8. A sample consisting of 0.50 moles of AB(g) was placed in a 10.0 L reaction chamber at 500oC. After equilibrium was reached, it was found that 20% of the AB had dissociated according to the reaction 2AB(g)  2A(g) + B2(g) Calculate the equilibrium constant KC for this reaction at this temperature. A. 4.1 x 10 –7 B. 3.5 x 10 –6 C. 3.1 x 10 –5 D. 3.1 x 10 –4 E. 4.4 x 10 –39. A 10.0 L flask was filled to a pressure of 1.00 atm with A2(g) at 700oC. The dissociation reaction A2(g)  2A(g) was allowed to reach equilibrium, at which point the total pressure was found to be 1.31 atm. Calculate KC, the equilibrium constant in terms of concentrations, for this reaction at this temperature. A. 9.7 x 10 –4 B. 2.8 x 10 –3 C. 7.0 x 10 –3 D. 7.8 x 10 –2 E. 5.6 x 10 –110. Consider the gas-phase dissociation process shown below. A2(g)  2A(g) KP = 0.250 @ 727°C At 727°C, the total pressure of the system at equilibrium is found to be 1.00 atm. Calculate the equilibrium partial pressure of A. A. 0.39 atm B. 0.50 atm C. 0.59 atm D. 0.67 atm E. 0.75 atm13. Estimate the pH of a 250 mL aqueous solution prepared by dissolving 10. g of acetic acid (CH3COOH) in water. [Ka = 1.8 x 10–5 for acetic acid] A. 2.16 B. 2.46 C. 2.76 D. 3.06 E. 3.3635. Radon-222 decays by alpha emission. What is its decay product ?36. The only stable isotopes of carbon are C-12 and C-13. Predict the mode of decay of C.38. Uranium-238 decays with a half-life of 4.5 ´ 109 yr to eventually form the stable isotope lead-206.What would the atom ratio of 206Pb to 238U be in a uranium mineral from a rock that is 3.0 ´ 109 years old? (You may assume that there was no 206Pb in the rock when it was formed.) A. 0.17 B. 0.26 C. 0.36 D. 0.47 E. 0.59CHEM 188 – Spring, 2012Final Exam (Red)May 11, 2012Instructions:Your scantron answer sheet must show your NAME, STUDENT ID NUMBER, and LAB SECTION. (Begin these entries at the LEFT end of the space provided.)In answering the questions, be careful to fill in the corresponding circles on the answer sheet according to the number of the question on the exam. USE A SOFT (No. 2) PENCIL.Note that a periodic table of the elements is attached at the end of the exam.You are allowed to use a one page (8½”x11”) study guide that you have prepared.Useful information: Gas constant R = 0.0821 Latm/Kmol = 8.314 J/KmolAvogadro’s constant NAvo = 6.02  1023 mol1 Speed of light c = 3.00  108 m∙s1 Boltzmann’s constant k = 1.38  1023 JK1Planck’s constant h = 6.63  1034 Js1 atm = 760 mmHg = 760 torr 1 Latm = 101.3 J 1 kg = 6.02  1026 amu1. Consider the reaction N2(g) + 3H2(g) 2NH3(g)Suppose that the rate of loss of molecular hydrogen, d[H2]/dt, is 0.060 M/s at a particular time during the reaction. What is the rate of loss of molecular nitrogen, d[N2]/dt? A. 0.020 M/s B. 0.030 M/s C. 0.060 M/s D. 0.12 M/s E. 0.18 M/s2. Consider the hypothetical reaction A + 2B  products. Use the following data to determine the rate of the reaction when [A] = 0.25 M and [B] = 0.15 M. Expt. # [A]0 [B]0 Initial rate 1 0.20 0.20 0.76 M/s 2 0.20 0.40 0.76 M/s 3 0.40 0.20 1.52 M/s1A. 0.19 M/s B. 0.38 M/s C. 0.57 M/s D. 0.95 M/s E. none of these3. The rate constant for the second-order reaction 2NOBr(g)  2NO(g) + Br2(g)is 0.80 M 1s 1 at 10oC. If the initial concentration of NOBr was 0.50 M, what would the concentration of NOBr be after 2.0 sec?A. 0.068 M B. 0.12 M C. 0.19 M D. 0.28 M E. 0.36 M4. The isomerization of cyclopropane follows first-order kinetics. The rate constant at 600 K is 2.72 x 107 min1, and the activation energy for the reaction is 270 kJ/mol. Calculate the value of the rate constant (in min1) at 800 K.A. 6.20 x104 B. 2.05 x 101 C. 1.86 x 101D. 6.87 x 102 E. 1.32 x 104 5. If a catalyst could be found that would lower the activation energy by 20.0 kJ/ mol for a particular reaction, by what factor would the rate constant for this reaction be increased at 25oC? (Assume the frequency factor remains the same.)A. 5.66 x 101 B. 4.26 x 102 C. 3.20 x 103 D. 2.41 x 104 E. 1.81 x 105 6. The equilibrium constant (Kc) for the reaction shown below is 4.17 10-34 at 25°C. 2 HCl(g)  H2(g) + Cl2(g)What is the equilibrium constant for the following reaction at the same temperature? H2(g) + Cl2(g)  2HCl(g)A. 4.17  10-34 B. 2.04  10-17 C. 4.90  1016 D. 2.40  1033 E. none of these7. For the reaction H2(g) + I2(g)  2HI(g) Kc = 50.2 at 445oC.If [H2] = [I2] = [HI] = 1.75 x 10 3 M at 445oC, which one of the following statements is(are) true? 1. The system is at equilibrium; no change will occur. 2. The concentration of H2 will increase as the system approaches equilibrium. 3. The concentration of HI will increase as the system approaches equilibrium. 4. The concentration of I2 will decrease as the system approaches equilibrium. 5. The concentration of HI will decrease as the system approaches equilibriumA. 1 only B. 2 only C. 3 only D. 2 & 5 E. 3 & 4 28. A sample consisting of 0.50 moles of AB(g) was placed in a 10.0 L reaction chamber at 500oC. After equilibrium was reached, it was found that 20% of the AB had dissociated according to the reaction 2AB(g)  2A(g) + B2(g)Calculate the equilibrium constant KC for this reaction at this temperature.A. 4.1 x 10 –7 B. 3.5 x 10 –6 C. 3.1 x 10 –5D. 3.1 x 10 –4E. 4.4 x 10 –3 9. A 10.0 L flask was filled to a pressure of 1.00 atm with A2(g) at 700oC. The dissociation reactionA2(g)  2A(g)was allowed to reach equilibrium, at which point the total pressure was found to be 1.31 atm. Calculate KC, the equilibrium constant in terms of concentrations, for this reaction at this temperature.A. 9.7 x 10 –4 B. 2.8 x 10 –3 C. 7.0 x 10 –3 D. 7.8 x 10 –2E. 5.6 x 10 –110. Consider the gas-phase dissociation process shown below. A2(g)  2A(g) KP = 0.250 @ 727°CAt 727°C, the total pressure of the system at equilibrium is found to be 1.00 atm. Calculatethe equilibrium partial pressure of A.A. 0.39 atm B. 0.50 atm C. 0.59 atm D. 0.67 atm E. 0.75 atm 11. Which of the following statements is true for a 0.10 M solution of a weak acid, HA?A. The pH > 1.00B. [H+] < [HA]C. [H+] = [A-]D. all