Equilibrium slideshare. WN.2011

Two opposing reactions ,[object Object],[object Object],[object Object]

A  B ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

a A + b B  c C + dD ,[object Object],Equil. Constant  K c = ‘ products over reactants’ (s) and (l) not included in K express b/c their 'concentration' is constant, so is not a fcn of the equil. (security) Homogeneous equil = all same phase Heterogeneous equil = different phases

K needs a BCE. Why? ,[object Object],If reverse BCE, K’ = 1/K If multiply BCE by  , K’ = K  1 CaSO 4 (s)  1 Ca 2+ (aq) + 1 SO 4 2- (aq) Kc = 2.4 x 10 -5

All rxns have a K ,[object Object],[object Object],P H2 , P O2 insignificant compared to initial amounts  'gone to completion' 1CaCO 3(s)  1CaO (s) + 1CO 2(g) K p = P CO2 (only) = 1.9 x 10 -23 essentially no CO 2  'no reaction' = 1.4 x 10 83

Gases ,[object Object],Gases are not usually measured in molarity, usually P Use K p (g), K c for (aq) You can convert Kc ↔ Kp Works b/c system in same T, V Where Δ n = n (g) product – n (g) reactant ; R = 0.08206 (no units) P always in atm

Calculate K ,[object Object],How many people shopping in mall if there are 45 people in the shoe store? Mall shopper ↔ Shoe shopper

Calculating K values ,[object Object],[object Object],[object Object]

Using K ,[object Object],[object Object]

The Shoe Store Mall Shoe store 90 10 equil (K = 0.11) 140 10 NOT equil -  +  shift 135 15 equil (K =0.11) Mall shopper ↔ Shoe shopper

Which way? ,[object Object],[object Object],[object Object],[object Object],[object Object],Mall shopper ↔ Shoe shopper

Which way did it go? ,[object Object],0.15 0.31 .11  initial P (atm) 0.15 0.20 5.1 x 10 -3  P I (atm)

2 HI (g)  1H 2(g) + 1I 2(g) K p = .0200 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

2 NO 2 (g)  1 N 2 O 4(g) ,[object Object],decrease V  all P go up same amount (PV=nRT) Which way shift? was happy at old P then we disturbed it  shift changes # moles, hence P changed

SALE SALE SALE! ,[object Object],[object Object],[object Object]

Temperature Shift (sale!) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],YouTube!

Which way? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Mall shopper  shoe shopper K c =0.037 (no sale) ,[object Object]

A flask is filled with 3.2 atm of NH 3 (g) and 1.5 atm of HCl(g) at 300 K. @ equilibrium, P total = 2.50 atm. Calculate K p . NH 4 Cl(s)  1 NH 3 (g) + 1 HCl(g)

2 ‘flavors’ of equilibrium problems Initial conditions Calc K given final conditions given K Calc final conditions In both cases, set up an ‘ICE’ table to do the problem

Equil Calc Short Cut (ECSC) ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

2 NH 3(g)  1N 2(g) + 3 H 2(g) Kc=6.56 x 10 -3 ,[object Object]

N 2 O 4(g)  2NO 2(g) K p = 0.25 @ 298 K ,[object Object]

Acid/base defined ,[object Object],Arrehenius : Acid = H+ ; base OH-

Acid-base pairings ,[object Object],Base Accepts H + Acid Donates H + Acid Donates H + Base Accepts H + Conjugate pair Conjugate pair NH 3(aq) ; NH 4 + (aq) are ‘conjugates’ K c = =K b = Base dissociation constant K b means “dissolved, reacts (as a base) with water”

[object Object],Base Accepts H + Acid Donates H + Acid Donates H + Base Accepts H + Conjugate pair Conjugate pair H 2 O is amphoteric  can act as a base or an acid K c = =K a = Acid dissociation constant H 3 O + (aq) ≡ H + (aq) K a means “dissolved, reacts (as an acid) with water”

Predicting Ka, Kb eqns, expressions ,[object Object],[object Object]

[object Object],[object Object]

HOCl vs HOI E.N. = 3.0 E.N. = 2.5 K a = 3.0 x 10 -8 K a = 2.3 x 10 -11 CCl 3 COOH or CH 3 COOH ? Cl O H I O H

Relative strengths of conjugates ,[object Object],STRONG Acid WEAK base If acid (or base) is ‘strong’  conjugate is ‘weak’ Which is stronger? HF or HI ? HF smaller, higher bond E : HI Stronger acid

relative conjugate strength 1 HNO 3(aq)  1 NO 3 - (aq) + 1 H 3 O + (aq) K a big 1 NO 3 - (aq)  1 HNO 3(aq) + 1 OH - (aq) K b ? 1 Na + (aq)  1 NaOH (aq) + 1 H 3 O + (aq) K a ?

Ka, Kb of conjugates 1 HA (aq) + 1 H 2 O (l)  1 A - (aq) + 1 H 3 O + (aq) K a 1 A - (aq) + 1 H 2 O (l)  1 HA (aq) + 1 OH - (aq) K b 2 H 2 O (l)  1 OH - (aq) + 1 H 3 O + (aq) K w = K a K b

[object Object],1 H 2 O (l) + 1 H 2 O (l)  1 OH - (aq) + 1 H 3 O + (aq) K c =  K w = ionization product of water K w =1.0 x 10 -14 @ 25 C H 2 O (and H 2 O soln’s) ALWAYS have a little of both ‘ Neutral’ water -> [H 3 O + ] = [OH - ] = 1.0 x 10 -7

0.35 M HBr (aq) , calc [H 3 O + ], [OH - ] ,[object Object]

p  = -log  ,[object Object],[object Object],[object Object],[object Object],[object Object],pH pOH [H 3 O + ] [OH - ] pH + pOH = 14

pH 0.356 M Formic acid (HCOOH) ,[object Object]

pH of 0.150 M pyridine (C 5 H 5 N) ,[object Object]

Polyprotic acids ,[object Object],[object Object],[object Object],[object Object]

Calc pH of 0.170 M H 2 SO 4 sol’n ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Just rxn 2 ,[object Object],0.170 M 0.170 M 0.0M I 0.170– x 0.170+x x E We will use successive approximation Kc is small, so x probably << 0.170, pH = -log(0.180) = 0.744 Plug back in Make sense? YUP.

NaCN is basic, why? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

CH 3 COONa (aq) acidic or basic? ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

[object Object],[object Object],[object Object],[object Object]

pH 0.15 M NH 3(aq) ,[object Object],What happens if I throw in some NH 4 Cl into the solution?

pH 0.15 M NH 3(aq) in 0.035 M NH 4 Cl (aq) ,[object Object],Common Ion Effect  A shift of an equilibrium induced by an ion common to the equil.

Lewis Acid/Base defn ,[object Object],[object Object],[object Object],[object Object],6 e- around B, can hold 2 more (acid)

pH of a soln 0.452 M in HCOOH (formic acid )and 0.073 M in HCOONa. Ka formic acid = 1.8 x 10 -4 ,[object Object]

Buffer  sol'n whose pH changes minimally w/ acid or base added ,[object Object],[object Object],CH 3 COOH (aq) + H 2 O  H 3 O + + CH 3 COO - w/ large amounts of acid and base @ same time, equil ‘stuck’, can neutralize small amounts easily. Common ion effect

Henderson-Hasselbach Eqn ,[object Object],‘ Good’ buffers have [base] ≈ [acid] ≥ 1.0 To make an effective pH 4 buffer, pick a weak acid whose pK a is ~4

Acid + base  a salt + water ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Solubility ,[object Object],[object Object],[object Object]

'Saturated' solution  ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

0.064 g of PbF 2 dissolves in 100. mL of H 2 O. Calc K sp ,[object Object]

How much AgCl will dissolve in 100. mL of H 2 O. Ksp = 1.8 x 10 -10 ,[object Object]

1AgCl (s)  1 Ag +1 (aq) +1Cl - (aq) ,[object Object],[object Object],[object Object],[object Object],[object Object]

How much AgCl dissolves in 100. mL of 0.030 M NaCl? Ksp = 1.8 x 10 -10 ,[object Object]

Precipitation  2 (aq) ions together, form insoluble salt. ,[object Object],[object Object],[object Object],[object Object],[object Object], ’ shift left’, precipitate forms

Calc [Mg +2 ] from saturated Mg(OH) 2 in a buffer sol’n of pH = 12.00 ,[object Object]

Complex Ions ,[object Object],[object Object],[object Object],[object Object],[object Object],= 4.1 x 10 8 = K f  formation constant of complex ion

Calc [Zn +2 ] in sol’n initially 0.10 M Zn(NO 3 ) 2 , 6.0 M NH 3 ,[object Object]

Separate ions ,[object Object],[object Object],[object Object]

Equilibrium slideshare. WN.2011

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