Equilibrium & Further Aspects of Equilibria . ☒Pine Equilibrium GhÑvihF⑨£psFs⑧B Gonville ~ Reversible Reaction Equilibrium A : A reaction in which the products can react to : stage where the rate of forward reaction is equal to the rate of backward reaction ① Closed system is none of the in which one reactants or products escape from the system ④ Concentration of Products d. Reactants remain constant It is dynamic i e • - • • • Eg H2 . i + is called a reversible reaction reform the original reactants Is = In a closed system . (Reaction mixture ) . - Reactants & products are continoosly reacting Where of forward reaction rate is rate of backward equal to & concentration of products { reactants in a closed system reaction constant remain . . . 2 HI :-| ( . . lntial Cone § I Hz Iz HI 5 5 0 -4.32 -432 ☐ 7 = - 6- jѧ S kami " Gmt " " " " ° " " 432×2=8 " "8 ' ' - ÷: : 3 O - j Time 1sec { g. 21-1--1 2 10 = 1-12 + Iz A ÉÉ$ ? | ^ G ro = s EEEof a 3 - - - §É÷ ; : # he • If - Chatelier 's Principle more factors that one or affect " "" " "" " "" Equilibrium - - o - 8.64 ° ° " ° [-1-36] I -3612 1.3612 = 0-68 = 0-68 • equilibrium Equilibrium Shifts to Right field of Products Increase are changed the system will ohhose the Equilibrium Shifts to Left Cone / Yield of Reactants increases . change to restore equilibrium FACTORS AFFECTING EQUILIBRIUM ② ① Temperature Pressure Exothermic 9 Temp to Product Rmde > Pmoie + Pressure Product ( doesn't like heat b. Temp 9 Product Rmde - Pmole $ Pressure ↳ Product R mole = Pmole NO EFFECT Endothermic ( likes heat) * Temp - Product ↳ Temh ↳ Product ③ Concentration ④ Catalyst Product eg . to concentration 9 Yield Helps active T Concentration Ha Yield affect equilibrium A Concentration T Yield b- Concentration * Yield equilibrium faster , because it forward reaction affect Doesn't PRESSURE AT EQUILIBRIUM Fez , Vzos backward & Reactant OF yield but doesn't sheds uh both equaby . . Equilibrium 9 Industry → Haters Process CNH }) ☐ H= -92 KJMOI N2 (9) 31-12 + (9) ' - - l 2MHz → Contact Process ① S -102 ② → (1-12504) 502 exo 802+02 ÷ [ V20 5 (g) 2503 at 1-2 atm ] 4502 450C 200 atm Fe Increasing field decrease temperature 2) Increase 3) Condense } it is formed D b. 3 Pressure Remove To Increase field NHz as soon : temperature 2) F Pressure ; 1) as Catalyst 503 , although unnecessary as yield is + ④ Hasaoz Hasoa + Hao > → enough at 2 atm (expensive 1-125207 ( oleum 21-12504 ( 2 moles) EQUILIBRIUM CONSTANT A relationship that Equilibrium expression Eg Nz + 31-12 . KC links to equilibrium Kc [NH }] = " " eg 3 ( Molden-3 ) solid Kc Fe + , " included weather " Fe ( magma , , Kc concentration remains constant ( aq ) or product Agtcaq) Fe of in change caqi " concentration [ HIT Kc = no - pressure N2 + lvtemh 0 = & temperature Temperature 2 HI = , Kc . 2MHz tmtemh equilibrium to RHS - Nz -131-12 unit ÷÷ 31-12 on > conc to : • • . concentration does not affect Kc Pressure doesn't affect > Nat 31-12 conc to conctn Kc . equilibrium to LHS - 2MHz 2MHz To Restore temp • . Cag , [ Hz ] [ Is ] . Zdmb , Hat Iz • >+ 3) , reactant in Concentration } Pressure • Fe + - solid because its ignore we MOI = concentration equilibrium Cmd dm- I 4 " ] } = Ag Cs) on caq) is not = Effect • expression " Agtcaq) . ( MddM→ ) = " [ [Not [Hz] In equilibrium of reactants d. products is called stoichiometry 2MHz Equilibrium expression [Nz] [1-12] concentration c- . = [ Ntts] & concentration ( KC) To restore Kc Increases conct temp Decreases Kc concentration of [ HI] 9- equilibrium shifts to LHS . ! ONLY 0 TEMPRATURE As change in to ooieerse change . its AFFECTS conc Kc & Pressure forces the reaction to original state , only temperature causes EQUILIBRIUM CONSTANT ( KP) Partial : The pressure Pressure The total Ptotal Equilibrium EXAMPLE exerted by one any pressure of a gas equals Pa = + expression PB P in the gas to the sum is mixture of the Pressure - partial pressures of the individual gases Pc + partial pressures involving . N2 (9) + 31-12 = (9) (21-1419) EXAMPLE 2MHz ' H2O (9) + Kh (9) O_0 Equilibrium md ? ? 1-0 Initial md 20 Equilibrium md , #mafia Total pressure ( 214g OH (9) - 2022am:* o_0 to 1-0 1.0 Pot equilibrium at \ y Eq is 4014Pa formed . moles calculate Mole Fraction ( X) ✗ ✗ " " " Hao ( 21-14 = ( 21-14 = ^ total Hao = total £9M" brim moles ( 21-14 ( of ) I = ( zgiiiiiln.ua?0ksat) = 1+1+1 \ § 1-3 / n CZHSOH CZHSOH = n § = total Calculate Partial Pressures Pcztlq ✗ ↳ Hq ✗ Ptotal = 113 P Hao = ✗ ✗ Ptotal = V3 pgµ , , = ygµµ, xp,, , , , Kp = Hao Pc2Hs0H Pcztlq ✗ PHD = Use moles to Calculate mole fraction . = 13-33 13-33×13-33 y, Paid = pa×pa3 Calculate mol left left Mol ' H2O (9) + °Fkh ( 21-1,01-1 ' 20 ↳ Ha (9) "" Pnyggx Piz 20 SOLUTION ÑNHZ = Initial mol ✗ called it 's partial pressure ✗ 40 ✗ 40 ✗ = go ¥3 13.33 = 13.33 KPa , y.gg } KPa = ypa -1 = 0.0750 Pa Calculate payygygyeggwy } Find Kp . Calculate KP . . . Acids & Bases Strong Acid into HCl + Hao + A Base - NaOH which 21-120 • + + 21-130 H2O " said + Nat Caq) > Base : Is a is proton a donor . ' H2O + CH , COO - + . 1-130-1 - disassociates into ions OH + - . Caq) - An acid partially disassociates into ions Acid which CHZCOOH Bronsted dowry Acids and Bases Acid : An - Ci completely which Acid . 1-130-1 • Weak completely disassociates added to Water when ions 1-12504 Strong Base Acid An - Aap ( Ht) Ctlz COOH Cag) Acid Base H2O A) -1 Conjugate conjugate Base 1-130-1 Cag) - → cuz Caql + COO • proton accentor ( Ht) I 1 conjugate H+ donated Acid Base - Pair Measuring PH The PH scale lower than We can 7 use be used to can acidic is pH a indicate weather a , meter solution a with solution 7 is neutral pH is the fit universal indicator or is acidic, alkali , , a or neutral solution . At 25C pH with (or 298 K ) - greater a than solution 7 is with alkali hH . . - 3IV 0 I 2345 6 7 8 9 10 11 13 ' 12 4- concentration pH 1-120 H+ caq) A) Equilibrium 1kW ) [ H+ = can log ,o[ an aqeous base solution Ht Caa! it has ] [ OH OH + - , - > can ] a Degree very of small has dissociation is so [1-12014] remains constant = [Htcaa ] [ OH can of value 1-0 ✗ 10 " at 298k pure water the concentrations of H+ & OH [ Htcaq) ] =/ OH law] 1-0×10-14 mddm→ - = - ions are equal so , . of the 10 hydrogen . Caq ) - Kw ( Ionic Product) kw In - [1-12014] Constant : = in to Dissociation of water €1B Yhe ☐ negative logarithm no units . ion Calculating the pH of strong Bases Strong Acids Weak Bases } Weak Acids , Strong acids of acid concentration molecule eg ionised ( For . / monohrotic monobasic acids , of concentration in solution so H+ ions is approximately the same which contain only acids replaceable H+ one as the ion her like HU ) pH . completing are of (o / mddm- 3) HCl . Htlaql + a- caq) HCl lag)→ [1-14]=[1-1] mddm-3.i.PH -109,010.17--1 = 0.1 = logo [ 1-17 - strong bases : eg Calculate NaOH (ag) Kw = 0.05 100×10+4 = [ OH ] pH OH Cag) + 0-05 [1-1+1101-5] = [1-1-1] - Nat Caq) → 0.05 0.0500 mddm? solution of NaOH of concentration fit of a the . Kw = = - = - log - 2×10 = " -3 moldm 0.05 [1-1-1] ,o log ,o [2×10-3] 12-7 = Weak Acids Weak Acids dissociate ( Hz COOH Cag ) Dissociation of HA 1991 only ' - a partially in aq CH , COO weak Acid = CHZCOO - caqj is law . solution + : Carbonic Acid (1-12103) pKa : Ht can = [ Ht can ] [ A- can ] (Acid Dissociation ) [ HA law constant = (1-1+49) The Value [ " A ""] . of the acid . acid of ka indicates the extent OF Ka form constants by taking the can be expressed negative logarithm of dissociation hka = - . tog oka , The higher the value of Ka The lower the value , of pKa the , the in a to base " (Equilibrium Expression for ) monroprotic (CH} COOH) Ythanoic Acid Acid dissociation → convienent more 10 Ka , Ht Caq) represented by + Examples . stronger the acid . stronger the acid . Eg 50M . . Calculate pH = - 2. 9 = [1-1-1] = . Sol ? of 0.010 motdm } ethanol log ,o[Htcaq )] 1-26×10-3 molding } [1-1-1] ? [1-1-1] = [ HA ] Eg A solution . 10-2-9 = i. acid log,o[Htcaql] - = Ka for ethanol of Ka the value Ka = Calculate 2 = [CHZCOOH ] I -59 ✗ Ka = = of = 0.100 0.100 molars = [1-1-1] ( Ht ) } mo1dm→ acid methanol 'c [1-1-1] = . ( Ka pH = - ? = - 1-32×10-3 motocross log ,o[ Ht ] log ,o[ 1.32×10-3] 0.100 = 1-74×10 smokers } ) - = 2 [ HA ] 1.74×10-5 1-59×10-4 mddm→ ' 0.010 10-4 moldrri } the pH [1-11001-1] (1-26×10-3) 2-88 acid , has a pH of 2^90 . Buffer Solution A Buffer solution is one types of Buffer solutions There are two : small amounts of acid d. alkali changes in pH when that resists are added . - 1) Acidic Buffer Solution It from commonly made is It has anti Grande weak acid & it's a conjugate base (salt) 7 - Ethan oic acid of Sodium Ethanoate CH } COOK G- 1300 Nat I can ↳ ( 143 COOH 2) Alkaline Buffer It is & Cag ) 4-13 COO - Caq ) . Conjugate base lodnanoate Ion) caq ) Solution commonly made from a & weak base conjugate it's (its salt) acid It has a pH > 7 Example NH } Cag ) Bd & HOW A Buffer solution Works Consider The of Ammonia solution Ammonium chloride solution . N buffer equilibrium that exsists HA lag ) Ht Caq) ' - If some HCl + containing this in solution is That means equilibrium shifts IF some HA Caq) a The OH is NaOH + OH - Equilibrium shifts - weak acid added is Can to ' and its conjugate base A- . - to LHS , the . • from acid The H+ added with A- can + in the solution is " mopped up " the OH the weak acid & once - by reaction with the conjugate ' by reaction reacts with tweak acid) HA Hao CD with base and also because in the solution . Pitts mopped up . weak acid extra , A- (conjugate base) HA Can pH doesn't change it remains stable added to the solution → Ht reacts H+ lag) + conjugate base is tweak acid ) HA solution , extra A- (aq) HA formed added * are A- Cag) added to this is Equal Molar Proportions : general acidic a * Hata lag) - again pH doesn't change , remains stable . A common buffer used in the study of biochemical ( Hzpoci ) the acid And Hpoa 1-12 POI caqj . , " acts ions as a prostrate buffer containing dihydrogen phosphate reactions is a conjugate base ions . HPOÉ lag) + Ht Cag, Blood hH In order for the between & 7.35 body to function properly human 7.45 . , 7-8 greater than Values the human or less than body needs to maintain a often result in death 6.8 tAHADmAAEA Nor I 6- 6 6-8 I 1 ¥ 7.2 1 74 I 7.678 To prevent acidosis or alkalosis the body relies on the The cells in our body produce CO2 interaction of a chemical buffer system . CGH 206 , 602 + > 602 61-120 + glucose oxidation via ?⃝ * help of carbonic anhydrase enzyme Hzcozcaq) In blood carbonic acid [weak acid Ht caq) Hcoj , Hzcoz lag) Now if [1-1-1] increases if [ Ht ] decreases , , ' - equilibrium equilibrium shifts shifts to + Cag ) to LHS and hit remains RHS and keeps the pH constant constant . . in the blood . . fit Calculating the hit of a Buffer solution 6cg I calculate . and moldnri 0-250 solution hit of a the } containing ethanoate sodium . 0-200 moldrñ} ethanoic acid . Solution CH 199) COOH } Ctlzcoo ( aq) ' - 0.200 Mol Ka [ CH , COO ] [1-1-1] = 0 - + H+ faq ) 250m01 [ CH } COOH] [1-1-1] Ka [ CH > COOH] = = 1.74×10-5 ✗ 0250 [ CH } COO ] - pH = = . * The Henderson htl So = - pKa : - log ,o[ Ht] - logo [1-39×10-5] pH of this Hasselbach + 109,0 ( = - - log ,, Ka + = Equation 10-5 mddrri } 200 4.86 hka + logo [ conjugate base or salt] [ weak acid] [ HA ] above we can use this formula logo (0-2-50) (0-200) + 1-39 ✗ . [A ] logic (1-74×10-5) = 4.86 buffer system is for the same question hH= 0 - logo (0-250) (0-200) = 4-86 as follows : ( Ka - 1-74×10-5 moidms ) Titration curves ① Calculate initial pH using strong Acid a " - concentration whenever possible :eg ?⃝ a 13 " HH " 10-5 equivalence at 7.0 - o,;oÉ i - • sharp fall at • Graf is 3- 5- inverted Weak Acid HH equivalence - 4 Curve flattens at ñ 3. g Volume of NaOH % s 10 's Y 's: ' so as added 1cm - pH - ÷ S 4 - 2 " ' s to i do 's ? z 's Jo 3 56 " as added 1cm Volume of NaOH • Titration Sharp Fall No Strong Acid Curve Final equivalence pH approaches s } to * ? , gradual Change No Indicator , suitable - Titration Curve Weak Base Y " ios , " ' s to i 's do z 9 h" - 7- 70 - - 8 - - - 4 - - z Volume of NaOH → to 3 - 10.5 's at a 56 } added 1cm I Ss 1 Dt equivalence at 7.0 5.0 ñ curve flattens at ñ 3. g I - , so , - 2 Required fÉÉ f h" - g 's 70 - - 6 Volume of NaOH Shark fall from , 10 l i - ,, at a g. o Curve flattens at ñ 13 13 : - - o - o . Éu - - - do ~ - , ¥ - 7- 's point " - 9 7. I 5 - Strong Base - do - 2 Required → at - 3 vertically when strong acid is - BAM - b I 12 i - g ? " 13 g curve - ?⃝ - added to base instead pH logo (0-1)=1 - " - Volume of NaOH o - - a , 11 hH= Titration Weak base - , " h" - - - i ñ till "" " "" " " - 4 " Weak acid - 6- z "" / " 7- s 0-100 mddrris Ht is FG Iat g . Concentration . Titration Curve Strong base " pH { Indicators l , - " 0 ' s to i 's ? do to Volume of NaOH ME Sharp Fall 3.5 - 7-0 3 's at a 56 } added 1cm ? s 's , so Bt Indicators Equivalence Point : The acid & alkali colour pH Range because Indicator methyl 3. I bromo phenol - not of an indicator Yellow 4- 6 Yellow Blue 6.3 Red Yellow phenol 6- 8- 8. a Yellow Red 82 Colorless Pink Example : - 10.0 weak Acid - - same pH values ionised & ionised Strong Base titration since Addition of a base will single dsoh of change pH it is important to use suitable indicator r . point at which equivalent of a titration is the point between indicator numbers of at which the indicator moles forms which the are present . has intermediate of changes . Alkali Color Acid Color 4.2 Phenol hhtalein un Red - the is the of the is The end point necessarily mothy red red . 4.4 2- 9- blue are compare able amounts pH Range orange have been added these - The pH range : point of a titration equivalence colours,