Exam 5 - Chemistry Reactions and Thermodynamics - Prof. Stephen P. Mccord, Exams of Chemistry

Download Exam 5 - Chemistry Reactions and Thermodynamics - Prof. Stephen P. Mccord and more Exams Chemistry in PDF only on Docsity! 048 version last name first name signature McCord CH301 Exam 5 Dec 5, 2017 50085 BUR 106 Tuesday TTh 11 am - 12:30 pm Remember to refer to the Periodic Table handout that is separate from this exam copy. NOTE: Please keep this exam copy intact (all pages still stapled - including this cover page). You must turn in ALL the mate- rials that were distributed. This means that you turn in your exam copy (name and signature included), bubble sheet, pe- riodic table handout, and all scratch paper. Please also have your UT ID card ready to show as well. hanagandi (kvh343) – Exam 5 - F17 – mccord – (50085) 2 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 5.0 points Determine the boiling point of acetone using the following thermodynamic phase data. ∆Hvap = 31.3 kJ mol ∆Svap = 95.0 J mol ·K 1. 0.329 K 2. 329 K correct 3. 659 K 4. -329 K 5. -659 K 6. 165 K Explanation: A phase change is in physical equilibrium where ∆G = 0. Therefore, you can rearrange the free energy equation so that: T = ∆H/∆S T = (31.3 kJmol × 1000 J kJ)/95.0 J mol·K T = 329 K 002 5.0 points A theoretical reaction has the following ther- modynamic data: ∆H◦ = 12 kJ/mol ∆S◦ = 68 J/mol What is the ∆G◦ for this reaction? Is this reaction spontaneous or non-spontaneous? Assume this reaction is run at 25◦C. 1. -2.26 kJ, spontaneous 2. -8.26 kJ, spontaneous correct 3. -8264 kJ, non-spontaneous 4. −20250 kJ, spontaneous 5. -8.26 kJ, non-spontaneous 6. 10.3 kJ, non-spontaneous Explanation: Using the free energy equation: ∆G = ∆H − T∆S −8.26 = 12− (298)(68/1000) If ∆G = -8.26 kJ, the reaction is spontaneous. 003 5.0 points You have one mole of each of the following substances at room temperature: Cmercury = 0.140 J g◦C Ciron = 0.450 J g◦C Assuming no phase change occurs, which sample will have the lowest final temperature after 22.7 J heat is added to each? 1. Mercury correct 2. Iron 3. The temperature will be equal for both samples Explanation: Heat capacity is a substance’s ability to re- sist changes in temperature. Therefore, the substance with the higher heat capacity will have the lower change in temperature. The data given is the specific heat capacities of mercury and iron. Although mercury has a lower specific heat, one mole will have a much greater mass than one mole of iron. As an extensive property, the heat capacity of mer- cury is greater for one mole (28.22 J/◦C) than the heat capacity of iron (25.13 J/◦C). Mer- cury will have a lower final temperature. Iron will have a higher temperature. 004 5.0 points A chemical reaction releases 78.9 kJ heat while 27.7 kJ work is done by the system hanagandi (kvh343) – Exam 5 - F17 – mccord – (50085) 5 011 5.0 points Calculate the ∆Ssurr for the following reaction at 9.0◦C and 1 atm. CH3F(ℓ) → CH3F(g) ∆H ◦ rxn = +16.9 kJ 1. 59.9 J/K 2. -939 J/K 3. -1.88 J/K 4. -59.9 J/K correct 5. 1880 J/K 6. 1.88 J/K Explanation: In general for any process: ∆Ssurr = −∆Hsys Tsurr This is because the heat flow in the surround- ings is just the opposite of the heat flow for the system (qsurr = −qsys) and at constant pressure the heat is equal to ∆H. therefore ∆Ssurr = −16900/(9.0 + 273.15) = −59.9 J/K 012 5.0 points For which of the following chemical equations would ∆H◦rxn = ∆H ◦ f ? 1. 1 2 N2(g) + 3 2 H2(g) → NH3(g) correct 2. CH4(g) → C(s, graphite) + 2H2(g) 3. N2(ℓ) + 3Cl2(g) → 2NCl3(ℓ) 4. CH4(g) + 2Cl2(g) → CCl4(ℓ) + 2H2(g) 5. 2C(s, graphite) + O2(ℓ) → 2CO(g) Explanation: If ∆H◦rxn = ∆H ◦ f , the reaction written must be a formation reaction. A formation reaction is a single mole of products formed from its constituent elements in their standard states. 013 5.0 points When a 0.401 g sample of a clean-burning hy- drocarbon (molecular weight = 42.05 g/mol) is combusted in a rigid container, the temper- ature increases from 24.987◦C to 28.623◦C. The total volume of water is 0.746 L. The sum of all hardware components of the calorimeter have a heat capacity of 2.55 kJ/◦C. Calcu- late the internal energy of combustion for one mole of this hydrocarbon. 1. -3.71799 kJ/mol 2. +9277.90 kJ/mol 3. +51.4232 kJ/mol 4. +2162.35 kJ/mol 5. -20.6207 kJ/mol 6. +3.71799 kJ/mol 7. +20.6207 kJ/mol 8. -2162.35 kJ/mol correct Explanation: Heat released by the combustion reaction: q = mC∆T+ C∆T qcal = (746)(4.184)(3.636) + (2.55× 1000)(3.636) qcal = 20620.7J For the system, qv = -qcal: qv = −20.6207kJ This is how much heat was released when 0.401 g of hydrocarbon was combusted (we consider it to be negative). Convert to this to how much heat would be released per mole of ethylene combusted. −20.6207 kJ 0.401 g × 42.05 g 1mol = −2162.35 kJ/mol 014 5.0 points The unknown element X is a metallic solid that oxidizes at room temperature and pres- sure to form X2O3. Given the following ther- modynamic data, calculate the ∆G◦f . hanagandi (kvh343) – Exam 5 - F17 – mccord – (50085) 6 Substance ∆H◦f S ◦ kJ/ mol J/ mol K X(s) − 38.3 O2(g) − 205 O(g) 249 161 X2O3(s) −653 87.8 1. 6.09× 106 kJ mol−1 2. 1260 kJ mol−1 3. 565 kJ mol−1 4. −742.35 kJ mol−1 5. −5420 kJ mol−1 6. −1260 kJ mol−1 7. -565 kJ mol−1 correct 8. 742.35 kJ mol−1 Explanation: For the formation of one mole of X2O3, you can use the chemical equation shown below: 2X(s) + 3 2 O2(g) → X2O3(s) ∆G = ∆H − T∆S Using ∆H◦f and S ◦ values: ∆S = 87.8− [(1.5× 205) + (2× 38.3)] ∆S = −296.3J/K ∆H = −653kJ ∆G = −565kJ This was solved per mole based on the forma- tion reaction. 015 5.0 points When an endothermic reaction has a positive internal energy change for the system, I. the internal energy of the system is the sum of all energy (potential and kinetic) in the system II. ∆Usys = −∆Usurr III. there is an overall energy change in the universe IV. heat enters the system from the sur- roundings 1. I, II, and IV only correct 2. I only 3. I, II, III and IV 4. I and II only 5. I and IV only 6. III and IV only 7. II only Explanation: The first law of thermodynamics can be summed up with the idea that energy is al- ways conserved in the universe. Mathemati- cally, this means that: ∆Usys = −∆Usurr ∆Uuniverse = ∆Usys +∆Usurr ∆Uuniverse = 0 Internal energy is defined as the total en- ergy content of a system. This can be repre- sented as the sum of heat and work or as the sum of potential and kinetic energy. For an endothermic reaction, the heat flows from the surroundings to the system. 016 5.0 points A 23 mL sample of liquid water at 39◦C is frozen and cooled to a final temperature of -18◦C. Calculate the heat of this process. 1. -8547 J 2. 12300 J hanagandi (kvh343) – Exam 5 - F17 – mccord – (50085) 7 3. −12300 J correct 4. -534.8 J 5. 11440 J Explanation: You will need to make three different calcula- tions for this problem: 1) Cool liquid water from 39◦C to 0◦C: q = mC∆T = (23 g)(4.184 J g◦C )[0◦C− (39 ◦C)] = −3753 J 2) Freeze the ice at constant temperature: q = (m)(−∆Hfus) = (23 g)(334 J g ) = −7682 J 3) Cool the ice from 0◦C to -18◦C: q = mC∆T = (23 g)(2.09 J g◦C )[−18◦C− 0◦C] = −865.3 J Total heat required: (−3753 J) + (−7682 J) + (−865.3 J) = −12300 J 017 5.0 points Consider the following balanced chemical re- action: H2SO4(l) → SO3(g) + H2O(g) To solve for the ∆H◦rxn for this reaction, you collect the following data in the lab: H2S(g) + 2O2(g) → H2SO4(l), ∆H◦rxn = -241.3 kJ/mol 1 2 H2S(g) + O2(g) → 1 2 SO3(g) + 1 2 H2O(l), ∆H◦rxn = -105.7 kJ/mol H2O(l) → H2O(g), ∆H◦rxn = 41.9 kJ/mol What is the ∆H◦rxn of the overall balanced reaction? 1. 178 kJ/mol 2. 71.8 kJ/mol correct 3. 495 kJ/mol 4. -411 kJ/mol 5. -546 kJ/mol Explanation: The objective is to arrange the three reactions so that the sum of the reactions gives you the overall balanced chemical reaction. You will need to make two modifications. The first re- action must be reversed, which means you will flip the sign of the enthalpy of reaction. The coefficients of the second reaction must be doubled, which means you will need to mul- tiple the enthalpy of reaction by two. From there: ∆H◦rxn = −(−241.3)+(2×−105.7)+(41.9) = 71.8kJ/mol 018 5.0 points Consider a reaction that has a negative change in entropy and a positive change in enthalpy. Which of the following conditions will favor spontaneity? 1. This reaction is spontaneous at all tem- peratures 2. This reaction is non-spontaneous at all temperatures correct 3. High Temperatures 4. Low Temperatures Explanation: For this reaction, both the entropy and enthalpy terms are unfavorable. This reaction will never be spontaneous 019 5.0 points