CHEMISTRY 102B Hour Exam II. Dr. D. DeCoste T.A.

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1 CHEMISTRY 10B Hour Exam II March 19, 015 Dr. D. DeCoste Name Signature T.A. This exam contains questions on 9 numbered pages. Check now to make sure you have a complete exam. You have one hour and thirty minutes to complete the exam. Determine the best answer to the first 0 questions and enter these on the special answer sheet. Also, circle your responses in this exam booklet. Show all of your work and provide complete answers to questions 1 and. The answers must be kept within the spaces provided. 1-0 (40 pts.) 1 ( pts.) (18 pts.) Total (80 pts) Useful Information: Always assume ideal behavior for gases (unless explicitly told otherwise). PV = nrt {R = Latm/molK} Root mean square velocity = 3RT M = s m ΔT q K = C + 73 E= λ hc ΔE =.178 x J [ ] n f n i c =.998 x 10 8 m/s 1 nm = 10-9 m h = x Js

2 Hour Exam II Page No You have neon gas in the left side, and helium gas in the right side, of a two-bulb container connected by a valve as shown below. Initially the valve is closed. Ne He 3.00 L 1.00 L 1.00 atm 3.00 atm The left bulb has a volume of 3.00 L and the Ne gas is at a pressure of 1.00 atm. The right bulb has a volume of 1.00 L and the He gas is at a pressure of 3.00 atm. After the valve is opened, what is true about the relative partial pressures of helium and neon? Assume constant temperature. a) The partial pressure of neon is 3.00 times as great as the partial pressure of helium. b) The partial pressure of neon is 1.33 times as great as the partial pressure of helium. c) The partial pressure of helium is 3.00 times as great as the partial pressure of neon. d) The partial pressure of helium is 1.33 times as great as the partial pressure of neon. e) The partial pressures of helium and neon are equal.. For which of the following constant volume cases is the number of collisions of gas particles with the walls of the container the greatest? Volume is the same in all cases. a) 1.0 mol of helium gas at 5 C. b) 1.0 mol of helium gas at 15 C. c) 1.0 mol of oxygen gas at 5 C. d) 1.0 mol of oxygen gas at 15 C. e) Choices a and b have equal collision rates, and these are higher than those in c and d. 3. For metals that form a + charge in an ionic compound, the general equation for such a metal reacting with hydrochloric acid is M(s) + HCl(aq) MCl (aq) + H (g) 3.7 g of a metal is reacted completely with an excess of HCl and all of the hydrogen gas produced is collected in a balloon at 5 C and 1.00 atm. The volume of the balloon is measured and found to be 3.74 L. Identify the metal that was added to the HCl. a) Ni b) Ba c) Ca d) Zn e) Mg

3 Hour Exam II Page No. 4. What mass of helium gas will exert twice the pressure of g of nitrogen gas at the same conditions of volume and temperature? a) 1.43 g b) 4.86 g c) g d) 85.7 g e) g 5. For a particular process, q = 31 kj and w = 5 kj. How many of the following statements must be true? I. Heat flows from the surroundings to the system. II. The surroundings do work on the system. III. E = 6 kj IV. H = 31 kj a) 0 b) 1 c) d) 3 e) 4 6. A 1.00-g sample of potato chips is burned in a bomb calorimeter. The heat capacity of the bomb calorimeter is 11.3 kj/ C and the temperature rises by 1.98 C. If a serving of potato chips is considered to be 8.4 g, determine the caloric content of 1 serving of potato chips. 1 food Calorie = 4184 J a) 5.35 Cal. b) 84. Cal. c) 134 Cal. d) 15 Cal. e) 68 Cal. 7. The goal of the Chemistry 103 lab this week was to determine the identity of an unknown salt by determining the ΔH value when it dissolved in water. For this problem, you are to determine the final temperature of the solution when you dissolve a known salt in water. Consider the dissolution of ammonium nitrate: NH 4 NO 3 (s) NH 4 + (aq) + NO 3 (aq) ΔH = 5.7 kj A g sample of NH 4 NO 3 (s) is dissolved in ml of water with both substances initially at room temperature (4.7 C). Calculate the final temperature of the solution. You may make the following assumptions: no heat loss to the surroundings; the specific heat capacity of the solution is 4.18 J/g C; and the density of the solution is 1.00 g/ml. a) 17.0 C b) 17.7 C c) 4.7 C d) 31.7 C e 3.4 C 8. Use the following enthalpies of combustion to calculate the enthalpy of formation, ΔH o f, of methanol (CH 3 OH) from its elements. CH 3 OH(l) + 3O (g) CO (g) + 4H O(l) C(graphite) + O (g) CO (g) H (g) + O (g) H O(l) ΔH rxn = kj ΔH rxn = kj ΔH rxn = kj a) 38.7 kj/mol b) +43. kj/mol c) kj/mol d) kj/mol e) kj/mol

4 Hour Exam II Page No What color of light is emitted when an excited electron in the hydrogen atom falls from the 4 th energy level to the nd energy level? Visible light spectrum red (65 nm to 700 nm) orange (585 nm to 65 nm) yellow (565 nm to 585 nm) green (470 nm to 565 nm) blue (40 nm to 470 nm) a) Red b) Orange c) Yellow d) Green e) Blue 10. How many of the following statements is/are true? I) Bohr s atomic model works for hydrogen and helium, but no other atoms. II) An excited atom can return to its ground state by absorbing electromagnetic radiation. III) The 1s orbitals for different elements have different radii. IV) Each energy level has only one type of orbital. V) The ground state phosphorus atom has one unpaired electron. a) 1 b) c) 3 d) 4 e) A ground state electron in the hydrogen atom absorbs enough energy to get to n=. Which orbital will the electron occupy? a) The s orbital. b) The p x orbital. c) The p y orbital. d) The p z orbital. e) Each of the above is equally likely. 1. How many electrons can be described by the quantum numbers n = 3, l = 1? a) 0 b) c) 6 d) 10 e) Which of the following statements (a-c) is false? a) An electron configuration for an excited state of the carbon atom could be 1s s p 3. b) The ground state electron configuration for the most stable ion of sodium in a compound is 1s s p 6. c) The ground state electron configuration for the valence electrons of the halogens (Group 7A) is ns np 5. d) At least two of the above statements (a-c) are false. e) All of the above statements (a-c) are true.

5 Hour Exam II Page No Which of the following concerning second ionization energy values of K and Ca is true? a) The second ionization energies are equal for K and Ca since the elements are in the same row of the periodic table. b) That of Ca is higher than that of K because as you go across a row on the periodic table ionization energy increases. c) That of Ca is lower than that of K because Ca wants to lose the second electron, so it is easier to take the second electron away. d) That of Ca is higher than that of K because the Ca atom has one more proton than the K atom and thus has greater attraction for the electrons. e) That of Ca is lower than that of K because the second electron taken from K is from a lower energy level. 15. Which of the following best evaluates the statement The 1st ionization energy for an oxygen atom is lower than the 1st ionization energy for a nitrogen atom? a) It is consistent with the general trend relating changes in ionization energy across a period from left to right because it is easier to take an electron from an oxygen atom than from a nitrogen atom. b) It is inconsistent with the general trend relating changes in ionization energy across a period from left to right and due to the fact that oxygen has one doubly occupied p orbital and nitrogen has all unpaired electrons. c) It is consistent with the general trend relating changes in ionization energy across a period from left to right because it is harder to take an electron from an oxygen atom than from a nitrogen atom. d) It is inconsistent with the general trend relating changes in ionization energy across a period from left to right and it is due to the fact that the oxygen atom has two doubly occupied p orbitals and nitrogen has only one. e) The given statement is incorrect. 16. Many decades ago, a chemist at UIUC reported the discovery of a new element and named it Illinium. Unfortunately, he could not substantiate its existence and many years later another chemist claimed it and it is now named Promethium, Pm. What is the expected ground state electron configuration for the element formerly known as Illinium? a) [Xe] 6s 5d 10 4f 4 b) [Xe] 6s 5d 1 f 4 c) [Xe] 6s 5d 1 4f 4 d) [Xe] 6s 6d 1 5f 4 e) [Xe] 6s 6d 1 6f Which of the following correctly ranks the atoms from smallest ionization energy to greatest ionization energy? a) Li, B, C, N, Na b) Fr, N, P, O, F c) K, Na, S, Cl, F d) Be, Mg, Ca, Sr, Cs e) At least two of the above (a-d) correctly ranks the atoms from smallest ionization energy to greatest ionization energy.

6 Hour Exam II Page No The following graph plots the first, second, and third ionization energies for Mg, Al, and Si. A B C Which of the following corrects matches the plot to the element? a) A = Mg B = Al C = Si b) A = Si B = Al C = Mg c) A = Si B = Mg C = Al d) A = Mg B = Si C = Al e) A = Al B = Si C = Mg 19. In order to remove an electron from the potassium atom, energy is, and in order to remove an electron from a chlorine atom, energy is. a) required, required b) required, released c) released, required d) released, released 0. Which of the following is true about the general trend of atomic size across a row on the periodic table? a) Atomic size increases to the right because with a greater number of electrons, there is more electron repulsion. b) Atomic size increases because with a greater number of protons, the nucleus is bigger thus the atom is bigger. c) Atomic size decreases to the right because with a greater number of protons, the electrons are more attracted to the nucleus. d) Atomic size decreases to the right because with a greater number of electrons, there is more attraction to the nucleus. e) Atomic size is constant since all atoms in a row have the same valence energy level.

7 Hour Exam II Page No Consider a steel, rigid tank at 1.00 atm and 5 C that contains equal masses of hydrogen gas and nitrogen gas. a. Circle one of the following choices: The partial pressure of hydrogen gas is greater than less than equal to the partial pressure of nitrogen gas in the steel tank. [1 point] b. Explain your answer to part a above using the ideal gas law. [3 points] V and T are constant; P and n change PV=nRT P = n[rt/v]; thus, for larger n, P is larger. We have equal masses of H and N ; since H has a smaller molar mass than N, we have more moles of H than of N, thus the partial pressure of H is greater than that of N. c. Explain your answer to part a above using the kinetic molecular theory. [4 points] Pressure is due to gas particle collisions with the walls of the container. Since T is constant, the molecules have the same average kinetic energy, and since the volume is the same the molecules have the same average distance to travel to hits the walls of the container. The pressure is due, then to the number of hits and the force of these hits. With the same T and V, these factors cancel for a given molecule (that is, the H molecules will hit more often because they are faster and the N will hit with more force because they are heavier), resulting in the same pressure for the same number of molecule. Because of this, the pressure will depend solely on the relative number of molecules (more hits ); since we have equal masses and since H has a smaller molar mass than N, we have more moles of H than of N, thus the partial pressure of H is greater than that of N. d. Determine the ratio of Partial pressure of Partial pressure of hydrogen gas. Show all work. [3 points] nitrogen gas P. pressure of P. pressure of H N = mol H mol N = MM N MM H 8.0 = =

8 Hour Exam II Page No (con t) e. Determine the density of the mixture of nitrogen and hydrogen gases in g/l. Show all work. [4 points] METHOD I Assume 8.0 g of each N and H : g total This gives 1 mol of N and mol H = mol gas. V = nrt/p = (14.90 mol)( Latm/molK)(98 K)/(1 atm) = L d = mass/vol = g/ L = g/l METHOD II Assume 1 L n = PV/RT = (1 atm)(1 L)/{( Latm/molK)(98K)} = mol (mol H )/(mol N ) = mol; mol H + mol N = mol Solve for moles and then mass get g each or 0.154g/L f. You initiate a reaction between the nitrogen and hydrogen gases to form ammonia (NH 3 ) gas and let the reaction run to completion at constant temperature. i. Calculate the new pressure in the tank after the reaction is complete. Show all work. [4 points] The new pressure is compared to the original pressure of 1 atm, and due solely to the number of moles of gas after the reaction compared with before the reaction. Assume the 1:13.90 mol ratio as before: N (g) + 3H (g) NH 3 (g) I C E We have 1.90 moles after as compared with moles before. The pressure, then, goes down by a factor of 1.90/14.90, so the new pressure is atm. ii. How does density compare to that you found in part e? Determine the ratio of density of gases after the reaction. Support your answer/show all work. density of gases before the reaction [3 points] The density must be the same (that is the ratio = 1). Mass is conserved in a chemical reaction and the volume of the vessel does not change. Thus, the density remains constant.

9 Hour Exam II Page No. 8. Consider the combustion of ethanol as represented by the following balanced equation: C H 5 OH(l) + 3O (g) 3H O(l) + CO (g) a. Using the enthalpies of formation given below, determine H for the combustion of ethanol in units of kj. Show all work. [3 points] H f [C H 5 OH(l)] = 78 kj/mol H f [H O(l)] = 86 kj/mol H f [CO (g)] = 394 kj/mol 3mol( 86 kj/mol) + mol( 394 kj/mol) {1mol( 78 kj/mol)} = 1368 kj b. Explain your work above. Do not just state what you did, but explain the conceptual reasons behind why what you did works. In your explanation, explain what H f values represent and why you were not given the H f value for O (g). [7 points] H f values represent the change in enthalpy when we form 1 mole of a compound from its elements in their standard states. We aren t given the H f value for O (g) since this is in its standard state (the value for H f is taken as 0). We keep the sign the same as what is given for the products since we are forming the products. We change the sign for the reactants from what is given since we are using the reactants (opposite of forming) We have to multiply the values by the coefficients in the balanced equation since the values are in kj/mol.

10 Hour Exam II Page No. 9. (con t) c. You combust 1.50 g of ethanol in an excess of oxygen and all of the energy produced as heat is transferred to g of ice [H O(s)] at 10 C. Determine the final state (solid/liquid mixture, solid, or liquid) and temperature of the H O. Use the information below and show all work. [8 points] H fusion (water) = 6.0 kj/mol Specific heat capacity of ice =.03 J/g C Specific heat capacity of water = 4.18 J/g C Heat lost by ethanol = 1368 kj/mol (see part a): 1.50 g C H 5 OH x 1mol x g 1368 kj 1mol = kj available Heat gained by the ice: [raises temperature of ice to 0 C] + [melts ice to water] + [raises temperature of water] [raises temperature of ice to 0 C] = (100.0g)(.03 J/g C)(10 C) =.03 kj (thus at least some will melt) [melts ice to water] = (100.0 g x 1mol g )(6.0 kj/mol) = kj (thus all melts) [raises temperature of water] = (100.0 g)( 4.18 J/g C)(ΔT).03 kj kj + (100.0 g)( 4.18 J/g C)(ΔT) = kj ΔT = 1.8 C Thus, we are left with water at a temperature of 1.8 C.