Chem 115 Final Formula sheet PDF

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Chem 115 Final- Formulas

Average atomic mass= (mass1)(% abundance) + (mass2)(% abundance) *convert % to decimal or use 1-x and x if % of one mass is not given

Avagadro’s number 6.022x1023

Mass % of element= (mass of element (in compound)/total weight of compound) x 100 Molarity= moles/volume. (mol/L) Dilution of solution: M1V1= M2V2 or C1V1= C2V2 *use stoic ratio to figure out what to divide by

Limiting reagent- bring each reactant to a product with 1 element from each reactant- one that produces least amount of moles is limiting Density= mass/volume mass=(density)(volume)

% yield= (actual yield/ theoretical yield) x 100

Rate of Reactions: aA + bB  cC + dD

Rate =

1 ∆[ A] a ∆t

−¿

=

−1 ∆ [ B ] b ∆t

=+

1 ∆ [C ] c ∆t

=+

Rate= k[A]n or k[A]n[B]m n 0 1 2 3

k (units) Ms-1 s-1 M-1s-1 M-2s-1

Overall Order 0 1 2 3

In Charts: exp 4 : used for rate and exp 2 concentrations n R

Predicting Rates: lnk= -

ln

Ea 1 ( ) RT T

+ lnA

R= 8.314 J/mol K (check Ea for kJ)

1 k1 Ea 1 ) ( − = k2 R T1 T 2

k2= k1 e ^

(

1 Ea 1 − R T1 T2

K= ˚C + 273.15

[ ( )]

k1 k2 Ea= [ ]xR 1 1 ( ) − T1 T 2 ln

)

Arrhenius Equation: k= A e ^ -

Ea RT

T

: Ea

ex (x is Frequency Factor:

0 Order:

k=

A=

rate= k[A]0

At =−kt A0

¿−

)

Ea RT

[A]t= -kt[A]0 t1/2=

A0= initial concentration

0.639 k

or lnAt= -kt + ln A0 or At= A0 e^ -kt + lnA0

2nd Order: rate= k[A]2.

1 [ A]0

than Rxn Rate

(faster rxn)

Rate [ A ]n

1st Order: rate= k[A]1. ln

:T

k e

Half Life:

Ea

t1/2=

1 k[ A]0

1 [ A ]t = kt +

1 ∆[ D ] d ∆t

Reaction Mechanisms: slow step is always the rate determining step - ignore rates after slow step - in slow step Ea is larger When slow step is first: - Rate law is the same as overall rxn of 1st step - Ex) rate= k[A]n[B]m When slow step is second: Step 1: Rforward= Rreverse : k1 [A]n = k-1[B]m [B]m = k1/k-1 [A]n Step 2: R= k2[C][B]m now sub (k1/k-1 [A]n )for [B]m Rxn Intermediates are crossed out!!!!

Equilibrium aA + bB  cC + dD -

K= [C]c [D]d/[A]a [B]b (products/reactants)

Reactant have larger rate constant= exothermic Products have larger rate constant= endothermic Final rates are equal Final volumes do not have to be equal Small K= more reactants Large K= more products K does not change when there are changes in concentrations K will change when there is a change in temperature

Kbackwards= 1/Kforwards Kbackwards= reactants(now products)/products(now reactants) : when rxn flips n= the coefficients added to rxn Knew= K1 x K2 x K3… Knew=Knorig Kp= partial pressures Pproducts/Preactants Kp=Kc(RT)∆n Kc= Kp/(RT)∆n aA + bB  cC + dD

ONLY GASES ∆n= stoich coefficients of (product-reactants) R=0.0821 L atm/mol K

Qc= [C]ic [D]id/[A]ia [B]ib

only in (g) and (aq), no solid or liquid

QK -rxn goes in reverse direction towards reactants, shifts left

Adding Reactant: rxn increases and shifts to the right Taking away products: shifts to right - System adjusts by forming more reactants Taking away reactants: shifts to left Adding Products: rxn decreases and shifts to the left Right Shift: towards products - System adjusts to form more products Left Shift: towards reactants Volume and Pressure: As pressure :volume - Shifts to side with fewer moles of gas As pressure : volume - Shifts to side with more moles of gas Addition of inert gas at constant pressure: shift to right Addition of inert gas at constant volume: no shift No change when adding solids or liquids Temperature: - Exothermic - ∆Hrxn (heat on products): shifts to left - Endothermic + ∆Hrxn (heat on reactants): shifts to right Catalyst:

-

Equilibrium is reached faster with a catalyst, but no effect on position of K c or Kp

Acids and Bases Arrhenius: Acids provide H+. Base provides OH-

Bronsted: Acids donate H+. Bases accept H+.

Conjugate Acids and Bases HNO2 + H2O  H3O+ + NO2- HNO2 conjugate base is NO2- H2O conjugate acid is H3O+

Strong Acids HBr, HCl, HI, HNO3, HClO4, H2SO4

Weak Acids HF, HClO, HNO2, HCOOH, CH3COOH, NH4+

pH of a weak acid: HA + H2O (l)  A- + H3O+ RICE Tables

−b ± √ b −4 ac 2a 2

x=

Lewis: Acids accept eBases donate eConjugate Strength: SA= very weak base SB= very weak acid WA= WB WB= WA

Strong Bases LiOH, NaOH, KOH, Ca(OH)2, Ba(OH)2, Sr(OH)2

Weak Bases NH3, C6H5NH2, (CH3)2NH2, CH3NH2, C5H5N

pH of a weak base: B + H2O (l)  HB+ + OHR: reaction I: initial concentration C: change (stoich coefficients) E: equilibrium values

or neglect x by these reasons: - 10-2 > K > 102 check: (approx. value of x/ initial value of x) x 100 > 5% - Ma/Ka or Mb/Kb > 100 In rice table liquids and solids are not used (H2O not accounted for)

Acid Base conversions

[H+] [OH-] = 1x10-14 pH + pOH = 14 Kw= Ka x Kb = 1x10-14 Ka=Kw/Kb pKa= -log Ka Ka=10-pKa - Weak acid: large pKa and small Ka - Strong acid: large Ka and small pKa pKb= -log Kb Ka=10-pKb - Weak base: large pKb and small Kb - Strong base: large Kb and small pKb

Acidic H3O > OH pH is 1-6

Basic H3O < OH pH is 8-14

Neutral H 3O = OH pH is 7

% ionization Molarity of ionized acids x 100 ex) [H3O]equilibrium x 100 Initial molarity [HA]i

Amphoteric/ Polyprotic -

Function as acid or base Need H+ and (-) charge

ex) H3PO42- or H2O Step 1 (Ka1): H3PO42- + H2O (l)  H2PO4- + H3O+ Step 2 (Ka2): H2PO4- + H2O (l)  HPO4 + H3O+ Step 3 (Ka3): HPO4 + H2O (l)  PO4+ + H3O+

Ka1 > Ka2 > Ka3 H3O+ will equal what x equals in previous step You know when H+ is used up when x equal Ka value

Buffers: Must contain both weak acid and conjugate base - Change pH slightly A−¿ Hasselbalch Equation: ¿ pH= pKa + log ( ) -

¿ ¿

-

Only use when both species are present, cannot be used for weak acids and weak bases by themselves

Titrations: -

Equivalence point: point at which both acid and base have been consumed Endpoint: point at which we stop adding reactant (titration is over)- use of a colour indicator

Ksp Ksp= [product]n[product]m Qsp= [product]i[product]I

ex) AB2(s)  A(aq) + 2B(aq) Ksp= [A][B]2 Qsp= [A]i[B]i2

Solubility: g/L. Molar Solubility: mol/L Ksp = less soluble Solubility

as pH

- Solve for x to determine solubility Ex) A3B4 (s)  3A(aq) + 4B(aq) Ksp= [A]3[B]4  (3x)3(4x)4  (9x3)(16x4)  432x7 - Divide both sides by 432 the 7th root it.

If: Qsp > Ksp -super saturated - Precipitate forms Qsp = Ksp -should be saturated - No pericipatate Qsp < Ksp -under saturated - No percipitate

Phase Change:

Enthalpy Change:

Thermochemistry W= -P∆V *-w: work done BY system (expand) *+w: work done ON system (shrink)

E=q + w q: (+) system gains thermal energy (-) system loses thermal energy ∆E= Ef (products)- Ei (reactants)

101.325J= 1L atm H= ∆E + P∆V *+∆H: endothermic (absorb energy)- gets hotter  increase energy. ∆H>0 - feels cold *-∆H= exothermic (release energy)- gets colder decrease in energy. ∆H0)  more disorder (less organized) -∆S (∆S 0 ∆S= qrev/T - Then spontaneous qrev= flow of heat ∆S : T and ∆S : T Phase Change Sublimation Deposition Freezing Condensation Melting Evaporation

States solidgas gas solid liquid solid gas liquid solid liquid liquidgas

∆H (+) endothermic (-) exothermic (-) exothermic (-) exothermic (+) endothermic (+) endothermic

∆Ssys >0 0 >0

>0...