Equilibrium Continued Introduction to Acid-Base Concepts Solutions

Chem Worksheet - Week 9 - Solutions Equilibrium Continued Introduction to Acid-Base Concepts Key Questions 1. a.[ For the reaction H2(g) + I2(g) º 2 HI(g) at 425 oC, Kc = 54.8. Are the following mixtures of H2, I2, and HI at 425 oC at equilibrium? If not, in which direction must the reaction proceed to achieve equilibrium? H2] = [I2] = 0.360 mol/L, [HI] = 2.50 mol/L The system is not at equilibrium, because Q < Kc. In Q, the denominator is too big and/or the numerator is too small. Thus, if the reaction proceeds to form HI while using up H2 and I2, equilibrium can be reached; i.e., shift right. b. [H2] = 0.120 mol/L, [I2] = 0.560 mol/L, [HI] = 2.20 mol/L The system is not at equilibrium, because Q > Kc. In Q, the denominator is too small and/or the numerator is too big. Thus, if the reaction proceeds to form H2 and I2 while using up HI, equilibrium can be reached; i.e., shift left. c. [H2] = 0.120 mol/L, [I2] = 0.219 mol/L, [HI] = 1.20 mol/L The system is at equilibrium, because Q = Kc. 2. For the reaction (NH3)B(CH3)3(g) º NH3(g) + B(CH3)3(g) at 100 EC Kp = 4.62 atm. If the partial pressures of NH3(g) and B(CH3)3(g) in an equilibrium mixture at 100 EC are both 1.52 atm, what is the partial pressure of (NH3)B(CH3)3(g) in the mixture? For simplicity, let us write the reaction as AB(g) º A(g) + B(g), for which Kp is defined as Let x = pAB. Then substituting into Kp, 3. At 425oC, 1.00 mol of H2(g) and 1.00 mol of I2(g) are mixed in a one liter vessel. What will be the concentrations of H2(g), I2(g), and HI(g) at equilibrium. K = 54.8 at 425 oC for the reaction H2(g) + I2(g) º 2 HI(g) Let x be the amount of H2(g) lost to reach equilibrium. Then, from the stoichiometry of the reaction, the amount of I2(g) lost is also x. Likewise from the stoichiometry, the amount of HI(g) gained will be 2x. H2(g) + I2(g) º 2 HI(g) Add 1.00 1.00 0 Change –x –x +2x At equilibrium 1.00 - x 1.00 - x 2x Taking the square root of both sides 2x = 7.40 - 7.40x 9.40x = 7.40 x = 7.40/9.40 = 0.787 mol/L [H2] = [I2] = 1.00 - 0.79 = 0.21 mol/L [HI] = (2)(0.787) = 1.57 mol/L 4. Check the values you found in Key Question 3 by calculating the value of Q. Does your value agree with Kc = 54.8? This is acceptable agreement. 5. Suppose 0.800 mol H2(g), 0.900 mol I2(g), and 0.100 mol HI(g) are mixed in a one liter vessel at 425oC. Kc = 54.8 at 425 oC for the reaction H2(g) + I2(g) º 2 HI(g) a. In which direction must the reaction run (forward or backwards) to achieve equilibrium? Calculate Q and compare to K. Q << K, so this reaction needs to run to the right to achieve equilibrium. b. What are the concentrations of all species at equilibrium? Check your final answers with a Q calculation. Let x be the number of mol/L of H2 or I2 that is lost to reach equilibrium. Add Change At equilibrium H2(g) + 0.800 –x 0.800 - x I2(g) º 0.900 –x 0.900 - x 2 HI(g) 0.100 +2x 0.100 + 2x 0.0100 + 0.400x + 4x2 = 39.46 - 93.16x + 54.8x2 50.8x2 - 93.56x + 39.45 = 0 a b c This is a quadratic equation. Use a programmable calculator to solve this; otherwise, the manual solution is given by Reject the root x = 1.19, because [H2] = 0.800 - x and [I2] = 0.900 - x would be negative, which is impossible for a concentration. Thus, x = 0.654. From this it follows [HI] = 0.100 + (2)(0.6536) = 1.407 mol/L = 1.41 mol/L [H2] = 0.800 - 0.6536 = 0.1464 mol/L = 0.146 mol/L [I2] = 0.900 - 0.664 = 0.2464 mol/L = 0.246 mol/L Check: 8. For each of the following reactions at equilibrium, predict the effect (if any) the indicated stress would have on the position of the equilibrium. Note whether or not the value of the equilibrium constant changes. a. H2(g) + I2(g) º 2 HI(g) More HI(g) is added. The reaction shifts left to use up the excess HI(g) by reforming H2(g) and I2(g). The value of the equilibrium constant (Kc or Kp) remains the same. b. N2(g) + 3 H2(g) º 2 NH3(g) NH3(g) is removed as it forms. The reaction shifts right to replace the lost product NH3(g). The value of the equilibrium constant (Kc or Kp) remains the same. c. 2 NO(g) + Cl2(g) º 2 NOCl(g) Overall pressure is increased. The difference, ∆n, between the sum of coefficients of gas products and the sum of the coefficient of gas reactants is ∆n = 2 – (2 + 1) = –1; i.e., smaller on the product side. Increased pressure can be mitigated by shifting toward the side with fewer gas molecules. Therefore, the reaction shifts right. The value of the equilibrium constant (Kc or Kp) remains the same. d. 2 NO(g) + Cl2(g) º 2 NOCl(g) ∆Ho = –75.5 kJ/mol Temperature is increased. The endothermic process consumes heat, so raising the temperature drives that direction. The forward direction is exothermic, so the reverse direction is the endothermic process. Raising the temperature will cause a shift left to form more NO(g) and Cl2(g). The value of Kp at the higher temperature will be smaller, because reactant amounts are now greater than product amounts, relative to the lower temperature Kp. e. H2O(g) + C(s) º H2(g) + CO(g) Overall pressure is increase. Count only the coefficients of gases; i.e., ignore the C(s). Thus, ∆n = 2 - 1 = +1. Increasing pressure can be mitigated by shifting left to use up some of the H2(g) and CO(g) and form more H2O(g). The value of the equilibrium constant (Kc or Kp) remains the same. f. C(s) + O2(g) º CO2(g) Overall pressure is decreased. Again, count only coefficients of gases. These are the same on both sides, so ∆n = 0. Changing the pressure will have no effect on the position of the equilibrium. Of course, the value of the equilibrium constant (Kc or Kp) remains the same. g. N2O4(g) º 2 NO2(g) N2(g) is added, increasing overall pressure. Adding N2(g) will increase the overall pressure, but it will not change the partial pressures of the reactants or products. Because N2(g) is neither a reactant nor a product in this reaction, it has the same non-effect as adding an inert gas (e.g., He, Ne, Ar). Again, the value of the equilibrium constant (Kc or Kp) remains the same. h. N2(g) + 3 H2(g) º 2 NH3(g) Iron powder is added as a catalyst. Adding a catalyst has no effect on the position of the equilibrium. In only affects the rate at which equilibrium is achieved. Again, the value of the equilibrium constant (Kc or Kp) remains the same. 9. An equilibrium mixture of H2(g), I2(g), and HI(g) in a one-liter vessel at 425 oC is found to have the following concentrations: [H2] = 0.146 mol/L, [I2] = 0.246 mol/L, [HI] = 1.41 mol/L. If 0.59 mol HI is added to the vessel, what will be the concentrations of all species once equilibrium is reestablished? Kc = 54.8 at 425 oC for the reaction H2(g) + I2(g) º 2 HI(g) Because more product has been added, the equilibrium will need to shift left to reform reactants. H2 Original equilibrium Stress Initial Change New equilibrium 0.146 + I2 0.246 0.146 +x 0.146 + x 0.246 +x 0.246 + x º 2 HI 1.41 0.59 2.00 –2x 2.00 – 2x 50.8x2 + 29.4816x – 2.0318 = 0 Solving the quadratic equation gives x = 0.062242, –0.6425. We reject the negative root, because it would make the concentrations of H2 and I2 smaller and the concentration of HI larger, changes in the opposite direction to what we know must happen by LeChatelier’s Principle. Using the positive root, we obtain [HI] = 2.00 – (2)(0.062242) = 1.8755 mol/L = 1.88 mol/L [H2] = 0.146 + 0.062242 = 0.2082 mol/L = 2.08 mol/L [I2] = 0.246 + 0.062242 = 0.3082 mol/L = 0.308 mol/L Check: 10. In the spaces provided, write the formulas of the conjugate bases of the given BrønstedLowry acids. Remember to write the proper charge, if any, for the conjugate base in each case. Acid HNO3 NH4+ HSO4– H2SO4 HC2H3O2 Conjugate base NO3– NH3 SO42– HSO4– C2H3O2– 11. In the spaces provided, write the formulas of the conjugate acids of the given BrønstedLowry bases. Remember to write the proper charge, if any, for the conjugate acid in each case. Base OH– NH3 HS– SO42– CH3NH2 Conjugate acid H2O NH4+ H2S HSO4– CH3NH3+ 12. Using the notation acid1/base1 and base2/acid2 (as shown above), identify the Brønsted-Lowry conjugate acid-base pairs in the following reactions. HS–(aq) + HC2H3O2(aq) º H2S + C2H3O2– base1 acid2 acid1 base2 3– – HF(aq) + PO4 (aq) º F (aq) + HPO42–(aq) acid1 base2 base1 acid2 HNO2(aq) + H2O(l) º NO2–(aq) + H3O+ acid1 base2 base1 acid2 CH3NH2(aq) + H2O(l) º CH3NH3+(aq) + OH–(aq) base1 acid2 acid1 base2 13. Given the following concentrations of H3O+ or OH–, calculate the concentration of OH– or H3O+, and indicate whether the solution is acidic or basic. [H3O+] [OH–] Acidic or basic? 1.0 x 10–9 M 1.0 x 10–5 M basic 2.5 x 10–6 M 4.0 x 10–9 M acidic 1.2 x 10–8 M 8.3 x 10–7 M basic 14. Complete the following table by calculating the missing entries and indicate whether the solution is acidic or basic. Be sure your answers are expressed to the proper number of significant figures. [H3O+] [OH–] pH pOH acidic or basic? 5.8 x 10–5 M 1.7 x 10–10 M 4.24 9.76 acidic 1.6 x 10–9 M 6.2 x 10–6 M 8.79 5.21 basic 5.6 x 10–12 M 1.7 x 10–3 M 11.24 2.76 basic 5.0 x 10–5 M 2.0 x 10–10 M 4.30 9.70 acidic 2.0 M 5.0 x 10–15 M –0.30 14.30 acidic (very!) 15. Determine the concentrations of H3O+ and OH– in the following solutions of strong acids or bases in water. [Caution: Think about how the acid or base dissociates in water.] Solution [H3O+] [OH–] 2.5 x 10–3 M HCl 2.5 x 10–3 M 4.0 x 10–12 M 0.0010 M NaOH 1.0 x 10–11 M 1.0 x 10–3 M 7.5 x 10–5 M Ca(OH)2 6.7 x 10–11 M 1.5 x 10–4 M In the last case, realize that two moles of OH– are produced for every one mole of Ca(OH)2 by the dissociation reaction, Ca(OH)2(aq) ÷ Ca2+(aq) + 2 OH–(aq).