Chemical Equilibrium. Chapter 17: Keeping the chemical themes straight. Chapter 17 Homework Problems. N2O4 (g) 2NO2 (g)

Transcription

1 Chapter 17: Chemical Equilibrium Equilibrium: The Extent of Chemical Reactions 17.1 The Dynamic Nature of the Equilibrium State 17. The Reaction Quotient and the Equilibrium Constant 17.3 Expressing Equilibria with Pressure Terms: Relation between Kc and Kp 17.4 Reaction Direction: Comparing Q and K 17.5 How to Solve Equilibrium Problems 17.6 Reaction Conditions and the Equilibrium State: LeChatelier s Principle Chapter 17 Homework Problems Keeping the chemical themes straight Problems have been updated for the Principles of Chemistry, Silberberg, nd Edition Textbook. Kinetic theory tells us how fast or slow a chemical reaction will occurs (reaction rates). 1,,3,9,13,17,19,3,5,8,30,3,35,39,4,44,46, 50,51,54,59,64,66,69,7,80,81,86,89,90. Thermodynamics tells us about the energetics of a process (enthalpy and internal energy). Equilibrium theory tells us the extent that a chemical reaction will occur (must be experimentally determined..i.e. pre-determined by the big guy). All chemical reactions are reversible with reactions occurring simultaneously. NO4 (g) NO (g) double-arrow symbolic representation of reversiblity 1) What happens to a reaction after it reacts? ) Can we predict the concentrations after the reaction stops? 3) What does it mean to react and stop reacting?

2 Chemical is a property of all reactions, and occurs when the when the rate of the forward and reverse chemical reactions are equal! N O 4 (g) Start with pure NO Start with pure N O 4 Start with a mixture Concentration k f A + B AB k NO (g) r k r Concentration Concentration Equilibrium occurs when the rate of the forward reaction and the rate of the reverse reaction are equal! k f rate forward = k f [A][B] rate reverse = k r [AB ] rate f = rate r k f [A][B] = k r [AB ] Assume a reaction that is also an elementary reaction Because the reaction are elementary we can write rate laws from balanced equation At the reactions continue but at the same rate! Time Time Time There is no change in concentration after time at k f [AB ] = K k c = r [A][B] K c is defined as the ratio of the rate constants and is tied to stoichiometry of the elementary event! Let s play with data and attempt to find the right equation by trial and error. Experiment N O 4 (g) Initial Concentrations [N O 4 ] [NO ] NO (g) Inital Ratio (Q) [NO ] [N O 4 ] Equilibrium Concentration [N O 4 ] eq [NO ] eq CONSTANT! Equilibrium Ratio (K) [NO ] eq [N O 4 ] eq NOTICE: By taking the ration of [products] to [reactants] raised to their stoichiometric factors we get a constant number. N O 4 (g) NO (g) K c = [NO ] [N O 4 ] x ! 9.4x x x x CONSTANT! The constant, Kc is a unitless number that summarizes the extent of a chemical reaction. It is related to the concentrations of reactants and products. Kc gives us information on whether a reaction is product-favored or reactant-favored or both (extent) aa + bb <==> cc + dd N O 4 (g) Unitless Equilibrium Constant NO (g) K c = [NO ] [N O 4 ] Balanced chemical equation [Products] raised to stoichiometric coefficient [Reactants] raised to stoichiometric coefficient Kc is temperature dependent and does not depend on the initial concentrations of either products or reactants. VERY SMALL Equilibrium Constant Reactant Favored Kc = [C] c [D] d [A] a [B] b Kc Both reactants and products [Products] raised to stoichiometric coefficient [Reactants] raised to stoichiometric coefficient VERY LARGE Product-favored

3 A large Kc means the reaction favors the products; a small Kc means the reaction favors the reactants (no reaction); intermediate means both are reactants and products are present. Kc is a experimentally derived number that characterizes the extent of a reaction under question. Here are some example observations! small K intermediate K large K Pure solids and pure liquids are excluded from the constant expression. C(s) + H O(g) <==> CO(g) + H (g) Solids and pure liquid solvents do not change concentration (Just as density of a material does not change whether you have a little or a lot of the substance) [CO][H K c = ] [H O] = P CO P H (RT) (-1) P H O T = 650 C Same [CO] Same Kc CH 3 COOH(aq) + H O(l) K c = [CH 3 COO- ][H 3 O + ] [CH 3 COOH] CH 3 COO - (aq) + H 3 O + (aq) Different CaO and CaCO3 We distinguish two types of expressions: homogeneous and hetereogeneous. 1. Homogenous ---reactants and products are have the same phase. (i.e. all liquids, or all gases). All reactants and products are included in the expression. 3ClO - (aq) ClO3(aq) + Cl - (aq) K c = [ClO 3] [Cl ] [ClO 3 ] 3 CH 4 (g) + HS (g). Heterogenous --reactants and products have different phases...solids reacting with gases or liquids. We exclude solids and pure liquids from the expression. CaCO 3 (s) CS (g) + 4H (g) CaO(s) + CO (g) We can use molarity or partial pressures. Pure liquids and solids are excluded from K c and K p Determine whether the following reactions are homogeneous or heterogeneous, and write equation for these reactions: K c = [CO] (a) CO (g) + C(s) <==> CO(g) [CO ] (b) Hg(l) + Hg + (aq) <==> Hg + (aq) K c = [Hg+ ] [Hg + ] (c) Fe(s) + 3H O(g) <==> Fe O 3 (s) + 3 H (g) (d) H O(l) <==> H O(g) (e) N (g) + 3 H (g) <==> NH 3 (g) (f) NH 3 (g) <==> N (g) + 3 H (g) K c = [H ]3 [H O] 3 K c = [NH 3 ] [H ] 3 [N ] Suppose for reaction a Kp > what does this imply? Suppose for reaction e Kc < what does this imply

4 The constant for reaction involving only gases can be written two ways. Kc is used for solutions, Kp is used for gas phase. They are related! N(g) + 3H(g) <==> NH3(g) K c = [NH 3 ] [N ][[H ] 3 K p = (P NH 3 ) P N (P H ) 3 Using Molarity Values Using partial pressure in atmospheres, not torr or mmhg! Kc = Kp RT -!n or Kp = Kc RT!n Kc is related to Kp through the ideal gas law PV = nrt K c = [C]c [D] d [A] a [B] b [C] = n V = P C RT aa + bb <==> cc + dd [D] = n V = P D RT K c = [C]c [D] d [A] a [B] b = [P C/RT] c [P D /RT] d [P A /RT] a [P B /RT] b Kc = Kp RT -[(c+d)-(a+b)] = Kp RT -!n Rearranging: PV = nrt and substituting for [A], [B], [C], [D] in K c K c = (P C) c (P D ) d (P A ) a (P B ) b (RT)a+b (RT) c+d = (P C) c (P D ) d (P A ) a (P B ) b (RT)(a+b) (c+d) regrouping n = (total moles of gaseous products) (total moles of gaseous reactants) Consider the following at 95 K: NH 4 HS (s) NH 3 (g) + H S (g) The partial pressure of each gas at is 0.65 atm. Calculate K p and K c for the reaction? Consider the following at 95 K: NH 4 HS (s) NH 3 (g) + H S (g) The partial pressure of each gas at is 0.65 atm. Calculate K p and K c for the reaction? K p = P NH3 P H S = 0.65 x 0.65 = Write the K p expression. K c = K p (RT) -!n Remember this equation!!n = total mol product - total mol reactant!n = 0 = T = 95 K R = K c = x (0.081 x 95) - = 1.0 x 10-4 Plug and chug The concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl (g) at 74 C are [CO] = 0.01 M, [Cl ] = M, and [COCl ] = 0.14 M. Calculate the constants K c and K p. CO (g) + Cl (g) K c = [COCl ] [CO][Cl ] = COCl (g) x = 16 R = K p = K c (RT)!n T = = 347 K!n = 1 = -1 The reaction quotient Q is used to predict the direction of a chemical reaction. N O 4 (g) NO (g) Q c = [NO ] [N O 4 ] If Q c > K c reaction proceeds from right to left If Q c = K c the system is at If Q c < K c reaction proceeds from left to right At K p = 16 x (0.081 x 347) -1 = 7.6

5 A reaction will start out at non- conditions and move towards until the reaction concentrations satisfy the number Kc (the God number). N O 4 (g) Qc Not at Kc At NO (g) Q c = [NO ] [N O 4 ] Qc uses non- concentrations Kc uses concentrations Q c is calculated by substituting the initial concentrations of the reactants and products into the constant (K c ) expression and comparing it to the value of Kc. If Q c > K c reaction proceeds from right to left If Q c = K c the system is at If Q c < K c reaction proceeds from left to right N O 4 (g) NO (g) Q c = [NO ] [N O 4 ] At Write the reaction quotient, Q c, for each of the following reactions: (a) The decomposition of dinitrogen pentoxide, N O 5 (g) NO (g) + O (g) (b) The combustion of propane gas, C 3 H 8 (g) + O (g) CO (g) + H O(g) Write the reaction quotient, Q c, for each of the following reactions: (a) The decomposition of dinitrogen pentoxide, N O 5 (g) NO (g) + O (g) (b) The combustion of propane gas, C 3 H 8 (g) + O (g) CO (g) + H O(g) SOLUTION: (a) N O 5 (g) 4NO (g) + O (g) Q c = [NO ] 4 [O ] [N O 5 ] (b) C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(g) Q c = [CO ] 3 [H O] 4 [C 3 H 8 ][O ] 5 Q c is the same as K c except Q is not at Using Qc to see if a reaction is at You are given a mixture containing 1.57 mol N, 1.9 mol H, and 8.13 mol NH 3 in a 0.0 L vessel at 500 K. The constant K c for the production of NH3 from N and H = 170 for the reaction. Is reaction mixture at? If not, what direction will the reaction occur? N + 3 H NH3 Given: Kc = 170 Initial [N ] = 1.57 mol/0.0 L = M; Initial [H ] t = 1.9 mol/0.0 L = M Initial [NH 3 ] t = 8.13mol/0.0 L = M Using Qc to see if a reaction is at 1. Balance the chemical equation N (g) + 3 H (g) NH3(g). Write the expression for the reaction K c = [NH 3 ] /[N ][H ] 3 3. Calculate Q c by substituting initial concentrations Q c = [NH 3 ] t /[N ] t [H ] t 3 = (0.46) /(0.0785)(0.096) 3 = Compare Q c/p to K c. Is Q > K or is Q < K? Q c > K c (370 >> 170) 5. Determine which way the reaction must go to attain. therefore the reaction mixture is not at and the reaction will proceed from right to left to reduce Q to 170.

6 Comparing Q and K To Determine Rxt Direction For the reaction N O 4 (g) NO (g), K c = 0.1 at 100 o C. At a point during the reaction, [N O 4 ] = 0.1 M and [NO ] = 0.55 M. Has the reaction reached? If not, in which direction is it progressing? Write an expression for Q c, substitute with the values given, and compare the Q c with the given K c. Comparing Q and K To Determine Reaction Direction For the reaction N O 4 (g) NO (g), K c = 0.1 at 100 o C. At a point during the reaction, [N O 4 ] = 0.1 M and [NO ] = 0.55 M. Has the reaction reached? If not, in which direction is it progressing? Q c = [NO ] [N O 4 ] = (0.55) =.5 > 0.1 (0.1) Q c > K c, therefore the reaction is not at and will proceed from right to left, from products to reactants, until Q c = K c. The constant K p for the reaction NO (g) NO (g) + O (g) is 158 at 1000K. What is the pressure of O if the P NO = atm and P NO = 0.70 atm? K p = P NO P O P NO Sample Converting Integrative Between Problem K c and K p A chemical engineer injects limestone (CaCO 3 ) into the hot flue gas of a coal-burning power plant for form lime (CaO), which scrubs SO from the gas and forms gypsum. Find K c for the following reaction, if CO pressure is in atmospheres. CaCO 3 (s) CaO(s) + CO (g) K p =.1x10-4 (at 1000K) P P NO O = K p P NO P O = 158 x (0.400) /(0.70) = 347 atm Sample Converting Integrative Between Problem K c and K p A chemical engineer injects limestone (CaCO 3 ) into the hot flue gas of a coal-burning power plant for form lime (CaO), which scrubs SO from the gas and forms gypsum. Find K c for the following reaction, if CO pressure is in atmospheres. CaCO 3 (s) CaO(s) + CO (g) K p =.1x10-4 (at 1000K) Computing K From Equilibrium Data Sample Integrative Problem Methane (CH 4 ) reacts with hydrogen sulfide to yield H and carbon disulfide, a solvent used in manufacturing. What is the value of K p at 1000 K if the partial pressures in an mixture at 1000 K are 0.0 atm of CH 4, 0.5 atm of H S, 0.5 atm of CS, and 0.10 atm of H? We know K p and can calculate K c after finding!n gas. R = L*atm/mol*K. SOLUTION:!n gas = 1-0 since there is only a gaseous product and no gaseous reactants. K p = K c (RT)!n K c = K p /(RT)!n = (.1x10-4 )(0.081 x 1000) -1 =.6x10-6

7 Computing K From Equilibrium Data Sample Integrative Problem Methane (CH 4 ) reacts with hydrogen sulfide to yield H and carbon disulfide, a solvent used in manufacturing. What is the value of K p at 1000 K if the partial pressures in an mixture at 1000 K are 0.0 atm of CH 4, 0.5 atm of H S, 0.5 atm of CS, and 0.10 atm of H? Balance equation Equilibrium laws can be reversed, scaled and added. 1. Kc of a reaction in the reverse direction is the reciprocal of the Kc in the forward reaction. N O 4 (g) NO (g) NO (g) N O 4 (g) K eq = [N O 4 ] [NO ] = 1 K eq = Products raised to stoichiometric coeff divided by reactants raised to stoich coeff!. Kc for a reaction multiplied by a constant is the constant raised to the power of the constant. K p = (0.5)(0.10) 4 /(0.0)(0.5) = 4.! 10-3 Plug in the numbers in ATM! N O 4 (g) NO (g) Now multiply the equation by K c is squared! Note that partial pressure must be in units of atmosphere! N O 4 (g) 4 NO (g) 3. Equilibrium If a reaction can in Multiple be expressed Reactions as the sum of two or more reactions, Kc for the overall reaction is given by the product of the constants of the individual reactions. A + B C + D A + B C + D E + F E + F K c K c K c K c `= [C][D] [A][B] K c `` = [E][F] [C][D] Summary What happens to Kc or Kp when we: Change the Direction of the Reaction. Multiply Stoichiometric Coefficients by a Constant 3. Sum Chemical Reactions (Hess s Law) aa + bb cc + dd K c = [C]c [D] d [A] a [B] b cc + dd aa + bb K = 1/K c K c = Kc K c = [E][F ] [A][B] = [E][F ] [C][D] [C][D] [A][B]. n aa + bb cc + dd K c = (K c ) n 3. For a sequence of equilibria, K overall = K 1 x K x K 3 x! Sample At 1565 K Integrative suppose we Problem have the following elementary reactions giving a net reation (3). What is the constant for the overall reaction? Sample At 1565 K Integrative suppose we Problem have the following elementary reactions giving a net reation (3). What is the constant for the overall reaction? CO(g) <==> CO(g) + O(g) HO(g) <==> H(g) + O(g) CO(g) + H <==> CO(g) + HO (g) K = 1.3 X CO(g) <==> CO(g) + O(g) HO(g) <==> H(g) + O(g) CO(g) <==> CO(g) + O(g) H(g) + O(g) <==> HO(g) K = Reverse ==> Take Inverse CO(g) + H <==> CO(g) + HO (g) K 3 = K1 X K = Divide equation by ============> Square root of K 3 CO(g) + H <==> CO(g) + HO (g) K 3 = = 11.1

8 Understanding reactions involving N and O, the most abundant gases in air, is essential for solving problems dealing with atmospheric pollution. Here is a reaction sequence between N and O to form nitrogen dioxide, a toxic pollutant that contributes to photochemical smog. (1) N (g) + O (g) NO(g) K c1 = 4.3 x 10-5 () NO(g) + O (g) NO (g) K c = 6.4 x 10 9 (a) Show that the Q c for the overall reaction sequence is the same as the product of the Q c s of the individual reactions. (b) Calculate the K c for the overall reaction. For the ammonia formation reaction the constant, K c, is.4x10-3 at 1000K. If we change the coefficients of the equation, which we ll call the reference (ref) equation, what are the values of K c for the following balanced equations? SOLUTION: N (g) + 3H (g) (a) 1/3N (g) + H (g) (b) NH 3 (g) /3NH 3 (g) 1/N (g) + 3/H (g) NH 3 (g) (a) The reference equation is multiplied by 1/3, so K c(ref) will be to the 1/3 power. K c = [K c(ref) ] 1/3 = (.4x10-3 ) 1/3 = 0.13 (b) The reference equation is reversed and halved, so K c(ref) is to the -1/ power. K c = [K c(ref) ] -1/ = (.4x10-3 ) -1/ = 0. Why is K Unitless? The real scoop: units of constants Equilibrium constants are really defined in terms of activity, not concentration. becomes: Dilute regime where concentration works fine in Kc Must use activities and activity coefficients as Kc is nonlinear Activity is unitless, so K is unitless. Review and Perspective in Equilibrium How Do We Use The Equilibrium Constant? Kc = [C]c [D] d [A] a [B] b 1. Write the Equilibrium Expression, Kc or Kp from equation.. Hess s Law with K 3. Predict The Direction of the Reaction: Qc vs Kc OR Kp = (Pc)c (Pd) d (Pa) a (Pb) b 3. Using Equilibrium Values, calculate Kc or one unknown 4. Given K, determining the concentrations The Van t Hoff equation relates the constant K1 at T1 to the equilbrium constant K at T. ln K = H rxn 1 1 K 1 R T T 1 R = J/mol K!H = enthalpy (watch units) K is a function of temperature! Any K needs a temperature specified.

9 The Arrhenius, Clausius-Claperyon and the Van Hoff t equations take similar mathematical forms. Stearic acid, nature's most common fatty acid, dimerizes when dissolved in hexane: ln P = H vap 1 1 P 1 R T T 1 ln k = E a 1 1 k 1 R T T 1 Clausius-Claperyon: vapor pressure and temperature Arrhenius: rate constant and temperature C 17 H 35 COOH <=> (C 17 H 35 COOH) ;!H rxn = -17 kj The constant for this reaction at 8 C is 900. Estimate the constant at 38 C. ln K = H rxn 1 1 K 1 R T T 1 ln K = H rxn 1 1 K 1 R T T 1 Van Hoff t: constant and temperature R = universal gas constant = J/mol K K 1 is the constant at T1 and K is the constant at T 17.5 Solving Equilibrium Problems Two basic kinds of chemical problems 1. Equilibrium quantities are given (concentrations or partial pressures) and we solve for Kc. This is easy--plug and chug.. More than one unknown. We use ICE table and algebra to calculate either Kc from initial quantities or we calculate concentrations given Kc. In either case we can use a book-keeping technique Initial, Change Equilibrium Method Let s try it in an example problem! RULES FOR ICE BOOK-KEEPING 1. Alway write a balanced chemical equation.. Make an ICE (initial, change, ) Table 3. Use as the unknown variable x for one substance and use the stoichiometry to determine relative concentrations of other reactants and products. Think, before choosing the x-variable. 6. Substitute the concentrations into the expression from the balanced equation. 7. Solve the equation for x using any math methods at your disposal (1. simple squares,. quadratic equation, 3. Simplifying assumption. 8. Answer the questions, remember significant figures. Determining concentrations from K c EASIEST TYPE PLUG AND CHUG A chemical engineer mixes gaseous CH 4 and H O in a 0.3 L flask at 100 K. At, the flask contains 0.6 mol of CO, mol of H, and mol of CH 4. What is the [H O] at? A Handbook states that K c = 0.6 for this reaction is: CH 4 (g) + H O(g) CO(g) + 3H (g) K c = 0.6 Set up a reaction table, write Kc CH 4 (g) + H O(g) concentration (M) initial change CH 4 (g) + H O(g)? = 0.13 M CO(g) + 3H (g)??? mol/0.3 L??? 0.6mol 0.3 L = 0.81 M K c = [CO][H ] 3 [CH 4 ][H O] CO(g) + 3H (g)?? mol 0.3 L = 0.8 M [H O] eq = [CO] eq[h ] eq 3 [CH 4 ] eq K c = (0.81)(0.8]3 (0.13)(0.6] = 0.53 M

10 Determining concentrations from initial concentrations and K c (Perfect Square) Fuel engineers use the extent of the change from CO and H O to CO and H to regulate the proportions of synthetic fuel mixtures. If 0.50 mol of CO and 0.50 mol of H O are placed in a 15 ml flask at 900K, what is the composition of the mixture at? At 900K, K c is 1.56 for this reaction. CO(g) + H O(g) CO (g) + H (g) K c = 1.56 Initial concentrations must be calculated as M, we have from the data given [CO] = [H O] = 0.50/0.15L =.00M. concentration CO(g) + H O(g) CO (g) + H (g) initial change -x -x +x +x.00 -x.00 -x x x K c = [CO ][H ] [CO][H O] = x (.00 x)(.00 x) = x (.00 x) = ) Balance equation and write expression ) Set up ICE table, 3) find the concentrations of all species at initial conditions or in the problem 4) use algebra to determine concentrations and then substitute into a K c expression. x.00 x = 1.56 = ±1.5 x = 1.5(.00 x) = x.5x =.50 x = 1.11M = [CO ] = [H ] It s a perfect square and easy to solve! We toss out the negative result! [CO] = [H O] =.00 - x = 0.89 M Same problem...but not a perfect square! Fuel engineers use the extent of the change from CO and H O to CO and H to regulate the proportions of synthetic fuel mixtures. If 0.50 mol of CO and 0.15 mol of H O are placed in a 15 ml flask at 900K, what is the composition of the mixture at? At 900K, K c is 1.56 for this reaction. CO(g) + H O(g) CO (g) + H (g) K c = )Balance equation, )write expression, 3) set up ICE table, 4)find the concentrations of all species at initial conditions or in the problem 5)use algebra to determine concentrations and then substitute into a K c expression. Initial concentrations must be calculated as M, we have from the data given [CO] = [H O] = 0.50/0.15L =.00M. concentration CO(g) + H O(g) CO (g) + H (g) initial change -x -x +x +x.00 -x x x x K c = [CO ][H ] [CO][H O] = x (.00 x)(1.00 x) = x x 3.00x +.00 = 1.56 x = 1.56(x 3.00x +.00) x = 1.56x 4.68x + 3.1) 0.56x x = 0 ax + bx + c =0 x = -b ± " b 4ac a K c = [CO ][H ] [CO][H O] = x (.00 x)(1.00 x) = x x 3.00x +.00 = 1.56 x = 1.56(x 3.00x +.00) x = 1.56x 4.68x + 3.1) 0.56x x = 0 ax + bx + c =0 x = - (-4.68 ± "(-4.68) 4(0.56)(3.1) x = (0.56) x = 7.6M and x = 0.73M -b ± " b 4ac a x = 7.6M makes.00 - x < 0 and makes no sense-toss it x = 0.73 = [CO] = [H] and [CO] = [HO] = Calculating K c from concentration data In a study of hydrogen halide decomposition, a researcher fills an evacuated.00 L flask with 0.00 mol of HI gas and allows the reaction to proceed at 453 C. At, it is found that the concentration of [HI] = M. What is the constant Kc? HI(g) H (g) + I (g) Find the molar concentration of the starting material (in this case [HI] and then use algebra and the balanced chemical equation to determine the amount of reactants and products at.

11 1. Write down balanced equation and K c HI(g) H (g) + I (g) K c = [H ] [I ] [HI]. Set up the ICE table (initial, change, ) Molarity HI(g) H (g) + I (g) initial.00 mol/.00 L change -x x x x x x The problem said that: [HI] eq = M therefore: x = [HI] eq = M Therefore: x = M Determining concentrations from K c If 1.0 mol H is allowed to react with 1.0 mol of Cl to form hydrogen chloride in a 10.0 L vessel at 700 K, what are concentrations of H, Br and HCl at? What is the composition of mixture if K c for the reaction of H and Cl at 700 K is 57.0? 0. What does the problem ask for ( values) 1. Write a balanced equation:. What information is provided? H + Cl HCl Initial concentrations [H ] = [Cl ] = 0.10 M K c = [H ] [I ] [HI] = [0.011][0.011] [0.078] = 0.00 Determining concentrations from K c, cont t 0. What does the problem ask for ( values) 1. Write a balanced equation:. What information is provided? H + I HI Initial [H ] = [I ] = 0.10 M H + I HI Initial (M) Change (M) -x -x +x Equilibrium (M) (0.10 x) (0.10 x) x K c = [HI] [H ][I ] = 57.0 K c = K c = K c = Text [HI] [H ][I ] = 57.0 [x] [0.10 x][0.10 x] = K c = 57.0 = [HI] /[H ][I ] = (x) /(0.10-x)(0.10-x) = (x/(0.10-x)) "57.0 = ± 7.55 = x/(0.10-x) [x] [0.10 x] = 57.0 = ±7.55 [x] = 57.0 [0.10 x] H + I HI Equilibrium concentrations from calculated value of x: [H ] = [I ] = 0.10 x = 0.01 M [HI] = x = ()(0.0791) = M Check result: K c = [HI] /[H ][I ] = (0.158) /(0.01)(0.01) = 57 Number of moles in mixture: moles of H = moles I = (0.01 mol/l)(10.0 L) = 0.1 mol moles of HI = (0.158 mol/l)(10.0 L) = 1.58 mol Solving for x: two solutions +7.55(0.10-x) = x x = 0.755/9.55 = M -7.55(0.10-x) = x Calculating x = concentration /5.55 = using M a simplfying Initial concentrations assumption are (no 0.10 quadratic M, so x can t equation) exceed 0.10 M The reaction of nitrogen with oxygen giving nitrogen oxide contributes to air pollution whenever a fuel is burnt at high temperature. At 1500K the K = 1.0 X Suppose a sample of air has [N] = 0.80 M, and [O] = 0.0M before a reaction occurs. Calculate the concentrations at. N(g) + O(g) <==> NO(g) Kc = 1.0 X 10-5 Set up ICE table and begin filling it in. Solving for x: two solutions +7.55(0.10-x) = x

12 1. Write down what we solve: Kc and what we know. [NO] K c = = 1.0 X 10 [N ] [O ] -5. Set up the ICE table Molarity N(g) + initial + change O(g) <=> NO(g) x -x +x x x x 3. Use expression and substitute values. [NO] [x] K c = = 1.0 X 10 = [N ] [O ] -5 [.80 -x] [.0 -x] = There are 3 cases: (1) perfect square, () simplifying assumption (both easy to solve) (3) quadratic equation (harder) 3. Use expression and substitute values. [NO] [x] = 1.0 X 10-5 = [.80 -x][.0 -x] K c = [N ] [O ] 4. We can use a simplifying assumption here: RULE: If 100 x K < [A]0 then.80 - x = X 10-5 = 10-3 < [A]0...it s ok to rid x. [NO] = 1.0 X 10-5 = [.80] K c = [N ] [O ] ( [x] [x] = [.0] [.16] x = (1.0 X 10-5 (.16) 1/ = 6.3 X [N] =.80 - x = X 10-4 = 0.80M [O] = x = X 10-4 = 0.0M [NO] = x = 6.3 X 10-4 X = 1.3 X 10-3 M ) Using the Quadratic Formula 1. Write Balanced Equation Br (g) Br (g) At 180 C the constant (K c ) for the reaction Br (g) Br (g) K c is 1.1 X 10-3 is 1.1 x If the initial concentrations are [Br ] = M and [Br] = 0.01 M, calculate the concentrations of these species at.. Write Equilibrium Expression 3. Set Up ICE Table Initial (M) Change (M) [Br] K c = = [Br ] Equilibrium (M) K c = [Br] K c = [Br ] Br (g) Br (g) [0.063] [0.01] -x +x x x ( x) x = 1.1 x 10-3 Let x be the change in concent ration of Br The expression is not a perfect square: we therefore must use quadratic equation to solve it ( x) K c = = 1.1 x x -3 4x x = x 4x x = 0 ax + bx + c =0 x = -b ± " b 4ac a x = x = ( x) K c = = 1.1 x x -3 4x x = x 4x x = 0 ax + bx + c =0 x = -b ± " b 4ac a x = x = Initial (M) Br (g) Br (g) [0.063] [0.01] Two solutions Initial (M) Br (g) Br (g) Change (M) -x +x Change (M) -x +x Equilibrium (M) x x Equilibrium (M) x x At, [Br] = x = M and M At, [Br ] = 0.06 x = M At, [Br] = x = M or M At, [Br ] = 0.06 x = M

13 Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction, COCl (g) CO(g) + Cl (g) K c = 8.3 x 10-4 = [CO] [Cl ] [COCl] COCl (g) CO(g) + Cl (g) K c = 8.3 x 10-4 (at 360 o C) COCl (g) CO(g) + Cl (g) Calculate [CO], [Cl ], and [COCl ] at when the initial amount of phosgene gas is mol in a 10.0 L flask. Initial (M) Change (M) Equilibrium (M) x +x +x x x x [COCl ] = mol/10.0 L = M K c = 8.3 x 10-4 (at 360 o C) K c = 8.3 x 10-4 = (x) (x) ( x) (continued) K c = 8.3 x 10-4 = a = 1 b = 8.3 x 10-4 c = x 10-6 (x) (x) ( x) K c = 8.3 x 10-4 ( x) = x Kc = 8.3 x x 10-4 x = x Kc = x x 10-4 x x 10-6 = 0 ax + bx + c =0 x = -b ± " b 4ac a Solving for x yields x =.5 x 10-3 M and x = 7.5 x 10-3 M. [CO] + [Cl ] [COCl ] (continued) K c = 8.3 x 10-4 = (x) (x) ( x) K c = 8.3 x 10-4 ( x) = x Kc = 8.3 x x 10-4 x = x Kc = x x 10-4 x x 10-6 = 0 ax + bx + c =0 " -b ± b 4ac x = a 4 a c = 4 x 1 x x 10-6 b = (8.3 x 10-4 ) = 6.89 X 10-7 c = x 10-6 c = x 10-6 a = 1 b = 8.3 x 10-4 c = x 10-6 Solving for x yields x =.5 x 10-3 M and x = 7.5 x 10-3 M. Predicting reaction direction and calculating concentrations The research and development unit of a chemical company is studying the reaction of CH 4 and H S, two components of natural gas. CH 4 (g) + H S(g) CS (g) + 4H (g) In one experiment, 1.00 mol of CH 4, 1.00 mol of CS,.00 mol of H S and.00 mol of H are mixed in a 50 ml vessel at 960 o C. At this temperature, K c = (a) In which direction will the reaction proceed to reach? SOLUTION: USE INITIAL CONDITIONS TO DETERMINE Q [CH 4 ] initial = 1.00 mol/0.5 L = 4.00 M [CS ] initial = 1.00 mol/0.5 L = 4.00 M [H S] initial =.00 mol/0.5 L = 8.00 M [H ] initial =.00 mol/0.5 L = 8.00 M Q c = [CS ][H ] 4 [CH 4 ][H S] = [4.0][8.0] 4 [4.0][8.0] = 64.0 (b) If [CH 4 ] = 5.56 M at, what are the concentrations of the other three substances? Q c of 64 is >> than K c = The reaction will progress to the left. Find the initial molar concentrations of all components and use these to calculate Q c. Compare Q c to K c, determine in which direction the reaction will progress, and draw up expressions for the concentrations. CH 4 (g) + H S(g) CS (g) + 4H (g)

14 SOLUTION: USE AN ICE TABLE TO SOLVE concentrations CH 4 (g) + H S(g) CS (g) + 4H (g) initial change + x + x - x - 4x x x x x At [CH 4 ] = 5.56 M, so 5.56 = x; thus, x = 1.56 M [H S] = x = 11.1 M Therefore: [H ] = x = 1.76 M Le Châtelier s Principle If a chemical system at experiences a change in concentration, temperature, volume, or partial pressure, then the shifts to counteract the imposed change according to the law of mass action. There are 4 Possible Disturbances to consider 1. Concentration changes. Pressure or Volume changes (gas) 3. Temperature Changes 4. Addition of a catalyst (not) [CS ] = x =.44 M Le Châtelier s Principle 1. Changes in Concentration aa + bb Change cc + dd Increase concentration of product(s) Decrease concentration of product(s) Increase concentration of reactant(s) Decrease concentration of reactant(s) Shifts the Equilibrium left right right left Le Chatlier Principles: Effect of a Change in Concentration To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O ; H S(g) + O (g) S(s) + H O(g) What happens to (a) [H O] if O is added? (b) [H S] if O is added? (c) [O ] if H S is removed? (d) [H S] if sulfur is added? Examine Q c to predict what will happen Qc = [C]c [D] d [A] a [B] b We can write and compare Q with K when the system is disturbed to see in which direction the reaction will progress. Q = [H O] [H S] [O ] Le Chatlier Principles: Effect of a Change in Concentration To improve air quality and obtain a useful product, chemists often remove sulfur from coal and natural gas by treating the fuel contaminant hydrogen sulfide with O ; H S(g) + O (g) S(s) + H O(g) (a) When O is added, Q decreases and the reaction progresses to the right to come back to K. So [H O] increases. (b) When O is added, Q decreases and the reaction progresses to the right to come back to K. So [H S] decreases. (c) When H S is removed, Q increases and the reaction progresses to the left to come back to K. So [O ] decreases. (d) Sulfur is not part of the Q (K) expression because it is a solid. Therefore, as long as some sulfur is present the reaction is unaffected. [H S] is unchanged.. Pressure and Volume Effects In Gases There are two cases for changes in pressure and volume: 1.!n = 0 (mol of product gases - mol of reactant gases = 0). "n # 0 Cl (g) + I (g) 1NO4 (g) ICl(g) NO (g) any change in P or V has no affect on the! If pressure is increased (volume decreased) the system will shift towards the side with fewer number of moles. If pressure is decreased (volume increased) then shifts towards side with greater moles.

15 Le Châtelier s Principle Summary: Changes in Volume and Pressure For anything change to occur in to there must be a difference in the number of moles on each side of the equation! A (g) + B (g) C (g) Change will occur A (g) + B (g) C (g) + D(g) No Change Stress or Change Increase pressure Decrease pressure Increase volume Decrease volume Shifts the Equilibrium Side with fewest moles of gas Side with most moles of gas Side with most moles of gas Side with fewest moles of gas NO4 (g) NO (g) Figure 17.8 The effect of pressure (volume) on an system. + lower P (higher V) Q p = K p = (P NO) P NO4 Q p = ( P NO) ( P NO4 ) = 4K p > K p Equilibrium will shift to the left (note it shifts to the side with fewer number of moles) more moles of gas higher P (lower V) fewer moles of gas Le Chatlierʼs Principle: Affect of volume (pressure) on the position How would you change the volume of each of the following reactions to increase the yield of products? Le Chatlierʼs Principle: Affect of volume (pressure) on the position How would you change the volume of each of the following reactions to increase the yield of products? (a) CaCO 3 (s) CaO(s) + CO (g) (a) CaCO 3 (s) CaO(s) + CO (g) (b) S(s) + 3F (g) (c) Cl (g) + I (g) SF 6 (g) ICl(g) (a) CO is the only gas present. To increase its yield, decrease the pressure which = increasing the volume. When gases are present, a change in volume will affect the concentration of the gas. If the volume decreases (pressure increases), the reaction will shift to fewer moles of gas and vice versa. (b) S(s) + 3F (g) (c) Cl (g) + I (g) SF 6 (g) (b) There are more moles of gaseous reactants than products, so decrease the volume (increase the pressure) to shift the reaction to the right. ICl(g) (c) There are an equal number of moles of gases on both sides of the reaction. Therefore, a change in volume will have no effect.

16 3. Changes in Temperature There are two cases. Think of heat as reactant or as product! 1. When H < 0 => Exothermic => think heat is a product A (g) + B (g). When H > 0 = Endothermic => think heat is reactant heat + A (g) + B (g) C (g) + D (g) Summary Changes in Temperature Change Exothermic Rx Increase temperature Decrease temperature C (g) + D (g) + heat What happens to K can be analyzed using mass-action! Kc = [C]c [D] d [A] a [B] b K decreases K increases Endothermic Rx K increases K decreases Is the amount of NO(g) formed from the given amounts of N and O greater at higher of lower temperatures? N (g) + O (g) NO (g) #H = kj TRANSLATION: "H = kj =>ENDO = HEAT IS REACTANT! Heat + N (g) + O (g) PRODUCT-FAVORED SHIFT NO (g) Increasing the temperature by adding more heat to this reaction will increase the amount of NO formed and decrease the reactant side. Kc will increase in value because NO increases! Kc = [NO] [N][O] How does an increase in temperature affect the concentration of the underlined substance and K c for the following reactions? (a) CaO(s) + H O(l) (b) CaCO 3 (s) (c) SO (g) Ca(OH) (aq)!h o = -8 kj CaO(s) + CO (g)!h o = 178 kj S(s) + O (g)!h o = 97 kj Express the heat of reaction as a reactant or a product. Then consider the increase in temperature and its effect on K c. How does an increase in temperature affect the concentration of the underlined substance and K c for the following reactions? (a) CaO(s) + H O(l) (a) CaO(s) + H O(l) (b) CaCO 3 (s) (c) SO (g) Ca(OH) (aq)!h o = - 8 kj Ca(OH) (aq) + heat Increase in temp will shift reaction left, decrease [Ca(OH) ], and decrease K c. (b) CaCO 3 (s) + heat CaO(s) + CO (g)!h o = 178 kj CaO(s) + CO (g) The reaction will shift right, resulting in an increase in [CO ] and increase in K c. (c) SO (g) + heat S(s) + O (g)!h o = 97 kj S(s) + O (g) The reaction will shift right, resulting in an decrease in [SO ] and increase in K c. For the endothermic reaction, X(g) + Y <==> XY(g) + Y(g) the following pictures depict molecular scenes reactions mixtures at various stages in the reaction. Green = X 1 3 1) If Kc = at the temperature of the reaction which scene represents the mixture at? ) What will have to happen to the other two scenes to reach (shift towards products or towards reactants?) 3) How will a rise in temperature alter the [Y]?

17 4. Adding a Catalyst to a system at does not shift the position of an system does not alter the value of Kc or Kp Le Châtelier s Principle 4. Adding a Catalyst does not change value of Kc does not shift the position of an system system will reach faster (kinetics) A catalyst lowers E a for both forward and reverse reactions and only achieves at a faster rate. Catalyzed Pathway Uncatalyzed Pathway uncatalyzed catalyzed Catalyst lowers E a for both forward and reverse reactions. Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change constant or shift. Summary of Le Châtelier s Principle Disturbance Concentration Pressure Volume Temperature Catalyst Shift Equilibrium yes yes yes yes no Change Kc or Kp no no no yes no