General Chemical Equilibrium, Study notes of Chemistry

Download General Chemical Equilibrium and more Study notes Chemistry in PDF only on Docsity! 2 7 General Chemical Equilibrium General Chemical Equilibrium Solving Equilibrium Problems Using the RICE-Table Method THE NATURE OF THE EQUILIBRIUM STATE Equilibrium is defined as the condition when the rate of the forward reaction equals the rate of the reverse reaction. Equilibrium reactions are reversible which is indicated by the presence of double arrows, <——, between the reactants and the products. The double arrows indicate that the reaction is proceeding in both the forward and reverse direction and once equilibrium is . oo. . 8) \— Forward established, the rate of each direction is equal (Figure 1). We é rate also say that once equilibrium is reached, it is dynamic. This 2 means reactants are forming products and products are forming 8 | Reverse Equibraim . : rate reactants simultaneously such that the net concentration of each Forward rate = Reverse rate remains constant. Notice, this does NOT say the concentrations of the products and reactants are equal! That is a very common tne misconception. Fig 1 Examine Figure 2 which represents the following generalized chemical reaction: o 5 [C] and [D] A+B @=C+D £ g Notice that equilibrium is reached once the concentrations of the 8 [A] and [B] reactants and products remain constant, but again, not necessarily t equal. Equilibrium is Time —— _ established What you should be able to interpret from the graphs in Figures | & 2: Fig. 2 1. Equilibrium is established as soon as the curves become “flat” on either graph. This indicates a constant value has been reached for the y-axis variable. e For Figure 1, the rates for the forward and reverse reactions converge and the rates are equal. e For Figure 2, the concentrations decrease for the reactants and increase for the products until a constant value is achieved, not necessarily an equal value. The time at which equilibrium is established is denoted as tequitibrium OF te. 2. Equilibrium is not necessarily established at some midpoint on the graph. e For Figure 1, notice that the two rates did not converge at the midpoint between the two original values. The forward reaction has a much larger rate constant than the reverse reaction so it “pushes” the equilibrium position down on the graph. e For Figure 2, the time at which equilibrium is established can occur anywhere along the x-axis. 588 Laying the Foundation in Chemistry General Chemical Equilibrium 2 7 THE EQUILIBRIUM POSITION Whether the reaction lies far to the right [products favored] or to the left [reactants favored] depends on three main factors: e = The initial concentrations of the reactants, the products, or both. The more molecules present, the more collisions that occur and the faster the reaction. e The relative energies of reactants and products. Nature goes to minimum energy, always. e The degree of organization of reactants and products. Nature goes to maximum disorder, always. THE EQUILIBRIUM EXPRESSION The first step in solving an equilibrium problem is to write the equilibrium constant expression. For the general reaction aA + bB @—cC+dD The equilibrium constant expression, K, (a.k.a. Law of Mass Action expression) is written as follows: _[ct or [aT [aT [ ] indicates concentration in molarity, M, (moles/liter) eK, indicates that all of the quantities in the problem are given in molarity. This applies to reactions in aqueous solution and sometimes gases. eK, indicates that all of the products and reactants are gases and the “p” is for partial pressures expressed in a pressure unit such as atmospheres. “K” values are always written without units. Pure solids do not appear in the expression Pure liquids do not appear in the expression Water does not appear in the expression since it is a pure liquid. Pure solids and liquids, including water, do not appear in the equilibrium expression since their concentrations do not appreciably change. In other words, they are so very concentrated compared to the reactants and products in the dilute aqueous phase. Never, ever forget that K is both a number and a relationship. You have to use both in order to be successful when working these problems. [products ] What is the significance of K? Always remember, that generally K= : [reactants] K > 1 means that the reaction favors the products at equilibrium. This means the concentration of the products is greater than the concentration of the reactants when the system reaches equilibrium. K <1 means that the reaction favors the reactants at equilibrium. This means that concentration of the reactants is greater than the concentration of the products when the system reaches equilibrium. Laying the Foundation in Chemistry 589 2 7 General Chemical Equilibrium Example 2 Consider the following reaction: Hoi) + CO2(g, Z— HO @ + COw® When H,(g) is mixed with CO2(g) at 1,000 K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H2] = [CO2] = 0.40 mol/L [H20] = 0.20 mol/L [CO] = 0.30 mol/L (a) Using the equilibrium concentrations given above, calculate the value of Ke, the equilibrium constant for the reaction. Solution: This one is an easy start. The equilibrium concentrations were given in the problem so, proceed directly to the equilibrium constant expression. (b) Determine Kp, in terms of Ke for this system. Solution: Time for the politically correct equation! K,=K(RT)™ Leatm 10 K, = 0.38 [oe .0008 2x) = 0.38 x 1.0=0.38 mol Remember, K, only equals K, when An is equal to zero since there is no change in the number of moles of gas between the products and reactants. It’s quite alright to simply state K, = K, as long as you also state that An is equal to ZERO. 592 Laying the Foundation in Chemistry Example 3 General Chemical Equilibrium 2 7 In a second experiment, the system in Example 2 is heated from 1,000 K to a higher temperature and the value of the equilibrium constant K,, changes. At this new temperature, K, = 1.5. Calculate the equilibrium concentrations for all of the reactants and products involved in the reaction if the initial concentrations of hydrogen and carbon dioxide are both equal to 0.50M. Solution: Construct a RICE table and insert the initial concentrations. R: Ho@ + — CO2() — I: 0.50 M 0.50M C: —x —x E: 0.50—x 0.50-x H,0 + co 20 (8) (g) Note: 6p 4 «CP = «RE o o for each reactant and product in * ** | the RICE table. 0 +x O+x Plug the new equilibrium values into the K expression and set it equal to the value of K given at this new temperature: [xIE] [0.50- x][0.50- x] vis _ [H,0][CO] _ [H, ][CO, ] [0.50 x = 0.50-x | x =1.2(0.50-x) 2.2x = 0.60 x = [0.27] =[H, and 2]=[co,]= 0] o]=[c 0.50-x = 0.50-0.27 =[0.23] Laying the Foundation in Chemistry [x} [0.50 - = 15 xf 593 2 7 General Chemical Equilibrium Example 4 K, for the decomposition of nitrogen monoxide is 2.5 x 10° at 1,700 K. 2 NO(g) <= Na(g) + O2(g) A.0.015 mol sample of nitrogen monoxide is placed in a 12.0-L flask and heated to 1,700 K. Calculate the equilibrium concentrations of all reactants and products once equilibrium has been reached. (a) R: 2 NO(g) = Ng) + Ong) 0.015 mol _ 9 go125 0 0 12.0L C: -2x +x +x E: 0.00125 -—2x x x xc UNO igs IR] [] > [No] [0.00125-2x] [0.00125-2x] 2.5x10° (0.00125 —2x)= x 3.125x10* —5.0x10°x =x" set the equation equal to zero with the value for x° being positive 0=x° +5.0x10°x —3.125x10 now you must use the quadratic formula to solve for x. It is suggested that you use a graphing calculator with a program that solves the quadratic formula. You will get two roots, usually one will be negative, which is ridiculous since x represents a concentration. Regardless, only one of the roots will “make sense” when inserted back into your RICE table. x =[N, -[0,]=1.53 x10 Mel [NO]=0.00125 —2x = 0.00122 =1.2x10 mel 504 Laying the Foundation in Chemistry General Chemical Equilibrium 2 7 (b) What is K for the following reaction that occurs at 125°C? 2CH3CO2C2H (ag) + 2H20 (¢) G— 2CH3COOH (ag) + 2C2HsOH faq) The equilibrium constant for the following reaction Ha(g) + L(g) ——= 2HI(g) is determined to be 57.85 at 450°C. (a) If 1.50 mol of each reactant is placed in a 2.00-L flask at 450°C what are the concentrations of Hp», Lb, and HI when equilibrium has been achieved? (b) What is K, for this reaction at 450°C? Ammonium hydrogen sulfide decomposes on heating. NHgHS(e) ——=. HaS¢e) + NH 3g) (a) If Kp is 0.11 at 25° C when the partial pressures are expressed in atmospheres, what is the total pressure in the flask at equilibrium? (b) What is K, for this reaction at 25°C? The following equilibrium constants are given at 500 K: H2(g) + Br(g) ——= 2 HBr(g) Ky =7.9 x 10"! H.(g) <= 2H(g) K, =4.8 x 10 Bro(g) <—= 2Br(g) K, =2.2 x 107° (a) Calculate K, for the reaction of H atoms and Br atoms to give HBr at 500 K. (b) Calculate K, for this reaction at SOOK. Laying the Foundation in Chemistry 597 2 7 General Chemical Equilibrium 7. The equilibrium constant for the dissociation of iodine molecules to iodine atoms is 3.76 x 10° at 1000 K. 1(g) = 2Kg) (a) Suppose 0.150 mole of l is placed in a 15.5-L flask at 1000K. What are the concentrations of I, and I when the system comes to equilibrium? (b) What is K, for this reaction at 1000 K? 8. K, for the decomposition of ammonium hydrogen sulfide is 1.8 x 10 at 15°C. NH,HS(s) —— NH3x(g) + H2S(g) (a) When 5.00 grams of the pure salt decomposes in a sealed 3.0-L flask at 15°C, what are the equilibrium concentrations of NH3 and H2S? (b) What are the equilibrium concentrations at 15°C if 10.0 grams of the pure salt decomposes in the sealed flask? 9. Hemoglobin (Hb) can form a complex with both O2 and CO. For the reaction HbO2(aq) + CO(g) <— HbCO (aq) + Or(g) at body temperature, K is about 2.0 x 10°. If the ratio [HbCO] H 2 comes close to 1.0, death is probable. (a) What partial pressure of CO in the air is likely to be fatal? Assume the partial pressure of O2 is 0.20 atm. (b) What is K for the reverse reaction? 598 Laying the Foundation in Chemistry General Chemical Equilibrium 2 7 10. Lexan is a plastic used to make compact disks, eyeglass lenses, and bullet-proof glass. One of the compounds used to make Lexan is phosgene (COClz), an extremely poisonous gas. Phosgene decomposes by the reaction COCl(g) <—= CO(g) + Ch(g) for which K, = 7.2 x 10"! at 80°C. a pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium If hi initial f 1.0 di culate th ilibri pressures of all species. (b) What is Ke for this reaction at 80°C? Laying the Foundation in Chemistry 599