Chem 30 Name Exam 3, Ch 7, 9, 4 November 9, 206 00 Points Please follow the instructions for each section of the exam. Show your work on all mathematical problems. Provide answers with the correct units and significant figures. Be concise in your answers to discussion questions. Part 0: Warmup. 4 points each. Reactions with a positive H o and a negative S o are a. spontaneous at all temperatures. b. non-spontaneous at all temperatures. c. spontaneous at low temperatures but non-spontaneous at Answer B high temperatures. d. non-spontaneous at low temperatures but spontaneous at high temperatures. 2. Which of the processes below DOES NOT result in an increase in entropy? a. 2 H 2 (g) + O 2 (g) 2 H 2 O(g) b. H 2 O(s) H 2 O(l) (The melting of ice.) c. CO 2 (s) CO 2 (g) (The sublimation of dry ice.) Answer A d. NH 4 NO 3 (s) N 2 O(g) + 2 H 2 O(l) 3. Consider the reaction below. If hydrogen chloride gas is being produced at.2 moles per liter per minute (M min - ), at what rate is hydrogen gas being consumed 2ICl(g) + H 2 (g) I 2 (g) + 2HCl(g) a..2 M min - b. 0.60 M min - c. 2.4 M min - Answer B d. Not enough information to determine. Part I: Complete all of problems 4-8 4. Describe what is meant when we call something a state function. How do the properties of a state function aid us in determining thermodynamic parameters and in predicting thermodynamic behavior? (6 points) Sate functions are only concerned with the starting and ending states or conditions of a process, but are not dependent on the pathway the process takes going from the start to the end. This allows us to model thermodynamic problems by building pathways from known components, like component reactions using Hess law or by using enthalpies, entropies and Gibbs energies of formation.
5. Concisely discuss the validity of each of the following statements. Clearly justify your reasoning. A few sentences should be sufficient. (4 points) a. Reactions with a positive H o and a positive S o can never be product-favored. Your answer should discuss how H o, S o and G o are related. Since G o = H o - T S o, as long as T S o is greater than H o, the reaction will be product favored. This can occur if S o is very large or at increased temperature. b. Free energy changes provide a good indication of which reactions are favorable and fast, as well as those that are unfavorable and slow. Free energy changes are thermodynamic quantities that tell us about the tendency for a reaction to proceed, they tell us nothing about the rate of the reaction. We need information about kinetics to do that. 6. Consider a first order reaction B Products. An experiment is performed and it is determined that, given a starting [B] = 0.00 M, it requires 56.2 seconds for the concentration to fall to 0.029 M. What is the rate constant for the reaction? How long would it take for the concentration to fall from 0.00 M to 0.000 M? (4 points) The integrated rate law for a first order reaction is helpful here: ln[a] t = -kt + ln[a] 0. From the problem, [A] t = 0.029 M, [A] 0 = 0.00 M and t = 56.2 s k = ln[a] t -ln[a] 0 = ln(0.029)-ln(0.00) = 0.022 s - = k -t -56.2 s Now that we have k, we can use the integrated rate law to solve the second part of the problem: t = ln[a] t -ln[a] 0 = ln(0.000)-ln(0.00) = 209 s = 20 s (2 sig figs) -k -0.022 s - Answer: Rate constant 0.022 s - Time to 0.000 M 20 s 2
7. How much heat energy is produced when 0.00 kg of gaseous ethane (C 2 H 6 ) undergoes a combustion reaction with excess oxygen gas to produce gaseous carbon dioxide and liquid water? (4 points) Species H o f, kj mol - Species H o f, kj mol - O(g) +249.2 H 2 O(l) -285.8 O 2 (g) 0 H 2 O(g) -24.8 H(g) +28.0 C 2 H 6 (g) -84.7 H 2 (g) 0 CO 2 (g) -393.5 2C 2 H 6 (g) + 7O 2 (g) 6H 2 O(l) + 4CO 2 (g) H o rxn = (6 H o f, H2O + 4 H o f,co2) (2 H o f, C2H6 + 7 H o f, O2) H o rxn = (6(-285.8 kj) + 4(-393.5 kj)) - (2(-84.7 kj) + 7(0 kj)) H o rxn = -39 kj So, 3288 kj of energy is released every time two moles of ethane is consumed. 00 g C 2 H 6 x mol C 2 H 6 x 39 kj = 588 kj 30.07 g C 2 H 6 2 mol C 2 H 6 Answer 588 kj 3
Part II. Answer three (3) of problems 8-. Clearly mark the problem you do not want graded. 4 points each. 8. Determine H o for the reaction N 2 H 4 (l) + 2H 2 O 2 (l) N 2 (g) + 4H 2 O(l) from these data: Reaction N 2 H 4 (l) + O 2 (g) N 2 (g) + 2H 2 O(l) H 2 (g) + ½ O 2 (g) H 2 O(l) N 2 (g) + O 2 (g) 2NO(g) H 2 (g) + O 2 (g) H 2 O 2 (l) H o -622.2 kj -285.8 kj +73. kj -87.8 kj Reaction H o The first reaction is left alone: N 2 H 4 (l) + O 2 (g) N 2 (g) + 2H 2 O(l) -622.2 kj Double the second reaction: 2(H 2 (g) + ½ O 2 (g) H 2 O(l)) 2(-285.8 kj) Reverse and double the fourth rxn: 2(H 2 O 2 (l) H 2 (g) + O 2 (g)) -2(-87.8 kj) The third reaction is not used Add the above reactions to get the overall reaction: N 2 H 4 (l) + O 2 (g) + 2H 2 (g) + O 2 (g) + 2H 2 O 2 (l) N 2 (g) + 2H 2 O(l) + 2H 2 O(l) + 2H 2 (g) + 2O 2 (g) N 2 H 4 (l) + 2H 2 O 2 (l) N 2 (g) + 4H 2 O(l) H o = -622.2 kj + (2(-285.8 kj)) +(-2(-87.8 kj)) = -88.2 kj Answer -88.2 kj 4
9. A coffee-cup calorimeter contains 00.0 ml of 0.300 M HCl at 20.3 o C. When.82 g zinc metal also at 20.3 o C is added and is allowed to react via the net ionic equation below, the temperature rises to 30.5 o C. What is the heat of reaction per mole of Zn? Assume no heat is lost to the environment during the course of the reaction and that the heat capacity and the density of the solution is the same as that of pure water (.00 g/ml and 4.84 J/g o C, respectively). Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) We first need to determine if all of the zinc will react. How many ml HCl will we need?.82 g Zn x mol Zn x 2 mol H + x 000 ml = 85 ml 65.39 g Zn mol Zn 0.300 mol H + So, HCl is the limiting reagent: q reaction = -q calorimeter n HCl H rxn = -mc T (0.00 L x 0.300 mol HCl/L) H rxn = -(0.82 g)(4.84 J/g o C)(30.5 o C -20.3 o C) H rxn = -(0.82 g)(4.84 J/g o C)(0.2 o C) = 44.8 kj x 2 mol HCl = -289.7 kj (0.00 L x 0.300 mol HCl/L) mol HCl mol Zn mol Zn (If you make the incorrect assumption that Zn is the limiting reagent, you find H = -56 kj/mol Zn.) Answer -290 kj/mol Zn 5
0. The reaction I - (aq) + OCl - (aq) IO - (aq) + Cl - (aq) was studied and the data below were obtained. Determine the rate law and the value of the rate constant for this reaction. [I - ] 0 (mol/l) [OCl - ] 0 (mol/l) Initial Rate (mol/ls) 0.2 0.8 0.079 0.060 0.8 0.0395 0.030 0.090 0.00988 0.24 0.090 0.079 Rate = k[i - ] x [OCl - ] y = 2 = k[0.2] x [0.8] y = 2 x x = Rate 2 k[i - ] x [OCl - ] y k[0.06] x [0.8] y Since there are no pairs or reactions in which only the OCl - concentration changes the I - concentration does not, we can use any of the pairs to determine the reaction order for OCl -. Rate 2 = k[i - ] x [OCl - ] y = 4 = k[0.06] [0.8] y Rate 3 k[i - ] x [OCl - ] y k[0.03] [0.09] y 2 = [0.8] y y = [0.09] y Rate = k[i - ][OCl - ] k = Rate = 0.079 M/s = 3.66 M - s - [I - ][OCl - ] (0.2 M)(0.8 M) Answer Rate = 3.7M - s - [I - ][OCl - ] 6
. Consider the reaction N 2 O(g) + 2H 2 O(l) NH 4 NO 3 (s) at 298K. Species H o f, kj mol - S o f, J mol - K - G o f, kj mol - O 2 (g) 0 205. 0 H 2 (g) 0 30.7 0 NH 4 NO 3 (s) -365.6 5. -83.9 N 2 O(g) 82.05 29.9 04.2 H 2 O(l) -285.8 69.9-237. H 2 O(g) -24.8 88.8-228.6 a. Is the forward reaction exothermic or endothermic? (5 points) H o rxn = H o f, NH4NO3 ( H o f, N2O + 2 H o f,h2o) H o rxn = -365.6 kj (82.05 kj+ 2(-285.8 kj)) H o rxn = +24 kj Since H o is positive, the reaction is endothermic b. What is the value of G o at 298 K? (5 points) S o rxn = S o f, NH4NO3 (S o f, N2O + 2S o f,h2o) S o rxn = 5.J/K (29.9J/K + 2(69.9J/K)) S o rxn = -209 J/K G o rxn = H o rxn - T S o rxn = +24kJ 298K(0.209kJ/K) = +86 kj Alternatively: G o rxn = G o f, NH4NO3 ( G o f, N2O + 2 G o f,h2o) G o rxn = -83.9 kj (04.2 kj+ 2(-237. kj)) G o rxn = +86 kj c. Does the reaction occur spontaneously at temperatures above 298 K, below 298 K, both, or neither? Justify your answer. (4 points) Since H o rxn is positive, and S o rxn is negative, G o rxn will be positive at all temperatures, since G o rxn = H o rxn - T S o rxn = (+) (+)(+) = (+) regardless of temperature. So, the reaction is non-spontaneous at all temperatures. (NOTE: If you did not calculate S o rxn in part b, you can infer that it is negative by looking at the reaction. Since there are fewer moles of gas on the products side of the reaction compared to the reactants, we expect S o rxn to be negative.) 7
Possibly Useful Information R = 0.08206 L atm mol - K - R = 8.34 J mol - K - K = o C + 273.5 y y slope m x x 2 2 y x q=mc T q=n LR H rxn q=m H S universe S system - S surr G H - T S S surr H sys /T rate = k[a] 0 [A] t = -kt + [A] 0 t /2 = [A] 0 /2k ln A kt ln A rate = k[a] t 0 t /2 = 0.693/k rate = k[a] 2 A t A 0 kt t /2 = /(k[a] 0 ) 8
9 To save some calculation time, you may round all atomic masses to two (2) decimal points.