Midterm II Notes

Example 1 What is the lattice energy of KI (Hsoln =+21.5 kJ/mol) and MgCl2 (Hsoln = –150 kJ/mol)? Use Table 13.1 to determine heat of hydration. Solution: The heat of hydration for KI is –336 kJ/mol (K+) + –247 kJ/mol (I–) = –583kJ/mol. The heat of hydration for MgCl2 is –1903 kJ/mol (Mg2+) + 2(–313 kJ/mol) (Cl–) = –2529kJ/mol Using Equation 13.2, Hsoln = –U + Hhydration For KI Hsoln = –U + Hhydration +21.5 kJ/mol = –U + –583 kJ/mol 604.5 kJ/mol = –U 604 kJ/mol = U = lattice energy Since this was addition and subtraction, the fewest decimal places were used to determine significant figures. For MgCl2 Hsoln = –U + Hhydration –150 kJ/mol = –U + –2529 kJ/mol 2379 kJ/mol = –U –2379 kJ/mol = U Example 2 What is the heat of solution and the dissolution reaction if when 1.893 g lithium fluoride dissolves in 250.0 mL water, the temperature changes from 23.69 °C to 19.59 °C? Solution: The ionic compound lithium fluoride is made from Li+ and F– ions, so the reaction is LiF Li+ + F– The moles of lithium fluoride that dissolved are 1.893 g LiF 1 mol 25.94 g  =  0.07298 mol LiF The heat evolved is determined from q = (mass solution)(specific heat water)(T) q = (251.9 g)(4.184 J/g•°C)(23.69 °C – 19.59 °C) q = (251.9)(4.184)(4.10) q = 4321 J (Note: only three digits are significant because of the value of T) The heat of solution will be positive because the temperature of the solution decreased. So H  =  +q mol  =  –(4321 J) 0.07298 mol  =  +59,211 J/mol  =  +5.92 x 104 J/mol If you prefer, you can convert J to kJ +5.92 x 104 J/mol 1 kJ 1000 J  =  +59.2 kJ/mol Example 3 What is the sign on S for the following reactions? 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) NaCl(s) Na+(aq) + Cl–(aq) 1 N2(g) + 3 H2(g) 2 NH3(g) Solution: In this example a liquid and a gas become gases. There is also an increase in number of moles (27 to 34). Both of these indicate an increase in randomness or +S. One mole of solid makes 2 moles of a mixture. This is also an increase in randomness and +S. In this case the physical state of all materials is a gas, so physical state is not a criterion that can be used. However, the reaction takes 4 moles of gas and creates 2. It also goes from a mixture (two substances) to a pure substance. Therefore randomness decreases and S is negative. Example 4 Under what temperature conditions are the following reactions spontaneous? NH4Cl(s) + OH–(aq) NH3(g) + H2O(l) + Cl–(aq)        endothermic 2 NO(g) + O2(g) 2 NO2(g)       exothermic CaCO3(s) CaO(s) + O2(g)       endothermic Solution: Because the reaction is endothermic, H is positive. Entropy increases as the solid becomes a gas and a liquid, so S is positive. Since S favors a spontaneous reaction but H does not, to maximize S, this reaction is spontaneous at high temperatures. Because this reaction is exothermic, the sign on H is negative. Going from more moles of gas to fewer decreases entropy, so S is negative. The enthalpy encourages spontaneity; entropy discourages it. To minimize entropy, this reaction will be spontaneous at low temperatures. Because this reaction is endothermic, H is positive. A solid turning to a gas results in an increase in entropy, so S is positive. While S is favorable for a spontaneous reaction, H is not. To maximize the effect of S, high temperatures are required. Therefore this reaction will be spontaneous at high temperatures. Example 5 According to the label, a serving of Reduced Fat TriscuitTM crackers has 3 g fat, 22 g carbohydrate, and 3 g protein. How many Calories are in a serving? Solution: The calories due to fat are 3 g fat 38 kJ 1 g 1 Cal 4.184 kJ  =  27.25 Cal 22 g carbohydrates 17 kJ 1 g 1 Cal 4.184 kJ  =  89.39 Cal 3 g protein 17 kJ 1 g 1 Cal 4.184 kJ  =  12.19 Cal The sum is 27.25 Cal + 89.39 Cal + 12.19 Cal = 128.83 Cal NOTE Average Rates, Equation 14.6 Average rates can be determined from any two initial and final time concentration values using Equation 14.6. rate  =  [X] t  =  ([X]f – [X]i) (tf – ti)          (Equation 14.6) Positive and negative values of rate will come about naturally. (Reactants decrease in concentration; products increase.) Like with thermodynamic measurements (see chapters 11 and 13), the sign indicates direction. aA + bB cC + dD –1 a [A] t  =  –1 b [B] t  =  +1 c [C] t  =  +1 d [D] t –1 a rateA  =  –1 b rateB  =  +1 c rateC  =  +1 d rateD Example 7 For the reaction, 2 H2 + O2 2 H2O, if the rate of disappearance of hydrogen gas is –0.012 M/s, what is the rate for the disappearance of oxygen and the appearance of water? Solution: Using the equation (–1/2)(rateH2) = (–1/1)(rateO2) = (1/2)(rateH2O) so (–1/2)(–0.012 M/s) = (–rateO2) –0.0060 M/s = rate of O2 and (–1/2)(–0.012 M/s) = (1/2)(rateH2O) 0.012 M/s = rate of water Example 8 For the reaction P4 + 6 Cl2 4 PCl3, if the rate of disappearance of chlorine gas is -0.237 M/s, what is the rate of the disappearance of phosphorus and the appearance of phosphorus trichloride? Solution: Use the relationship –(1/a)(rateA) = –(1/b)(rateB) = +(1/c)(rateC) = +(1/d)(rateD) and make it specific for –(rate P4) = –(1/6)(rate Cl2) = (1/4)(rate PCl3) Since the ate given is for a reactant, it is (by definition) negative. Using the value given in the problem for the rate of Cl2 disappearance, –(rate P4) = –(1/6)(–0.237 M/s) rate P4 = –0.0395 M/s and (1/4)(rate PCl3) = –(1/6)(–0.237 M/s) rate PCl3 = (4/6)(0.237 M/s) rate PCl3 = 0.158 M/s Example 9 What is the average rate based on the following data? The concentration of reactant changes from 0.50 M to 0.01 M in 1.5 minutes. The concentration of product is 0.056 M after 75 seconds. The [X]0 = 1.00 M at t = 0 and [X] = 0.34 M at t = 50 s. Solution: The average rates are determined from initial and final concentration and time data. The tradition is to subtract final from initial value. Initial concentration = 0.50 M, final concentration = 0.01 M, so [X] = 0.01 – 0.50 = –0.49 M. The time change is 1.5 minutes (or 90 s). So the rate can be either rate  =  –0.49 M 1.5 min  =  –0.33 M/min rate  =  –0.49 M 90 s  =  –5.4 x 10–3 M/s rate = –0.49 M/1.5 min = –0.33 M/min rate = –0.49 M/90 s = 5.4 x 10–3 M/s It is implied that there was no product to start, so the initial concentration = 0 M. Thus [X] = 0.056 M. The change in time is 75 s. So rate = 7.5 x 10–4 M/s. [X]0 represents initial concentration. Therefore average rate = (0.34 – 1.00)/(50 – 0) = 0.013 M/s. Example 10 What are the units on rate constant for the following rate laws, assuming the rate is M/s? rate = k[A]3 rate = k rate = [R][S] Solution: The order of the reaction is 3. Using the equation M1–order × time–1, M1–3× s–1 or M–2•s–1. The order of the reaction is zero. One minus zero is 1, so the units are M•s–1. The overall reaction order is 2. (One in each R and S.) Using 1 – order, the result is 1–2 = –1. So the units are M–1•s–1. Example 11 What is the order of the reaction for the following rate constants? k = 7.5 s–1 k = 66.2 M–2•min–1 k = 2.53 x 102 M–1•s–1 Solution: The equation of 1 – order = the exponent on concentration is used. Molarity does not appear in the units. This implies that its exponent is zero, since anything to the zero power is 1. 1 – order = 0; so order = 1. The rate law is first order. The exponent on molarity is –2. 1 – order = –2, so order = 3. The reaction is third order. The exponent on molarity is –1. 1 – order = –1, so order = 2. This is a second–order reaction. NOTE In each of the three graphs, time is graphed on the x-axis. To test for a zero-order reaction, graph concentration of reactant on the y-axis. To test for a first-order reaction, graph the natural log of reactant concentration on the y-axis. To test for a second-order reaction, graph the reciprocal of reactant concentration on the y-axis. Remember that the linear graph is the correct one. Example 12 The reaction of A has a rate constant of 1.4 x 10–4 s–1. If 1.0 M of reactant reacts for 25 minutes, how much is left? Solution: Because this is a first-order reaction (the units on rate constant give it away), the relevant integrated rate law is ln[A] = –kt + ln[A]0 The a nswer is determined by algebraically solving this equation for the unknown value, in this case, [A]. The units on k must match the time units. It is probably easier to change the units of time than to change those of the rate constant. Therefore t  =  25 min 60 s 1 min  =  1500 s ln[A] = –(1.4 x 10–4/s)(1500 s) + ln(1.0) ln[A] = –0.21 + 0 ln[A] = –0.21 [A] = 0.81 M Example 13 If the reactant concentration for a second-order reaction decreases from 0.10 M to 0.03 M in 1.00 hour, what is the rate constant for the reaction? Solution: This is a second-order reaction (the problem says so), so the relevant integrated rate law is 1/[A] = kt + 1/[A]0 Substituting the concentration values into this equation (and keeping units), 1/(0.03 M) = kt + 1/(0.10 M) 33.3 M–1 = kt + 10 M–1 23.3 M–1 = kt The units on the rate constant depend on the value used for time. The easiest thing is to use time as hours. So 23.3 M–1 = k(1.00 hr) 23.3 M–1•hr–1 = k If you use time in units of minutes, t = 60.0 min 23.3 M–1 = k (60.0 min) 0.39 M–1•min–1 = k If you use time in units of seconds, t = 3600 s 23.3 M–1 = k(3600 s) 6.5 x 10–3 M–1•s–1 = k Example 14 How long does it take 2.00 M of reactant to decrease in concentration to 0.75 M if the rate constant is 0.67 M/min? Solution: The units of rate constant imply that this is a zero-order reaction (rate = k). Therefore, the relevant integrated rate law is [A] = –kt + [A]0 In this example, the equation is solved for time. The units of time will be minutes, which is determined from the rate constant. (0.75 M) = –(0.67 M/min)t + (2.00 M) –1.25 M = (–0.67 M/min)t 1.9 min = t Example 15 Describe the molecularity of each step in the following mechanism. CH3OH + H+ CH3OH2       (slow) CH3OH2 CH3 + H2O       (fast) CH3COOH H+ + CH3CO2–        (fast) CH3CO2– + CH3 CH3CO2CH3       (fast) Solution: The first and last steps are bimolecular (two reactants). The two middle steps are unimolecular (one reactant). Example 16 Label the following energy profile with the H, Ea, and transition state. Solution: The H is the energy difference between products and reactants. Ea is the energy difference between reactants and transition state. The transition state is the highest energy between reactants and products. Example 17 What is the activation energy for a reaction with the following temperature and rate constant data? Temperature (K) Rate constant (1/s) 200 0.236 300 0.301 400 0.340 500 0.366 600 0.385 700 0.399 800 0.409 Solution: The linear relationship between rate constant and temperature is Equation 14.22, ln k  =  –Ea R 1 T  +  ln A Therefore a line would be a graph of ln(rate constant) on the y-axis and 1/temperature on the x-axis. The data for the graph is Temperature (K) Rate constant (1/s) 1/T ln k 200 0.236 0.00500 –1.445 300 0.301 0.00333 –1.200 400 0.340 0.00250 –1.078 500 0.366 0.00200 –1.004 600 0.385 0.00167 –0.955 700 0.399 0.00143 –0.920 800 0.409 0.00125 –0.894 The graph is shown in Figure 14.6. The equation of the line is ln k  =  –147 1 T  –  0.71 The activation energy is determined from the slope of this line, –147. The relationship between slope and activation energy is slope  =  –Ea R When R is the gas constant in units of J/mol•K, the activation energy is in units of joules. So –147  =  –Ea 8.314 1222 J = Ea Example 18 What is the equilibrium constant expression (as KC) for the following reactions? CH3NH2(aq) + H2O(l) CH3NH3+(aq) + OH–(aq) H+(aq) + OH–(aq) H2O(l) Fe2S3(s) 2 Fe3+(aq) + 3 S2–(aq) C(s) + 2 F2(g) CF4(g) Hg(l) + H2S(g) HgS(s) + H2(g) Solution: "(aq)" means the solute is dissolved in the solvent water. Since water is the solvent, it is not included in the equilibrium constant expression KC  =  [CH3NH3+][OH–] [CH3NH2] The reaction occurs in an aqueous solution, so the concentration of water is included in the constant. However, having no value on the top of a fraction is not mathematically appropriate, so a 1 is used. KC  =  1 [H+][OH–] In this expression, the solid on the bottom is included in the value of K instead of the concentration part of the expression. Since it is a more aesthetic expression to not use a 1 on the bottom, the expression is KC = [Fe3+]2 [S2–]3 It is equally appropriate to use the concentration of gases as the concentration of aqueous solutions, so the equilibrium constant expression is KC  =  [CF4] [F2]2 Mercury metal is a pure liquid, so the equilibrium constant expression is KC  =  [H2] [H2S] D. Converting Between KC and KP, Equation 15.17 The relationship between KC and KP is based on the ideal gas law. Rearranging the ideal gas law gives P = MRT. Substituting that into the KC and KP expressions gives Equation 15.17: KP = KC(RT)n In this equation, n = moles of product – moles reactant. The moles of products are calculated by adding the stoichiometric coefficients of the gaseous products, including the "understood 1s." Similarly, the moles of reactant are the sum of the stoichiometric coefficients of the gaseous reactants. Since solids and liquids are not included in the equilibrium constant expression, they are not included in n. Because this equation comes from the ideal gas law, the value for R is 0.08206 L•atm/mol•K, as it is in the ideal gas law. This is consistent with using units of atmosphere in KP and units of molarity (mol/L) in KC. This value also requires that temperature be in units of Kelvin. >> Example 19 What is the value of KP for the following equations? C2H4(g) + H2(g) C2H6(g)       KC = 0.99 at 500 K 2 NOCl(g) 2 NO(g) + Cl2(g)       KC = 4.4 x 10–6 at 250 °C Solution: n = (1) – (1 + 1) = –1, so KP = KC(RT)–1 = (0.99)[(0.08206)(500)]–1 = (0.99)[41.03]–1 = 0.99/41.03 = 0.024 n = (2 + 1) – (2) = 1, so KP = KC(RT)1 = 4.4 x 10–6[(0.08206)(523)] = 1.9 x 10–4 Example 21 What is the value of KC for the following equations? Si(s) + 2 F2(g) SiF4(g)       KP = 1.4 x 1082 at 1000 K CO2(g) + 2 Cl2(g) CCl4(g) + O2(g)       KP = 6.4 x 10–18 at 550 °C Solution: n = 1 – 2 = –1. The moles of silicon does not count because it is a solid. 1.4 x 1082 = KC[(0.08206)(1000)] 1.4 x 1082/82.06 = KC 1.7 x 1080 = KC n = 2 – 3 = –1. 6.4 x 10–18 = KC[(0.08206)(823)]–1 (6.4 x 10–18)(0.08206)(823) = KC 4.3 x 10–16 = KC If the equilibrium constant of the reaction, 2 NOCl(g) 2 NO(g) + Cl2(g), is KC = 4.4 x 10–6 at 250 °C, what is the equilibrium constant of 2 NO + Cl2 2 NOCl NOCl NO + 1/2 Cl2 6 NO + 3 Cl2 6 NOCl Solution: This reaction is the reverse of the original reaction. Therefore KCa  =  1 KC  =  1 4.4 x 10–6  =  2.3 x 105 This reaction is the original reaction multiplied by 1/2. Therefore KCb = (KC)1/2 = 2.1 x 10–3 This reaction is both reversed (compared with the original) and multiplied by 3. Therefore KCc  =  1 KC3  =  1 (4.4 x 10–6)3  =  1.2 x 1016 How does an increase in pressure affect the concentration of the first reactant in the following reactions? C2H4(g) + H2(g) C2H6(g) Xe(g) + 3 F2(g) XeF6(g) C(s) + 2 F2(g) CF4(g) H2S(g) + Hg(l) HgS(s) + H2(g) Solution: Since pressure is increasing, each reaction will shift to the side with fewer moles of gas. There are 2 moles of gas on the reactant side and 1 mole of gas on the product side. The equilibrium will shift toward products (right). This will use up reactants, so the concentration (and partial pressure) of C2H4 will decrease. There are 4 moles of gas on the reactant side and 1 mole of gas on the product side, so the equilibrium will shift toward the product (right) side. Therefore the concentration of xenon will decrease. There are 2 moles of gas on the reactant side and 1 mole of gas on the product side. The equilibrium will shift toward products (right), but since carbon is a solid, its concentration will not change. There is one mole of gas on the reactant side and there is 1 mole on the product side. A change is pressure will favor neither side, so the concentration of all products and reactants will not change. • The more general case in which a second order reaction has two different reactants involves a more complicated integration. • The complexity of this situation can be alleviated by ensuring that one of the reactants is in a large excess, resulting in a very small change in the concentration of that reactant during the reaction. This formulation is called a pseudo first order reaction because the integration of the rate law is the same as for a first order reaction, although it is still a second order reaction. A. G° = –RT ln K, Equation 15.19 >> Qualitative Relationships The key to qualitative relationships is to understand what the variables represent. G° is the free energy at standard state, which is 298 K and 100 kPa (about 1 atm), so it is dependent on the type of reaction rather than on reaction conditions. Recall that a negative value of G° represents a spontaneous reaction, and a positive value is a nonspontaneous reaction. The choices for equilibrium constant are greater than or less than 1. (There are no negative values of K.) The natural log of a fraction is negative and the natural log of a number greater than 1 is positive. Therefore reactions that are spontaneous at standard state have K > 1 and reactions that are nonspontaneous at standard state have K < 1. >> Example 1 Which reactions are spontaneous at standard state? 2 NOCl 2 NO + Cl2       KC = 4.4 x 10–6 at 400 K C2H4 + H2 C2H6       KC = 0.99 at 500 K Solution: That the K values given are not at standard state is not relevant, since the equilibrium constant value is related to standard-state conditions, not to nonstandard conditions. There is a relationship for nonstandard conditions [G = G° + RT ln Q], but since the question does not refer to it, it is not needed. Since K is a fraction, G° will be positive and the reaction is nonspontaneous. Since K is a fraction, G° will be positive and the reaction is nonspontaneous. >> Quantitative Relationships The key to doing quantitative relationships is units. Since G is a measure of energy, the gas constant value used is R = 8.314 J/mol•K. To make the units match this value, temperature must be in units of Kelvin and G° must be in units of joules. Most tables (including the one in the textbook's appendix) list G° with units of kJ. Therefore these values must be converted to joules. The value of K does not have units, but usually atmospheres and molarity apply. >> Example 2 What is the G° for the following reactions? 2 NOCl 2 NO + Cl2       KC = 4.4 x 10–6 at 400 K C2H4 + H2 C2H6       KC = 0.99 at 500 K Solution: Using the equation G° = –RT ln K, and including the values given G° = –(8.314 J/mol•K)(400 K) ln(4.4 x 10–6) G° = –(8.314 J/mol•K)(400 K)(–12.33) G° = +41,004 J/mol G° = +4.10 x 104 J/mol or 41.0 kJ/mol Using the equation G° = –(8.314)(500) ln(0.99) G° = –(8.314)(500)(–0.01) G° = 41.57 J/mol Since two decimal places in the natural log leave only one significant figure, the final answer is G° = 4 x 101 J/mol or 4 x 10–2 kJ/mol >> Example 3 What is the equilibrium constant for the following reactions? C2H4 + Cl2 CH2ClCH2Cl       G° = –148 kJ/mol at 500 K PbCl2 Pb2+ + 2 Cl–       G° = +28.1 kJ/mol at 25 °C Solution: Using the equation G° = –RT ln K –148,000 = –(8.314)(500) ln K 35.6 = ln K 3 x 1015 = K If you consider the significant-figure rule described in the previous example, the reverse is that number of decimal in the log term (35.6) becomes the number of significant figures in the antilog. Using the equation G° = –RT ln K +28,100 = –(8.314)(298) ln K –11.3 = ln K 1.19 x 10–4 = K     _>>_back_to_the_Top_of_the_Page_   B. Van't Hoff Equation The van't Hoff equation is used to determine the value of the equilibrium constant with temperature changes. The equation is ln K2 K1  =  –H° R ( 1 T2  –  1 T1 ) Because the gas constant is used with an energy term (H), the value R = 8.314 J/mol•K is appropriate. Because R uses units of joules, H should also be in units of joules, so that the units will cancel. As usual, temperature must be in units of Kelvin. >> Example 4 If the equilibrium constant is 1.5 x 10–8 at 298 K for a reaction with a H° of +77.2 kJ/mol, what is the equilibrium constant at 400 K? Solution: ln K2 K1  =  –H° R ( 1 T2  –  1 T1 ) If the value of equilibrium constant is called K1, then the values substituted into the equation are K1 = 1.5 x 10–8 H° = +77,200 J/mol (converted from kJ by 77.2 kJ/mol x (1000 J/1 kJ)) R = 8.314 J/mol•K T1 = 298 K T2 = 400 K K2 = ? Substituting these values into the equation, ln K2 1.5 x 10–8  =  –77,200 8.314 ( 1 400  –  1 298 ) ln K2 1.5 x 10–8  =  –9285.5 (2.5 x 10–3 – 3.36 x 10–3) ln K2 1.5 x 10–8  =  –9285.5 (–8.56 x 10–4) ln K2 1.5 x 10–8  =  7.95 K2 1.5 x 10–8  =  2823 K2 = 4.2 x 10–4 Example 2 What is the concentration for each substance at equilibrium for the following gaseous reaction C2H4 + H2 C2H6       KC = 0.99 if the initial concentration of ethene is 0.335 M and that of hydrogen is 0.526 M? Solution: The equilibrium constant expression for this reaction is 0.99  =  [C2H6] [C2H4][H2] Since the concentration values in this equation are equilibrium concentrations, the ICE table is used to determine those values. Each substance in the reaction is a column. The "initial" line will have the [C2H4] = 0.335 M and [H2] = 0.526 M, as described in the problem.   C2H4 H2 C2H6 Initial 0.335 0.526 0 Change       Equilibrium       Since the concentration of ethane, C2H6, is not mentioned in the problem, its initial concentration is zero. Because the ethane concentration is zero, its "change" must be an increase. Negative concentrations are impossible. Since ethane is a product, all other products, if there were any, would also increase in concentration. All reactants will decrease. Using x as the amount of change, the table becomes   C2H4 H2 C2H6 Initial 0.335 0.526 0 Change –x –x +x Equilibrium       The equilibrium line is determined by adding the "initial" and "change" lines for each column.   C2H4 H2 C2H6 Initial 0.335 0.526 0 Change –x –x +x Equilibrium 0.335 – x 0.526 – x x The values from the equilibrium line are substituted into the mass action expression. 0.99  =  [C2H6] [C2H4][H2]  =  x (0.335 – x)(0.526 – x) Solve the equation for x. 0.99  =  x (0.17621 – 0.861x + x2) 0.1744479 – 0.85239x + 0.99x2 = x 0.1744479 – 1.85239x + 0.99x2 = 0 Using the quadratic equation to solve for x, x  =  1.85239 ± [(1.85239)2 – 4(0.99)(0.1744479)]1/2 2(0.99) x  =  1.85239 ± [3.43135 – 0.69081]1/2 1.98 x  =  1.85239 ± [2.73325]1/2 1.98 x  =  1.85239 ± 1.65325 1.98 x  =  0.19913 1.98  or  3.5056 1.98 x = 0.1006 or 1.7705 Determine which value for x is correct by calculating equilibrium concentrations [C2H4] = 0.335 – x = 0.335 – 0.1006 = 0.234 M or [C2H4] = 0.335 – x = 0.335 – 1.7705 = –1.435 M Since a negative concentration is impossible, the x = 1.77 value must be incorrect. So [C2H4] = 0.234 M [H2] = 0.526 – x = 0.526 – 0.101 = 0.425 M [C2H6] = 0.101 M >> Example 3 What is the equilibrium concentration of silver ion in 1.00 L of solution with 0.010 mol AgCl and 0.010 mol Cl– in a solution with the equilibrium reaction of AgCl(s) Ag+(aq) + Cl–(aq) Solution: The equilibrium constant expression for the reaction is K = [Ag+][Cl–] = 1.8 x 10–10 Recall that as a solid, AgCl is not included in the expression. Therefore it need not be included in the ICE table either. Using the ICE table to determine equilibrium concentrations,   Ag+ Cl– Initial 0 0.010 mol/1.00 L = 0.010 M Change +x +x Equilibrium x 0.010 + x Since silver ion is not discussed, its initial concentration is zero. When the reaction approaches equilibrium, it must increase in concentration. The initial concentration of chloride ion is given. Recall that M = mol/L and that calculation is shown in the "initial" box. Because it is on the same side of the reaction as silver ion, it also increases in concentration. Using the equilibrium concentrations in the equilibrium constant expression 1.8 x 10–10 = [x][0.010 + x] Solve for x 1.8 x 10–10 = 0.010x + x2 0 = x2 + 0.010x – 1.8 X 10–10 Using the quadratic equation x  =  –0.010± [(0.010)2 – 4(1)(–1.8 x 10–10)]1/2 2 x = 0 if you round x = 1.8 x 10–8 if you don't An easier alternative! The value of K as 1.8 x 10–10 implies that the amount of product at equilibrium will be quite small. If you add a small number to a large number, you get the large number back. (This doesn't work for multiplication!) In this reaction, because of the equilibrium constant, the value for x must be small. Therefore if x is added or subtracted to a number, you get that number back. Using this assumption in the equilibrium constant expression, 1.8 x 10–10 = [x][0.010 + x] becomes 1.8 x 10–10 = [x][0.010] which is a lot easier to solve 1.8 x 10–8 = x The same answer! As a rule, if the equilibrium constant is small (<10–4 is a good cutoff), this assumption is valid. Nevertheless, you should always check that the assumption is valid by comparing the answer (x) to the value it was added or subtracted from. In this example, the assumption was 0.010 + x = 0.010 Using the value for x from the calculation 0.010 + 1.8 x 10–8 = 0.010 when rounding to the appropriate number of decimal places according the the significant figure rules, the statement is true. Therefore you can skip the quadratic and keep 1.8 x 10–8 as the final answer for x. More good news: Generally, even if the assumption turns out not to be valid, this quick calculation is a good way to estimate the answer. (The smaller the value of K, the better the estimate.) You can use it to check your calculation with the quadratic. >> Example 4 What are the equilibrium concentrations of all products and reactants for the following aqueous reaction: HSO4–(aq) + H2O(l) H3O+(aq) + SO42–(aq)       K = 0.012 where the initial concentrations are [HSO4–] = 0.50 M, [H3O+] = 0.020 M, [SO42–] = 0.060 M? Solution: The mass action expression for the reaction is K  =  [H3O+][SO42–] [HSO4–] Water, as the solvent, is not included. Setting up the ICE table,   HSO4– H3O+ SO42– Initial 0.50 0.020 0.060 Change       Equilibrium       Since there is no column with a zero, Q is used to determine the reaction direction. Q  =  [H3O+][SO42–] [HSO4–]  =  (0.020)(0.060) 0.50  =  0.0024 Note that the only difference between Q and K is that Q uses the initial line of the table and K uses the equilibrium line. Since Q < K, the concentration of products must increase and the reactants decrease. Therefore the table is   HSO4– H3O+ SO42– Initial 0.50 0.020 0.060 Change –x +x +x Equilibrium 0.50 – x 0.020 + x 0.060 + x Using the values in the equilibrium constant expression K  =  0.012  =  (0.020 + x)(0.060 + x) 0.50 – x  =  0.0012 + 0.080x + x2 0.5 – x Assuming x is small probably won't work, K = 0.012. In addition, all the x's would cancel, so unfortunately, the simplifying assumption won't work. 0.0060 – 0.012x = 0.0012 + 0.080x + x2 0 = x2 + 0.092x – 0.0048 x  =  –0.092± [8.464 x 10–3 – 4(1)(–0.0048)]1/2 2 x  =  –0.092± [0.027664]1/2 2 x = 0.0372 or –0.129 Calculating equilibrium concentrations, [HSO4–] = 0.50 – 0.0372 = 0.46 M or [HSO4–] = 0.50 – (–0.129) = 0.63 M and [H3O+] = 0.020 + 0.0372 = 0.057 M or [H3O+] = 0.020 + (–0.129) = –0.109 M Since this value is impossible, [H3O+] = 0.057 M and [HSO4–] = 0.046 M are correct and [SO42–] = 0.060 + 0.0372 = 0.097 M >> Example 5 What is the concentration of H+ if the initial concentration of H2CO3 is 0.14 M in the following reaction? H2CO3 2 H+ + CO32–       K = 2.4 x 10–17 Solution: The mass action expression for this reaction is K  =  [H+]2[CO32– [H2CO3] The ICE table is   H2CO3 H+ CO32– Initial 0.14 0 0 Change –x +2x +x Equilibrium 0.14 – x 2x x Note that the change for H+ is 2x. This is determined from the stoichiometry, where there are twice as many H+ ions as carbonate ions. In the change line, use the stoichiometric coefficient times x. Substituting the equilibrium concentrations into the equilibrium constant expression 2.4 x 10–17  =  (2x)2(x) 0.14 – x Since the equilibrium constant is very small, it is reasonable to assume that x is small. Therefore 0.14 – x = 0.14. This is fortunate because there is not an easy way to solve a cubic. 2.4 x 10–17  =  4x3 0.14 3.4 x 10–18 = 4x3 8.4 x 10–19 = x3 9.4 x 10–7 = x Checking the assumption, 0.14 – 9.4 x 10–7 = 0.14. Yes, x is small! The concentration of H+ is 2x. Therefore [H+] = 2x = 2(9.4 x 10–7) = 1.9 x 10–6 M.   22