Topics in CHM 1046Intermolecular forces (IMF)

Themodynamics

Chemical KineticsChemical Equilibrium

Combination of above:Unit 11 Intermolecular Forces, Liquids & Solids (T1)Unit 12 Properties of Solutions (T1)Unit 13 Thermochemistry (T1)Unit 14 Chemical Thermodynamics (T2)Unit 15 Chemical Kinetics (T2)Unit 16 Chem. Equilibrium Gases & Heterogeneous (T3)Unit 17.Acid Base Equilibria (T3)Unit 18 Acid Base Equilibria Buffers & Hydrolysis (T3)Unit 19 Acid Base Equilibria Titrations (T4)Unit 20 Aqueous Equilibria Solubility Product (T4)Unit 21 Electrochemistry (T4)

Unit 16Chemical Equilibrium: Gases & HeterogeneousDr. Jorge L. AlonsoMiami-Dade College Kendall CampusMiami, FLTextbook Reference: Chapter # 17Module # 5 (& Appendix 2)CHM 1046: General Chemistry and Qualitative Analysis

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

This does not mean chemicals are found in same concentration!The Concept of Equilibrium{Non-Equilibrium Reactions}: proceed in one direction, A B{Equilibrium Reactions}: proceed in both directions, A B2 NO (g) + O2 (g) 2 NO2 (g)(clear gases)(red-brown gas)(clear gas)(red-brown gas)2X

The Equilibrium Constant (Keq)Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

Rewriting this, it becomesThe Equilibrium Constant is the ratio of the rate constants at a particular temperature.Keq =

The Equilibrium Constant (K)To generalize this expression, consider the reactionThe equilibrium expression for this reaction would bewhere [X] = concentration of each chemical in moles/ liter (M).

What Does the Value of K Mean?If K > 1, the reaction is product-favored; product predominates at equilibrium.Kc =10 = 10Kc = 1 = 0.111010 x10 x

The Equilibrium Constant for GasesBecause pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be writtenwhere Kp = equilibrium constant for gases, and Px = partial pressure of each gas.Partial press: (75 torrs) (150 torrs) (300 torrs) (600 torrs) = total (1125 torrs)

Relationship between Kc and Kpideal gas law:PV = nRTPlugging this into the expression for KpKp = Kc (RT)nWheren = (moles of gaseous product) (moles of gaseous reactant)

=

Equilibrium Can Be Reached from Either DirectionThe ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.KC@ 100C

Equilibrium Can Be Reached from Either DirectionN2O4 (g)2 NO2 (g)

Manipulating Equilibrium ConstantsThe equilibrium constant of a reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.RULE #1: Reciprocal Rule = 4.72 at 100C

1= 0.212

Manipulating Equilibrium ConstantsIf the coefficients of a chemical equation are changed (increased or decreased) by a factor n, then the value of K is raised to that power.RULE #2: Coefficient Rule

Manipulating Equilibrium ConstantsWhen two or more equations are added to obtain a final resultant equation, the equilibrium constant for the resultant equation is product of the constants of the added equations.RULE #3: Multiple Equilibria Rule2 A(aq) + B(aq) 4 C(aq)4 C(aq) + E(aq) 2 F(aq)2 A(aq) + B(aq) + E 2 F(aq)

Homogeneous vs.Heterogeneous

EquilibriaReactions involving only gasses or only solutionsReactions involving different phases (s, l, g, aq) of matterCaCO3 (s)CO2 (g) + CaO(s) Kc = = 0.212 at 100C[NO2]2[N2O4]

The Concentrations of Solids and Liquids Are Essentially Constant The concentrations of solids and liquids do not appear in equilibrium expressions because their concentration does not change.Kc = [Pb2+] [Cl]2Concentration of both solids and liquids can be obtained by dividing the density of the substance by its molar massand both of these are constants at constant temperature. Write the equilibrium expression!

What are the Equilibrium Expressions for these Heterogeneous Equilibria?

As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.How does the concentration of CO2 differ in the two bell jars?

Equilibrium Calculations(2)Calculating Equilibrium Concentrations [ ]e when Kc & [ ]i are known(1)Calculating Kc from experimental data: [ ]i & [ ]eICE Tables:InitialChangeEquilibrium

I[A]i[B]i00C- change- change+ change+ changeE[A]e[B]e[C]e[D]e

(1) Calculating Kc from experimental data: [ ]i & [ ]eA closed system initially containing 1.000 x 103 M H2 and 2.000 x 103 M I2 at 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 103 M. Calculate Kc at 448C for the reaction taking place, which is[HI] Increases by 1.87 x 10-3 M Stoichiometry tells us [H2] and [I2] decrease by half as much

[H2], M + [I2], M 2 [HI], MInitially1.000 x 10-32.000 x 10-30Change-9.35 x 10-4-9.35 x 10-4+1.87 x 10-3At Equilibrium6.5 x 10-51.065 x 10-31.87 x 10-3

and, therefore, the equilibrium constant

[H2], M + [I2], M [HI], MInitially1.000 x 10-32.000 x 10-30Change-9.35 x 10-4-9.35 x 10-4+1.87 x 10-3At Equilibrium6.5 x 10-51.065 x 10-31.87 x 10-3

The ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.@ 100CCalculating Kc from experimental data: [ ]i & [ ]e- 2x- 2x- 2x+ 2x+ x+ x+ x- x xKc

I0.00.0200C+ x- 2xE0.00140.0200-2x

Practice Problem2006B Q5{ }

(2) Calculating Equilibrium Concentrations when K & [ ]i are known(1) Calculating Kc from experimental data: [ ]i & [ ]eEquilibrium CalculationsH2(g) + I2(g) 2HI(g) 51 =

[H2], M + [I2], M [HI], MInitially1.000 x 10-32.000 x 10-30ChangeAt Equilibrium1.87 x 10-3

I0.006230.004140.0224C- x- x+ 2xE0.00632- x0.00414- x0.0224 + 2x

(2) Calculating Equilibrium Concentrations [ ]e when K & [ ]i are knownSolve for x by using the quadratic equation:

I[A]i[B]i00C- ax- bx+ cx+ dxE[A]i- ax[B]i- bxcxdx

Calculating Equilibrium Concentrations when K & [ ]i are knownProblem: for the reaction H2(g) + I2(g) 2HI(g) , K= 54.3 @ 430C. Suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium.H2(g) + I2(g) 2HI(g)Deciding if x is negligible:

I0.006230.004140.0224C- x- x+ 2xE0.00632- x0.00414- x0.0224 + 2x

Problem: for the reaction H2(g) + I2(g) 2HI(g) , K= 54.3 @ 430C. Suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium.Deciding if x is negligible:Take the largest [A]i and multiply it by 100 and also divide it by 100.Compare the answers of above to the value of KIf comparison are not significant as compared to K, then x is negligible and you can delete the x that is subtracted from [A]IIf these calc values are significant compared to K, then x is not negligible and you must use the quadratic to solve for x.

YOU MUST USE QUADRATIC!

Calculating Equilibrium Concentrations when K & [ ]i are known[H2]e = 0.00623 0.00156) = 0.00467 M[I2]e = (0.00414 0.00156) = 0.00258 M[HI]e = (0.0242 + 2 x 0.00156) = 0.0255 MFirst answer is physically impossible, since conc. of H2 and I2 would be more than original conc.[H2]i = 0.00623[I2]i = 0.00414x calc without quadratic:

E0.00632- x0.00414- x0.0224 + 2x

(2) Calculating Equilibrium Concentrations when K & [ ]i are known(1) Calculating Kc from experimental data: [ ]i & [ ]eEquilibrium CalculationsH2(g) + I2(g) 2HI(g) 51 =Deciding if x is negligible:

[H2], M + [I2], M 2 [HI], MInitially1.000 x 10-32.000 x 10-30ChangeAt Equilibrium1.87 x 10-3

I0.006230.004140.0224C- x- x+ 2xE0.00632- x0.00414- x0.0224 + 2x

Le Chteliers PrincipleIf a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.Henri Louis Le Chatelier1850-1936 (clear gas)(red-brown gas){L.C.&PressVol}{Equi.Intro1}{Equi.Intro2}The effect of (2) Pressure & VolumeThe effect of (1) Concentration(3) TemperatureKC

H = -58.0 kJ{Le Chteliers&Temp}Le Chteliers Principle(clear gas)(red-brown gas)The effect of (3) TemperatureWhy does heat favor the formation of NO2 over N2O4? {Molecular Explanation}Keq at different Temp:

+ HEAT

The Effect of Temperature Changes{CoCl}(pink)(blue)H= +Add heat (), which way will the equlibrium be shifted?+ HEAT

Briggs-Rauscher reaction {ClockReaction}IO3 + 2H2O2 + CH2(COOH)2 + H+ ICH(COOH)2 + 2O2 + 3H2O

It was not until the early 20th century that this method was developed to harness the atmospheric abundance of nitrogen to create ammonia, which can then be oxidized to make the nitrates and nitrites essential for the production of fertilizers and ammunitions. H = -92.4 kJ/n @ 250 CNH3 + HNO3 NH4NO3 (explosives & fertilizers)Le Chteliers Principle & the Haber Process

Le Chteliers Principle & the Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid

N2 (g) + 3H2 (g) 2NH3 (g)FeFeShift equilibrium to the right by increasing press & conc. of N2 + H2 and by removing NH3 from systemRemoved by liquefaction

Subsb. pt.NH3-33CN2-196CH2-253C

Increase the rate of both the forward and reverse reactions.Equilibrium is achieved faster, but the equilibrium composition remains unaltered.The Effect of Catalysts on EquilibriumHow does the addition of a catalyst affect the equilibrium between subs. A and B?catalyst

Equilibrium Constant (K) vs. Reaction Quotient (Q)To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.At Equilibrium:At Non-Equilibrium:

If Q = K, the system is at equilibrium.If Q > K, there is too much product and the equilibrium shifts to the left.If Q < K, there is too much reactant, and the equilibrium shifts to the right.If Q < K?If Q > K?

Kc, Q and Spontaneity (G)@ Equilibrium K = Q, G = 0G= -G= -Reverse Process G = +Reverse Process G = +

Free Energy (G) & Reaction Quotient (Q) under non-Standard ConditionsG implies DIFFERENT CONCENTRATIONS of both reactants products and under non-standard conditions of concentration, temperature and pressure.G = - (spontaneity), implies COMPLETE conversion of all reactants to products @ standard conditions (25C, 1 atm, 1M).

If [Reactants], then G = - If [Products], then G = + If at Equilibrium , then G = 0G = G + RT ln Q = G + RT 2.303 log Q

Non-standard conditionsNon-standard conditions

G = - RT ln K

0 = G + RT ln K= - 2.303 RT log K

At equilibrium, G = 0 and Q = K soRearranging Free Energy (G) & Equilibrium Constant (K) at EquilibriumG = G + RT ln QK = eG/RT

GKProduct formation> 1Product-favored @ Equilibrium0= 1[Products] = [Reactants] @ Equilibrium (RARE)+< 1Reactant-favored @ Equilibrium

{G, Q & T NO2 N2O4}LowT{G, Q & TNO2 N2O4}HighTFree Energy (G), the Equilibrium Constant (K),and Temperature(clear gas)(red-brown gas)H = -58.0 kJLow Temperature:High Temperature:Products onlyReactants onlyQ = KEquilibrium mixtureLow Temp.High Temp.G = - RT ln K

G= -G= +@ T , K How would T or affect K? @ T , K G= -G= +

2003B Q12 HIH2 + I2

2003B Q1Kp = Kc(RT)n

2000A Q1

2000A Q1

2004B Q1

2007B Q1

2000 1

2003 B

2004 B

2006 A

2006 (B)

2007 (B)

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