P] = [R]

Le Chatelier's Principle: Temperature

Endo- T↑(favor products) T↓(favor reactants)

Exo- T↑(favor reactants) T↓ (favor products)

Le Chatelier's Principle: Pressure

P↑ V↓ → favor side w/ less gas

P↓ V↑ → favor side w/ more gas

Le Chatelier's Principle: Concentration

Bronsted Lowry acid/base

BL acid - donates H+

BL base - accepts H+

conjugate acid/base pairs

s acid → w. c. base

s base → w. c. acid

w acid → s c. base

w base → s c. acid

amphiprotic or amphoteric species

can act as acid or base (ex: H₂O)

Strong acids

1. HCl

2. HBr

3. HI

4. H₂SO₄

5. HNO₃

6. HClO₄

7. HClO₃

Decide favored side of reaction (relative strength table)

Strong → weak

Kw constant

Kw = 1.0×10^-7

Calc pH

pH=-log[H₃O+] → [H₃O+]=10^-pH

pOH=-log[OH-] → [OH-]=10^-pOH

pH+OH=14.00

[H₃o+][OH-]=Kw

wca + wcb = ?

neutral

sca + wcb = ?

acidic

wca +scb = ?

basic

sca + scb = ?

ka > kb → acidic

ka < kb → basic

ka = kb → neutral

what is the quadratic formula?

x= -b =-(-b² -4ac) // 2a