Essential Tool for Chemistry Concepts, Va riables & Equations, Including ~ S ample Problems, ~ Common P itfalls & Helpfu l Hints CONTENTS Pg Basic Skills, Math Review • Pg Statistics, Atomic Data, Chemical Formula Calculations • Pg Stoichiometry Pg Gas Law Calculations, Solids & Liquids • Pg Thermodynamics, Acid-Base Calculations • Pg Equilibrium Calculations, Kinetics Your success in chemistry depends on your ability to solve numerical problems BASIC SKILLS cont - cont MATH REVIEW The unit & numerical value are changed using a conversion factor or equation BASIC SKILLS f~m" CALCULATOR SURVIVAL 'B,wm, wi :2 CO + H20 CH4 and CO must have the same coefficient · I CH4 + :2 O2 => I CO + H 20 Now, determine the O coefficient: · I CH4 + O2=> I CO + H20 Check your work: I C, H and on each side It is balanced! BALANCING A REDOX EQUATION USING THE HALF-REACTION METHOD • Split the reaction into oxidation and reduction half­ reactions • You may need to add H20 and H+ for acidic, or H20 and OR- for basic reaction conditions • Balance these separately, then combine to balance the exchange of electrons ~Sample ' Balance the following for acidic solution: Mn04' + Fe 2+ => Mn2+ + Fe 3+ I In acidic solution: Add H+ to the left and H20 to the right side: Mn04' + Fe 2+ + H+ => H 20 + Mn2+ + Fe 3+ Identify the half-reactions: Fe2+ => Fe 3+ (oxidation) Mn04' + H+ => Mn2+ + H20 (reduction) Add electrons to account for valence changes: Fe 2+ =>Fe 3++ Ie' Fe(II) to Fe(III) 5e' + Mn04' + W =>Mn2++ H20 Mn(VII) to Mn(ll) Balance each half-reaction: a Oxidation: Multiply by a factor of to match electrons in reduction step: Fe 2+ => Fe+3 + e' Charge: +10 on each side, balanced! b Reduction: Balance 0, then W; check charge: 5e' + Mn04' + 8W =>Mn2+ + 4H20 Charge: +2 on each side, balanced! Combine half-reactions to eliminate the 5e': Fe 2+ + Mn04' + H+ => Mn2+ + H 20 + Fe3+ Check your work: Fe, Mn, and H on each side, atoms are balanced! Charge: + 17 on each side, charge is balanced! Lt, Pitfall: Make sure you use the H20, H+ (for acidic), or OH' (for basic), with the correct half-reaction Mass of a reactant is used to determine mass of product Given: Mass of reactant, balanced equation, molar masses of reactants and products n'ucl.tllI ma~ Calculate: Moles of reactant = 1'.'"('1111 molar rna" Calculate: Molar ratio = protlu,,~ ·qusti.on coern.ci' II( n'acranl fqu.lflOn cociflcienl Calculate: Moles of product = moles of reactant x molar ratio Calculate: Mass of product = moles of product x product molar mass ~Sample : Calculate the mass of Mg produced burning 1O.0g of Mg in excess oxygen Balanced equation: Mg + O2 => MgO Given: 10.0 grams of Mg, Mg at.wt = 24.305 g, MgO molar mass = 40.305 g Calculate: Moles of Mg = 1O.0g of Mgl24.305g Mglmole Mg = 0.411 moles of Mg Calculate: Molar ratio = 2/2 = I Calculate: Moles of MgO = 0.411 moles Mg x I mole MgO/mole Mg = 0.411 moles of MgO Calculate: Mass of MgO = 0.411 moles MgO x 40.305 g MgO/mole MgO = 16.6 gMgO a single 16.6 ~ '\lgO (nllte calle('INI unit, in Lt,Pitfall: If your balanced equation is wrong, your theoretical yield will usually be wrong HOW TO CALCULATE A LIMITING REAGENT FOR REACTANTS In a reaction with reactants, the mass ofproduct is constrained by the reactant in shortest supply, the limiting reagent Given: Balanced equation, mass of reactants, molar masses of reactants; specify reactant #s Calculate: Moles of each reactant = Calculate: Ideal reactant molar ",,,Ia "'." ratio coefficient of #lIcoefficient of reactant #2 Calculate: Actual reactant molar ratio = moles reactant # I /moles of reactant #2 Determine the limiting reagent: • If actual reactant molar ratio :5 ideal molar ratio, then reactant # is the limiting reagent • If actual reactant molar ratio > ideal molar ratio, then reactant #2 is the limiting reagent Calculate the theoretical yield based on the mass of the limiting reagent Hint: The reactant nwnbering is arbitrary, but you must stick with your choice for the entire calculation ~Sample: 1O.0g Mg reacts with 1O.0g O2; how much MgO is produced? Balanced equation: Mg + O => MgO Given: Mg molar mass = 24.305g; Reactant #1 O molar mass = 32.00g; Reactant #2 Calculate: Moles of Mg = 1O.0g Mglmol Mg = 0.411 mol Mg moles of O2 = 1O.0g 02/32.oog 02/mol O = 0.3125 mol O Calculate: Ideal reactant molar ratio = 2/ I = Calculate: Actual reactant molar ratio = moles MgI moles O2 = 0.41110.3125 = 1.31 Determine limiting reagent: 1.31 is less than 2.0, therefore, Mg is the limiting reagent Calculate the yield ofMgO based on 10.0 grams Mg (shown in the previous section) Lt, Pitfall: Make sure you distinguish between the ideal and actual molar ratios WORI(ING WITH GASES P\, = nRT Ideal Gas L3\\ Simple model for gas behavior Pressure, P; common units: atm, Pa, bar or mm Hg; the R given below is for P in atm Volume, V; common units: liter (L), m3; the R given below is for V in L Temperature, T; common units: Kelvin, °C or of; always convert °C and OF to Kelvin Number of moles n; moles = gas mass/gas molar mass Ideal Gas constant = R = 0.082 L atml(mol K) ,1,Pitfall: All data ~ust fit the units ofR HOW TO CALCULATE THE # OF MOLES OF A GAS SAMPLE Given: Mass of gas (g), molar mass of gas Calculate: Moles = mass of samplelmolar mass ~Sample: Determine # of moles in 5.0g ofH2 gas Given: 5.0g sample, H2 molar mass = 2.0 16 g/mole Calculate: Moles of H2 = 5.0 g H2 x I mole H212.016g H2 = 2.48 mol H If you are given density and volume, fIrst calculate massofthe gas; mass (g) = p(gIL) x V(L) HOW TO USE THE IDEAL GAS LAW ~Sample: Calculate the pressure for 2.5 moles of Ar gas occupying 3.5 liters at 25°C Make required changes to variables: T(K) = 25°C + 273.15 = 298.15K Calculate: P = nRTN = 2.5 moles x 0.082L atm/(mol K) x 298.15K13.5L = 17.5 atm (Other units cancel) HOW TO USE BOYLE' S LAW P & V are inversely proportional with constant T & n 10m ':I1."' WORKING WITH GASES SOLIDS Be LIQUIDS cont cont ~ HOW TO USE AVOGADRO'S LAW HOW TO CALCULATE MOLES OF REAGENTS IN SOLUTION V is proportional to n with constant T & P Va n This is a direct-proportionality problem: Given: The # of moles changes by a factor ofz Calculate: Vfin = Z x Vini" ~Sample : A 2.0 mole gas sample occupies 30.0L Determine V for 1.0 mole of the gas Given: n changes by a factor of 1/2 Calc ulate: Vfi n = 112 X 30.0 L = 15.0 L HOW TO CALCULATE THE SPEED OF A GAS MOLECULE Note: R = 8.314510 J/(K mol) y = 8.314510 kg m 21(s2 K mol) r ms =~3RT "1\1 ~Sample : Calculate the vrms for He at 300 K - Given: T = 300; HOW TO PREPARE SOLUTIONS He, M = 4.00 g/mol = 4.00 x10-3 kg/mol Calculate: vrrns = ,« RTf M) = '1 (3 x 8.31 45 kg m 2/s2/(K mol ) x 300 K74 00 x 10-T kg/mol ) = ,« 87 x 106 m2/s2) = 1,370 mls ,1, Pitfall: Watch the units on Rand M; the final unit is mls T must be in Kelvin PREPARING A DILUTE SOLUTION FROM A STOCK SOLUTION The rela tive rate of effusion for molecules of mass M I and M2 Key: Conserve mass and moles The molarity and volume of the stock and diluted solutions arc governed by: " k I" dll I ~ "mpl~ ' Prepare 50 mL of a 1.0 M soluti on ~Sample : Determine rel ative ~4 22 H2 diffuses times as fast as C02 ~8 ~O o 200 400 600 800 Pressure (mm Hg) Given: P changes by a factor of z Calculate: Vfin = lIz X Vini' ~ Sample: The pressure of a 4.0L sample changes from 2.5 atm to 5.0 atm What is the final V? Given: P changes 2-fold: 2.5 to 5.0 atm Calculate: Vfin = 1/2 x Vini' = 1/2 x 4.0L = 2.0L HOW TO USE CHARLES' LAW ,1,Pitfall : [t is easy to invert the M I /M2 ; the smaller -­ SOLIDS 8& LIQUIDS CALCULATING MOLES OF REAGENTS Determine the # of moles in "x-grams," or the mass needed to give a certain # of moles Required data: The molar mass ~ ample Calculate # of moles in 5.6 g of NaCI Given: NaCI molar mass = 58.44 g Calculate: Moles of NaCI = 5.6g/58.44g/mole = 0.096 moles NaCl One example is freezing point depression: AI f x on r ctor m: Molality; K( solvent constant; Ion factor: # of ions produced by solute: I for Molecular solute; or more for ionic salt ~ ~ampl' Calculate the freezing point depression for a solution of loog of NaCI in 500g of water Given: K f (water) = 1.86 °e /m; molar mass of salt = 58.44 g/mol Ion factor = Calculate: Mass of solvent = 500g x I kg/ I ,Ooog = 0.500 kg Calculate: m ofNael = (I OO.0g/58.44g/mol )l0.500kg = 3.42 In Calculate: AT = - 3.42 m x 1.86 "e lm x = -12.72 °C ~ Sample' Calculate mass of 0.25 moles of NaCI V and T are linearly proportional with constant n &P Calculate: Mass ofNaCI = 0.25 moles x 58.44g/mole = 14.61g NaCI Temperature (K) This is a direct-proportionality problem: V IX T: Given: T changes by a factor of z Calculate: V fin = from "a" mL of2.0 M stock Given: M,'ock = 2.0 M; V"ock ~ a; Mdi1u,c = 1.0 M; Voilu,c = 50 mL Calculate: "a" mL = 50mL x 1.0M/2.0M = 25mL HOW TO CALCULATE COLLIGATIVE PROPERTIES atom is always faster This is an inverse-proportionality problem: • P IX Il v Given: V changes by a factor of z Calculate: Pfin =' I z x Pinit • V IX IIp A solution is prepared by dissolvi ng a known mass of solid in a specific amount of solvent ~ a'"pl' Prepare one liter of 1.00 M NaCI Given: NaCI molar mass = 58.44 g/mole Step I Wei gh out 58.44 g of NaCI and transfer to a I L volumetric flask Step Dissolve the salt; fill with water to the I-L line Note: If you need a different M, change the mass of NaCI HOW TO USE GRAHAM'S LAW OF EFFUSION rate of effusion for H2 and CO2, Given: MI = MHl = g'mol; M2= M CO2 = 44 g'mol Calculate: Rate H 2/Rate CO = \ (44/2) = 4.7 :::!.6 Common units: Molarity (M): Moles of solute per liter of solution; molality (m), moles of solute per kg of solvent Multiply the solution volume by the molarity to calculate moles ofreagent Given: Solution molarity (M) Calculate: Moles = vol x M & Pitfall: Volume should be in L If you want to use "mL," denote M as "mmol/mL of sol ution" ; mmol = 0.001 mole ~ 3mph- Determine the moles and mass of NaCI in 25 mL of 2.35 M NaCI solution Given: NaCI molarity = 2.35 M; molar mass = 58.44 g Calculate: Volume = 25 mL x I U I000 mL = 0.D25 L Calculate: NaCI moles = 2.35 mollL x 0.025 L = 0.059 moles Calculate: NaCI mass = 0.059 moles x 58.44 glmole = 3.45g Z x Vini" ,1, Pitfall: T must be in Kelvin ; convert °C to K ~ Sample: A 3.5L sample of He gas is at 300 K; the T is raised to 900 K; what is the new V? Given: T increases 3-fold: 300 to 900 K Calculate: V fin = x Vinit = 3.0 x 3.5 L = 10.5 L HOW TO CALCULATE MASS OF L1aUIDS Density (p) has units of g/mL Pure reagent: Use p & volume to determine the mass vol p ma ~ Sample- Determine the mass of 30.0 mL of methanol Given: r = 0.790 glmL Calculate: Mass = vol x r = 30.0mL x 0.790 g/mL = 23.7 g Given: AGrO (Free Energy of Formation), in kJ/mole; AHrO (Enthalpy of Formation) in kJ/mole; So (Standard Entropy), J/(mole K) Calculate: AG = sum of product AGrO - sum reactant AG rO K ACID-BASE CHEMISTRY THERMODYNAMICS cont cont ~ ~H = sum of product ~HP - sum of reactant ~HP Calculate: ~S = sum of product So - sum of reactant So Calculate: HOW TO CALCULATE [H+]e • Kw = [OH-][W] = Ix10· 14 at 25 DC • For pure water: [OH-] = [H+] = I X 10- M ~Sample : Calculate DH for the reaction: ~(g) + 202 (g) =>C02(g) + 2H20(1) Given: ~HP -74.6 x 0.0 -393.5 x -285.8 Calculate: ~H = product ~HP - reactant ~Hp ~H = -393.5 -571.6 +74.6 = -890.5 kJ/mole Note: Combustion is an exothemic reaction Examine DH • Exothermic (releases heat) : ~H < • Endothermic (absorbs heat): ~H > - I kJ/mole at 25°C Given: ~G = -10 kJ; 'T = 25°C; R = 8.3145 J/(K mole) Calculate: T(K) = 25 °C + 273 I = 298 15K Calculate: ~G = -l kJ x 1,000 J/kJ = - I0,000 J Calcul ate: Keq = exp (- (-10,000 J/mole) 1(8.3145 J/(K mole) x 298.15 K) = e4.03 = 56.5 The equilibrium shifts to the right, a spontaneous reaction Calculate: pOH = 14 - 4.5 = 9.5 Calculate: [OH-] = 10(-9,5) = 3.2 x 10- 10 M • Ka = [W]eq [A-]et/[HA]eq • A- : Conjugate base; Kb(A-) = - B Init E'luil Kw/Ka(HA) Method: Substitute the "Equil" expressions into • pKb Solve this quadratic equation for "a" = [OH-]eq = Start with: Kb = a 2/([B]init - a) -Iog lo (Kb); weak bases have large pKb the experimental equilibrium concentrations into K [Wloq lA loq IIIAI" ~Sample ' Determine K and pK equilibrium concentration data for HA from [HA]eq = 1.0 M Calculate: K = I x I 0- X I x I 0-4/ 1.0 = I x I 0- Basic Salts react with water to form OH- Sample: Sodium acetate: Ac- + H20 HAc + OHAcidic Salts react with water to form H30+ Sample: Ammonium chloride: NH4+ + H20 NH3 + H30 + Neutral Salts not react with water Sample: NaCl (product of strong acid + strong base) Calculate: pKa = -log 10 (I x I 0-8 ) = 8.0 B => C ~H = 50 kJ/mole D => B+ F ~H = 43 kJ/mole reactions gives A + D => F DH = 50kJ/moie + 43kJ/moie 93kJ/moie What happens if you reverse a reaction ? Jfyou reverse the reaction, change the sign ofDH, ~G or ~S HOW TO CALCULATE Kb & pKb OF " CONJUGATE BASE (A-) OF AN ACID HA " K b ( " ' ) = K,,1K.( H ") ~Sample ' Determine Kb and pKb for the acetate ion, Ac-· G iven: Ka(HAc) = 1.7 Given : A + B => C ~H = 50 kJ/mole For C => A + B, the reverse of this reaction Calculate: ~H = - 50 kJ/mole X 2A + 2C => 20 Given: A + C => D ~H = -50 kJ For 2A + 2C => 0, "double" this reaction Calculate: ~H = x-50 kJ = -100 kJ Step I: Is the salt acidic, basic or neutral? Step 2: If acidic: Identify the weak acid, and the K.; ifbasic: Identify the weak base, and the % acidic or basic character 10-5 Calculate: Kb(Ac-) = KwfK.IHAc) = Ix 10-14/ 1.7xI0-5 = 5.9 Step 3: Set up the problem as a weak-base or X 10- 10 Calculate: p~(Ac-) = -loglO (5.9 X 10- 10 ) = 9.23 HOW TO CALCULATE DISSOCIATION OF AN ACID 00 W ~SampIe: Determine DH for the reaction: HOW TO CALCULATE THE [H+] OR [OH-] FOR A SALT If neutral: The solution will not have Identify the acid: Acetic acid C => A + B How equation coefficients impact the DH, ~G or flS? " Thermodynamic properties scale with the II coefficients ~ 181 1011- Kb and solve the quadratic equation: Given : [H +]eq = IxlO-4 M; [A-]eq = IxIO-4 M; ~ SampJe: Calculate ~H for the reaction A + D => F o + H 20 18Jlnl WHY DO SALTS HYDROLYZE? ~S ~ Sample: Determine DH for the reaction LL Pitfall: Watch out for round-off error when you solve for the roots of the quadratic equation • Kb = [OH-]cq [BW]eq;[B] eq • BH+: Conjugate acid; Ka(BH+) = Kw/Kb(B) Substitute Hess' Law: If you "sum" reactions, you also sum Given: A + Given: C + Sum of the Calculate: a = [W]eq = 0029 M Check your work: Ka = (0.0029 x 0.0029) 1(0.5 - 0.0029) = 1.7 x 10- HOW TO CALCULATE THE [OH-]eq B + H 20 BH + + 011 ­ 2, Base HOW TO HANDLE THE " ADDITION" OF REACTIONS Z _~ ~ • pKa = -Iog lo (Ka) Pitfall: T must be in K; make sure you are consistent with J and kJ ~ ~ the Ka expression and a + 1.7 x 10-5 x a - 8.5 x 10-6 = Solve the quadratic equation: Given: pH = 4.5 ~~ a + Ka x a - Ka x [HAc]init = a2 + 1.7 x 10-5 x a - 1.7 x 10-5 x 0.50 = \, ~Samplc Determine the [OH-] from pOH or pH ~Sample Calculate Keq if the ~G ofa reaction is Given K and [HA]init' use the quadratic formula to obtain "a," [H+]eq Substitute into quadratic equation : HOW TO CALCULATE pOH & [OH'] - ' a + Ka x a - Ka x [HA]init = Given: Ka = 1.7 x 10 - Given : pH = 8.5 Calculate: [H+] = 10(·8 5) = 3.2 x 10-9 M - - a ~Sample Calculate [H i ] for 0.5 M HAc Calculate: pH = -loglO [l.4 x 10-5] = 4.85 ~ a This rearranges to: Ka x ([HA] init - a) = a ~Samplc Determine [H+] from pH HOW TO WORK WITH Ka & Kb Acid IIA II' I- A­ HOW TO CALCULATE Keq ~H, ~G ,; pH = - Iog lo IWI III = lO·pH ~Samplc : Determine the pH for a specific [W] Given: IH+I = 1.4 x 10-5 M Ilnll-1I Start with: Ka = a /([HA]init - a) 10H-1 = lo-POIl pOH - - logJO 1011-1 p'OH + pH - 14 for an~ I!i>en solution Does the reaction proceed to completion? Is the reaction spontaneous? Examine ~G • ~G > not spontaneous Keq < I • ~G < spontaneous Keq > I • Use ~G to calculate Keq IH Ka and solve the quadratic equation: HOW TO CALCULATE pH & [H+] H20(l) => H20(g) Given: SO 70.0 188.8 Calculate: ~S = l88 - 70.0 = 118.8 J/(mole K) Notc: A gas has more entropy than a liquid H + + H\ n o IHAhnl! Method: Substitute the "Equil" expressions into • Acidic solution: [H+] > Ix10-7 M • Basic solution: [H+] < I x I 0- M ~Salllpic ' Calculate ~S for the phase change: Does the reaction release or absorb heat? Init Equtl weak-acid dissociation problem Given the initial salt concentration, calculate the equilibrium [W] or [OH-] ~Samplc Determine the [H" ] or [OH-] for a 0.40 M NaAc solution Step I: NaAc is a basic salt % dissociation = 100· ~ Sample' Determine the % diss for a 0.50 M HA that produces [H+]cq = 0.10 M and [HA]eq= 0.4 Given: [W]eq = 0.10 M; [HA]initial = 0.50 M Calculate: % diss = 100% x 0.1010.50 = 20% LL Pitfall: Be sure to use [HA]initial, not [HA] eq Step 2: Ac- is the base; Kb(Ac-) = KwfK (HAc) = 5.9 x 10- 10 Step 3: Solve as a "weak-base dissociation" problem [B]init = 0.40 M Ac-; calculate [OH-]equil LL Pitfall: You must correctly identify the acid or base formed by the salt ions, and determine the K or % I I(INETICS cont ACID-BASE CHEMISTRY -cont ~ HOW TO CALCULATE pH OF A BUFFER Z W ft IA Buffer of Weak Acid and Conjugate Base Start with both weak acid, [HAlini" and salt, [A-l ini,· The equilibrium concentrations are governed by K = [Wleq[A-leq/[HAleq· O HA H + A Init IHAlinlt IA Ilnll [quil IH I n ,-a a lA-lin ,+8 Method: Substitute the "Equil" expressions into and solve the quadratic equation: K = a x ([A-lini' + a)/([HAlini' - a) K X [HAlini' - a x Ka = a x [A-lini' + a ~ ~ K a + a x (K.+ [A-l ini ,) - Ka x [HAlini' = Given: K., [A-l ini , and [HAlini" solve for the roots of the quadratic, a = [H+leq ~~ampll' Detennine the pH of a buffer of O.S M HAc and 0.3 M Ac- Given: Ka (HAc) = \,7 x 10-5 Quadratic: = a2 + a x (\'7 x 10-5 + 0.3) -1.7 x 10-5 x O.S 0= a + a x (0.3) - 1.7 x 10- x O.S Solve quadratic: a = [H+l = 2.8 x 10-5 M Calculate: pH = - 10gIO (2.8 x 10- 5) = 4.SS HOW TO USE THE HENDERSON­ HASSELLBACH APPROXIMATION FOR BUFFER PH Assume that "a" in the previous problem ~ Z W ~ o­ ~ is « [HAclini' and [Ac-lini' Hendl'rson { IA I } Hassellbach: pH = pK + 10glO IHAI ~Samp l e ' Examine previous buffer problem: Given: tA-l = [Ac-l = 0.3 M; [HAl = [HAcl = O.S M pK = pK.(HAc) = 4.77 Calculate: pH = 4.77 + loglO(0.3/0 S) = 4.77 - 0.22 = 4.55 The approximation works HOW TO DO AN ACID-BASE TITRATION A systematic acid-base neutralization used to detennine the concentration of an unknown acid or base At the equivalence point, moles of acid = moles of base ~Samp l e The titration of SO.OO mL of an HCI solution requires 2S.00 mL of \.00 M NaOH Calculate the [HCI] Equation: HCI + NaOH => NaCI + H 20 This gives 1:1 molar ratio ofHCI: NaOH At the equivalence point: The moles balance, or more conveniently: Mmoles HCI = mmoles NaOH M(HCI) x vol-acid (mL) = vol-base (mL) x M (NaOH) M(HCI) = vol-base (mL) x M (NaOH)/vol-acid (mL) Calculate: M(HCI) = 2S 00 mL x \.00 M/SO.OO mL = 0.50 M HCI HOW TO DETERMINE IF THE REACTION IS AT EQUILIBRIUM Compare Qc with K ~Sa mpl e : For the reaction: A C, Kc = 0.60; the observed [Al = 0.1 and [Cl = 0.20 Is the reaction at equilibrium? Ifnot, predict the shift \ Qc = [C]/[Al = 20/0 10 = 2.0; Kc = 0.60 Qc > Kc; process is 1I0t at equilibrium, it will to the left HOW TO PREDICT EQUILIBRIUM CONCENTRATIONS ~Sam pl e : Calculate the equilibrium "orl ~p'nt"'ti()nd for the following gas-phase reaction data: 0.64; [COl init = [H20]ini' = 0.5 M: CO(g) + HzO(g) CO (el + H2 (e) I nit [COl init [H 20I init 0 Equi l [COl init-a IH 201 ini,-a a a N ote: The change "a" is the same for each because the I : I : I : I coefficients in the equation Identify equilibrium expression: Kc = [C0 2]eq[H 2]eq/[CO]eq [H20]eq Substitute "equil" values: K = a 2/ {([CO] init -a) x ([H20] init -all 0.64 = a 2/{ (0.SO -a)(O.SO -a)} Take square root of each side: 0.8 = a/(O.S-a) or -0.8 = a/(O.S-a) a = -1.2 a = 0.222 or Use the first option, since "a" must be positive [C0 2]eq = [H 2]eq = 0.222 M [CO]eq = [H 20]eq = O.SO - 222 = 0.278 M Check your work: Kc = (0.222 2)/(0.278 2) = 0.64 Kc = ill Pitfall: Watch out for round-off error; take the th at gives positive concentrations ~Sa m pl e : Determine the solubility limit for sil chloride, AgCl, given Ksp = 1.77 X 10- 10 Given: AgCI (s) Ag+ (aq) + CI- (aq) ; Ksp = [Ag'][CI-] = \,77 x 10- 10 Given: AgCI molar mass = 143.32 glmole At equilibrium, [Ag' ]eq = [Cl"Jeq = {(Ksp) Calculate: [Ag+]eq = -J(I.77xIO- IO ) = 1.33 x 10-5 M AgCI This is also [AgCIl eq, the molar solubility limit for AgCl Calculate: The AgCI giL solubility limit = [AgCl]eq x molar mass of AgCI = 1.33 x 10-5 moieslL AgCI x 143.32 glmole = 1.9 x 10-3 giL Pitfall: Watch the units on volume and molarity work with "L & mole" or "mL & mmole." Z W ft IA constant, Kc· (('Ie 101 d all other conditions, the process is Q, = I \1" IBI b described by the reaction quotient, Qc: if Q, ~ K c ' the reaction i at equilibrium if Q > K., the rcactinn "illel) tn the left if Q < tbe reactioll "ill eo to the right Gas-phase reactions may be described with K p, based on reagent partial pressures These calculations follow the same strategy as Kc O oiIIII "II1II "c Initial Rate Strategy: Step I: For [A] 10 measure the time required to produce Ll[S] of product Calculate: Rate I = Ll[B l/time Step 2: Measure the new reaction rate, rate2, for a different concentration, [A12­ The interplay of rate, [A] and x, are governed by the equation: ~"'aI1lJlk Determine "x" if doubling [A] also doubles the rate : I RateilRate2 = 2 [AMAh = = 2" x = I, this is a I st order process ~Samplc Detennine "x" if doubling [A] increases the rate by 4-fold: I RatellRate2 = [AMAh = = x, x = 2, this is a 2nd order process 2_ Integrated Rate Equation Strategy Analyze "[Al vs time" data for the reaction The reaction is 1st order if the "In [A] vs t" graph is linear The reaction is 2nd order if "1/[ A] vs t" graph is linear In each case, k is the slope of the line HOW TO DETERMINE THE ACTIVATION ENERGY, Ea Applications: a Predict kl at T 10 given k2 at aT and Ea· b Detennine E from k 1• k2' T I ,T 2; only have to worry about k 11k2 Sample: The rate constant doubles when the temperature changes from 2S.00C to SO.OOC What is E.? Given: k 1/k2 = Calculate: TI = SO.ooC+ 273.IS = 323.2 K Calculate: T2 = 2S.00C + 273.IS = 298.2 K Calculate: LlT =T 1-T = 2S.0 K Calculate: Ea = R In(k l /k2) TI x T 2/LlT = 8.314 l/mole K x In (2) x 323.2 K x 298.2 Kl2S.0 K = 22.200 l/mole = 22.2 kJ/mole illPitfall: T must be in Kelvin; if you use the equation with "liT I-liT2," beware of round-off error in calculating inverse T KINETICS & EQUILIBRIUM An equilibrium is characterized by competing forward and reverse reactions The forward and reverse rate constants (k r and kr ) are related to the equilibrium constant, Kcq At equilibrium: The forward and reverse reaction rates are equal k, ill • For a reaction that has not gone to completion: ~ a A + b B C C + d D At equilibrium, the process is " = IClOt,· I Dlcq d described by the equilibrium < IAlcq"IBlcq" Two common rate-law methods: "Initial rates" and "integrated rate equations." Consider the reaction "A=>S," with a rate law ofthe fann: Rate = k [A]x The goal ofkinetic study: Detennine "x," the order of the reaction rate rate appearance of product, Ll[Bl/Lltime; or, rate of loss reactant: - Ll[A]/Lltime ~ Sample How would you characterize the rate of: CaC0 (5) => CaO (5) + CO2 (g)? Answer: Focus on rate of CO2 production; Rate = Ll[C02]/Lltime HOW TO DETERMINE THE RATE LAW The rate law gives the order of the reaction based on the steps in the overall reaction "A + B => C." \ Rate = k [A], for a first-order reaction Rate = k [Af, or k [A][B], for a second-order reaction Rate = k [A1°, for a zero-order reaction ill Pitfall: Equation coefficients describe the balanced overall reaction, not the mechanism and rate law 6IW~W ~~IJll~ 14 CREDITS Author: Mark Jackson, PhD layout: Andrll Brisson CuS10mer Hotllnl' • , 8002309522 PRICE U.S.$5.95 CAN $8 95 0' Note: Due to the oondeosed nature this gutde, use as a quick re1~~ not ~ II rt!placemenl1or assigned OOIlrse WOOc All righl~ r enrvrd No p~rt oflhu publ>c.tion INI)' bel- rep.oduc:ttI o)fU'SI1SJDllcd./ in u,y rorm arby any nlt~r15 '!Iectronic Of nwchani~l incJUdJ", phOlotOp.l' r:~ofdh'l" or an)' t!lfomllluon~JC ami rc:tm,,'a l Sr~h:m withO\.ll wnuen pcr1l,iUWII from the: publ~he:r e 2003 200(i Ba.rCham Inc 10{\7 ISBN-13: 978-142320189-2 ISBN-10: 142320189-2 911~l1 ~ ~I I I ~Ilil~1111111111111Ilillll