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Kesetimbangan KimiaCopyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.1One way reactionsReaktan bereaksi membentuk produk.Reaksi akan terus berlangsung sampai salah satu reaktan habis.Produk tidak diijinkan untuk bereaksi membetuk reaktan kembali.Reaksi lebih jauh lagi tidak terjadi.Reversible ReactionsBeberapa reaksi berlangsung dalam dua arah reaktan ke produk dan produk ke reaktan.Contoh: CuSO4 .5H2O (s) CuSO4(s) + 5H2O(g)When heated the blue crystals change to white powders which change to blue when water is added.Reaksi dalam KesetimbanganReaktan bereaksi membentuk produk.Produk bereaksi membentuk Reaktan.Finally laju reaksi maju dan balik adalah setara. Reaksi berlangsung dalam dua arah, tapi lajunya setara.There is no overall change.Reaksi sekarang berada dalam keadaan kesetimbangan dinamis.Dynamic Equilibrium -characteristicsThere is no overall change: konsentrasi dari reaktan dan produk keseluruhan tetap sama.Rates of the forward and reverse reactions are equal.Kesetimbangan dapat dicapai dari arah yang berbeda.Kesetimbangan dapat dicapai pada suatu sistem yang tertutup.Kesetimbangan KimiaThe state where the concentrations of all reactants and products remain constant with time.On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation._______________________________ is a state in which there are no observable changes as time goes by.______________ ______________ is achieved when:the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constantPhysical equilibrium Chemical equilibrium 15.17N2O4 (g) 2NO2 (g)

Start with NO2Start with N2O4Start with NO2 & N2O4equilibriumequilibriumequilibrium15.18

15.1

constantThe NO2-N2O4 System at 25C9The Law of Mass ActionForjA + kB lC + mDThe law of mass action is represented by the equilibrium expression:

Konstanta Kesetimbangan konsentrasiAdalah: hasil perkalian konsentrasi zat hasil reaksi dibagi dengan perkalian konsentrasi zat pereaksi, dan masing-masing dipangkatkan dengan koefisien reaksinyaEquilibrium Expression4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

Quotient Reaksi. . . helps to determine the direction of the move toward equilibrium.The law of mass action is applied with initial concentrations.Reaction Quotient (continued)H2(g) + F2(g) 2HF(g)

In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q.

If Q = K, then system is at equilibrium.

Quotion Reaksi (Qc) adalah persamaan yang memiliki bentuk yang sama dengan persamaan konstanta kesetimbangan tetapi nilai konsentrasinya tidak perlu pada keadaan setimbangRemember that Kc = 3.93 for this reaction at 1200 K.Thus we have that Qc > KcFor Qc to become equal to Kc the reaction must shift to the left.Qc = [CH4]i . [H2O]i[CO]i . [H2]i3= 6.25=(0.00100) . (0.00100)(0.0200) . (0.0200)3CO (g) + 3H2 (g) CH4 (g) + H2O (g)If Qc > Kc, the reaction will go leftIf Qc < Kc, the reaction will go rightIf Qc = Kc, the reaction is at equilibriumIn general:RememberThis !!Solving Equilibrium Problems1.Setimbangkan Reaksi.2.Tuliskan persamaan kesetimbangannya.3.Tuliskan konsentrasi awal.4.Hitung Q dan tentukan pergerseran kesetimbangannya.

Solving Equilibrium Problems(continued)5.Tentukan konsentrasi pada keadaan setimbang.6.Masukkan konsentrasi pada keadaan setimbang dalam persamaan kesetimbangan dan selesaikan.7.cek konsentrasi yang dihitung dengan menghitung harga K.Determining K2 NOCl(g) 2 NO(g) + Cl2(g)2.00 mol NOCl dimasukkan dalam labu1.00 L. Pada kesetimbangan didapatkan 0.66 mol/L NO. HitunglahK.PenyelesaianSet of a table of concentrations [NOCl][NO][Cl2]Before 2.00 00ChangeEquilibrium 0.66Determining K2 NOCl(g) 2 NO(g) + Cl2(g)

PenyelesaianSet of a table of concentrations [NOCl] [NO][Cl2]Before 2.00 00Change-0.66 +0.66 +0.33Equilibrium1.340.660.332 NOCl(g) 2 NO(g) + Cl2(g) [NOCl] NO][Cl2]Before2.0000Change-0.66+0.66 +0.33Equilibrium 1.34 0.66 0.33

2 NOCl(g) 2 NO(g) + Cl2(g)[NOCl][NO][Cl2]Before2.0000Change-0.66+0.66 +0.33Equilibrium 1.34 0.660.33

K v. KpFor jA + kB lC + mDKp = K(RT)DnDn = sum of coefficients of gaseous products minus sum of coefficients of gaseous reactants.Konstanta kesetimbangan tekanan (Kp)Adalah hasil perkalian tekanan parsial produk dibagi dengan hasil perkalian reaktannya, dan masing-masing dipangkatkan dengan koeffisien reaksinya Kesetimbangan HeterogenA homogeneous equilibrium is an equilibrium that involves reactants and products in a single phase only.A heterogeneous equilibrium is an equilibrium involving reactants and products in more that one phaseHomogeneous EquilibriumCH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)Kc =[H2O] = constantKc = = KcGeneral practice not to include units for the equilibrium constant.

15.227Heterogeneous Equilibria. . . are equilibria that involve more than one phase.CaCO3(s) CaO(s) + CO2(g)K = [CO2]Posisi dari suatu kesetimbangan heterogen Tidak bergantung pada jumlah padatan atau cairan murni yang ada.

PCO2= KpCaCO3 (s) CaO (s) + CO2 (g)PCO2does not depend on the amount of15.229_______________ _______________ applies to reactions in which reactants and products are in different ____________.CaCO3 (s) CaO (s) + CO2 (g)Kc =[CaO][CO2][CaCO3][CaCO3] = [CaO] =Kc = [CO2] = Kc x[CaCO3][CaO]Kp = PCO2The concentration of ________ and ______________ are not included in the expression for the equilibrium constant.15.230The Meaning of K1.Dapat meramalkan apakah suatu reaksi cenderung ke arah produk atau reaktan.For N2(g) + 3 H2(g) 2 NH3(g)

Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

The Meaning of KFor AgCl(s) Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 1.8 x 10-5Conc. of products is much less than that of reactants at equilibrium. The reaction is strongly reactant-favored.

Ag+(aq) + Cl-(aq) AgCl(s)is product-favored.Pengaruh Perubahan Dalam Sistem1.Concentration: The system will shift away from the added component.2.Temperature: K will change depending upon the temperature (treat the energy change as a reactant).3.Pressure: a. Addition of inert gas does not affect the equilibrium position.b. Decreasing the volume shifts the equilibrium toward the side with fewer moles.

Effects of Volume and Pressure ChangesIf the pressure is increased by decreasing the volume of a reaction mixture, the reaction shifts in the direction of fewer moles of gas.In General:Effect of Temperature ChangeFor an exothermic reaction (H negative), the amounts of products are decreased at equilibrium by an increase in temperature.For an endothermic reaction (H positive), the amounts of products are increased by a increase in temperature.A + B C + D + heatA + B + heat C + DThe Effect of CatalystA catalyst is a substance that increases the rate of a reaction but is not consumed by it.A catalyst has no effect on the equilibrium composition of a reaction mixture. A catalyst merely speeds up the reaction to achieve equilibrium.Le Chteliers Principle. . . if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Le Chteliers Principle Changes in ConcentrationN2 (g) + 3H2 (g) 2NH3 (g)AddNH3Equilibrium shifts _________ to offset stress

15.539Le Chteliers Principle Changes in Concentration continuedChangeShifts the EquilibriumIncrease concentration of product(s)15.5aA + bB cC + dD Add40Le Chteliers Principle Changes in Concentration continuedChangeShifts the EquilibriumDecrease concentration of product(s)15.5aA + bB cC + dD Remove41Le Chteliers Principle Changes in Concentration continuedChangeShifts the EquilibriumIncrease concentration of reactant(s)15.5aA + bB cC + dD Add42Le Chteliers Principle Changes in Concentration continuedChangeShifts the EquilibriumDecrease concentration of reactant(s)15.5aA + bB cC + dD Remove43Le Chteliers Principle Changes in Volume and PressureA (g) + B (g) C (g)ChangeShifts the Equilibrium to theIncrease pressure Side with _______ moles of gasDecrease pressureSide with _______ moles of gasDecrease volumeIncrease volumeSide with _______ moles of gasSide with _______ moles of gas15.544Le Chteliers Principle Changes in TemperatureChangeExothermic RxIncrease temperature K ________Decrease temperatureK ________Endothermic RxK ________K ________ Adding a Catalyst does not change K does not shift the position of an equilibrium system system reaches equilibrium sooner15.545

uncatalyzedcatalyzed15.546The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)Kc = [COCl2][CO][Cl2]=0.140.012 x 0.054=Kp = Kc(RT)DnDn = 1 2 = -1R = 0.0821T = 273 + 74 = 347 KKp =15.247The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO2 = 0.400 atm and PNO = 0.270 atm?

22NO2 (g) 2NO (g) + O2 (g)15.2Kp = 2PNO PO2PNO22PO2= KpPNO22PNO2PO2(0.400)2= ________ atm(0.270)2=15848Consider the following equilibrium at 295 K:

The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction?

NH4HS (s) NH3 (g) + H2S (g)Kp = PNH3H2SP= 0.265 x 0.265 = _____________Kp = Kc(RT)DnKc = Kp(RT)-DnDn = 2 0 = 2T = 295 KKc = 0.0702 x (0.0821 x 295)-2 = ______________15.249A + B C + DC + D E + FA + B E + FKc =[C][D][A][B]Kc =[E][F][C][D][E][F][A][B]Kc = Kc Kc KcKc = Kc Kc xIf a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the _________________ of the equilibrium constants of the individual reactions.

15.250N2O4 (g) 2NO2 (g)= 4.63 x 10-3K = [NO2]2[N2O4]2NO2 (g) N2O4 (g)K = [N2O4][NO2]2=1K= 216When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the ________________ of the original equilibrium constant.

15.251Writing Equilibrium Constant ExpressionsThe concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm.The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions.The equilibrium constant is a dimensionless quantity.In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.15.25215.3Chemical Kinetics and Chemical EquilibriumA + 2B AB2kfkrratef = kf [A][B]2rater = kr [AB2]Equilibriumratef = raterkf [A][B]2 = kr [AB2]kfkr[AB2][A][B]2=Kc =53Calculating Equilibrium ConcentrationsExpress the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x.Having solved for x, calculate the equilibrium concentrations of all species.15.454At 12800C the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)Br2 (g) 2Br (g)Let x be the change in concentration of Br2Initial (M)Change (M)Equilibrium (M)0.0630.012-x+2x[Br]2[Br2]Kc = Kc = (0.012 + 2x)20.063 - x=ICESolve for x15.455Kc = (0.012 + 2x)20.063 - x= 1.1 x 10-34x2 + 0.048x + 0.000144 = 0.0000693 0.0011x4x2 + 0.0491x + 0.0000747 = 0ax2 + bx + c =0-b b2 4ac 2ax = Br2 (g) 2Br (g)Initial (M)Change (M)Equilibrium (M)0.0630.012-x+2x0.063 - x0.012 + 2xx = -0.00178x = -0.0105At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 MAt equilibrium, [Br2] = 0.062 x = 0.0648 M15.456Latihan soalPada suhu 440C kesetimbangan H2(gas) + I2(gas) 2 HI (gas) mempunyai Kc 49,5. jika kesetimbangan ini mengandung H2=0.8 mol, I2=0.2 mol dalam ruang 10 L, tentukan konsentrasi tiap komponen!Pada 1C, dan tekanan 20 atm, terdapat kesetimbangan C(s) + CO2(g)2CO dan tekanan parsial CO2 sebesar 2.5 atm.Hitung Kp dan berapa persen gas CO2 bila tekanan totalnya 40 atm?