Section A-Gases 1A. Circle the correct answer (18 pts total) (2 points each)

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1 Section A-Gases 1A. Circle the correct answer (18 pts total) (2 points each) (b) If a sample of gas is warmed in a rigid container (i.e., with fixed volume), which is true? i) the density of the gas remains the same ii) the pressure of the gas increases iii) the average distance between the gas particles remains the same iv) all of the above 2A. Write T (True) or F (False) for the following statements. If the statement is false correct it by changing, dropping, or adding a few words (6 points) c) For a fixed number of moles of an ideal gas at a constant temperature, when the pressure increases the volume will decrease and the molecular speed will increase. stay the same d) Gases are easier to compress than liquids because there is much more space between particles in a gas than in a liquid. T Free Response Questions--A 7A. (8 pts) A box of fixed volume contains two gases inside, A and B, at 35 C. Assuming the data in the plots are corrrect, state whether each statement below is true (T) or false (F). You do NOT need to correct it if it is false here!! _ F_(a) The particles of A have a greater average speed than those of B. _ T_(b) The partial pressure of A is greater than the partial pressure of B. _ T_(c) The molar mass of A is greater than the molar mass of B. _ F_(d) The particles of A have a greater average kinetic energy than the particles of B. speed 10A. (11 pts) (a) (7 pts) One industrial process for the removal of hydrogen sulfide from natural gas is its reaction with sulfur dioxide: 2 H 2 S(g) + SO 2 (g) 3 S(s) + 2 H 2 O(g) What volume of SO 2 at 5.0 atm and 252C will be used up in the production of 21 mol of S(s)? 1 mol SO 2 used 21 mol S 7.0 mol SO 2(g) used up. Assuming ideal gas behavior, 3 mol S produced this many moles of gas at 5.0 atm and 252 C will occupy a volume given by the ideal gas equation: # p a r t i c l e s A B T = 35C

2 V 7.0 mol L atm mol K ( ) K nrt L = 6.0 x 10 L P 5.0 atm (b) (4 pts) (not related to part (a) at all) A sample of a gas has a volume of 27.3 L at 134 K and 1.0 atm pressure. What will be the volume if the temperature becomes 222 K and the pressure becomes 2.5 atm? P T 1.0 atm 222 K V V P T (27.3 L) atm 134 K PV You could use T PV T L as well (equivalent mathematical statement

3 Section B- Thermochemistry 1B. (b) ii (c) ii It TAKES energy to break bonds; energy is RELEASED when bonds are made. If more energy is used up breaking bonds than the energy released making bonds, then the process overall will require energy (energy must flow IN for it to occur) and it will be endothermic. Since the number of bonds made in this case is two, and the number of bonds broken is also two, if the bonds in the reactants are stronger than the bonds in the products (i.e. bond energies of reactants are larger), it WILL take more energy to break the bonds in the reactants than the energy released upon making the bonds in the products, so the process will be endothermic. (e) v (2(90) 2(33) = 114); (f) ii 4B

4 6B. 10B. (a) (333 J/g)(9.2 g) = 3064 J needed (b) Answer: 60. C Energy flows out of the coffee in order to do TWO things: 1) melt the ice cube and 2) raise the T of the melted water at 0 C up to the final temperature. (It is the 2 nd one that I suspect many students would not think about, but if you did not include that part, I d certainly give you partial credit if you did the rest of the problem correctly.) The deltat of the coffee is related to the amount of energy that flows out of it. So, mathematically: q coffee = -(q flowing into ice to melt it + q flowing into the melted ice to raise its temperature to the final T ) (4.184 J/g C)(238g)(T f 65 C) = - [ (4.184 J/g C)(9.2g)(T f 0 C)] 996T f = T f 1034T f = T f = 60. C (which is reasonable (1 ice cube didn t cool down the coffee by that much)). NOTE: If you ignored the extra energy required to raise the T of the water from the melted ice [i.e., just assumed that the ice cube melted, and that was the only heat flow from the coffee], you d have gotten: (4.184 J/g C)(238g)(T f 65 C) = - [3064] (only approximately correct) 996T f = T f = = T f = 61.9 = 62 C

5 13B. Zero. The standard enthalpy of formation of any substance is the enthalpy change associated with forming ONE MOLE of a substance from its ELEMENTS (in their standard states and most stable forms). Since N 2 (g) is already an element (in its standard state and most stable form), then forming one mole of it from N 2 (g) is doing absolutely nothing (i.e., nothing changed), so there can be no change in enthalpy (H = 0) 13B (added) For each of the following substances, predict whether its standard enthalpy of formation is positive, negative, zero, or state that there not enough information to tell, and provide your reasoning. Strategy: In order to answer this, you need to write (or at least imagine ) what elements the substance is being formed from in the formation equation. So although it was not formally asked for in this problem, I have written those out in each case. If bonds are only broken in forming the substance from its elements (in their standard states and stable form), the enthalpy change is positive; if bonds are only made in forming the substance from its elements, the enthalpy change is negative. If both are involved, you cannot say without quantitative data. And if the substance is an element in its standard state, there is no process (no change at all physically) and thus the enthalpy change is obviously zero (see 13B above). Answers: (a) NO(g): ½ N 2 (g) + ½ O 2 (g) NO(g) Can t tell the sign b/c some bond making and some bond breaking. (b) Cl(g): ½ Cl 2 (g) Cl(g) Positive, b/c only thing happening is the breaking of a bond (c) Fe(s): Fe(s) Fe(s) (d) H 2 (l): H 2 (g) H 2 (l) Zero, b/c there is no change at all here; Fe(s) is the most stable form of iron in its standard state. Negative, b/c although H 2 (l) is an element, it is not in its standard state (at reasonable temperatures). This phase change (condensation) involves the bringing together of H 2 molecules (which attract one another), and so the process is energylowering/exothermic. (e) Mg(NO 3 ) 2 (s): Mg(s) + N 2 (g) + 3 O 2 (g) Mg(NO 3 ) 2 (s) Can t technically tell the sign (without some additional insight/knowledge) since there is both bond making and bond making occurring Section C-QM Model of the Atom (through electron configurations) 1C(a) iv (d) iii (photons of red light have lower energy than photons of green light, so the transition leading to the red emission must correspond to a smaller E between levels. The difference

6 between the 3 rd and 2 nd levels must be less than the difference between the 4 th and 2 nd levels since the higher the n value, the higher the (potential) energy. 2C(a) T The third ionization energy refers to removing the third electron out which would make Mg 3+. Many students don t think this through and think that changing third to second is correct. That is not true, so if you were thinking that, take some time to write out each ionization step and see if you can see why this is statement is true as stated. (b) F diamagnetic (c) T Blue light has photons with higher energy than yellow light. 5C. 7C. 8C. (a) (I ll give verbal descriptions here since I can t easily make boxes. (i) O: 2 electrons in the 1s; 2 electrons in the 2s; 2 electrons in one 2p; 1 electron in each of the other two 2p orbitals. (ii) O 2- : 2 electrons in the 1s; 2 electrons in the 2s; 2 electrons in all three 2p orbitals. (filled shell) (iii) Mg: 2 electrons in the 1s; 2 electrons in the 2s; 2 electrons in all three 2p orbitals. 2 electrons in the 3s. (iv) Mg 2+ : same as (b) (v) excited state of O: One of an infinite number is: 2 electrons in the 1s; 2 electrons in the 2s; 1 electron in each of the three 2p orbitals; 1 electron in the 3s (b) Ne

7 11C. 9C.(a) Small wavelength is associated with higher frequency and thus higher energy-per-photon light (E photon = h = hc/. What dictates the ability of light to eject electrons is its energy per photon (since it takes a certain minimum amount of energy to eject an electron from a surface).

ENTHALPY, INTERNAL ENERGY, AND CHEMICAL REACTIONS: AN OUTLINE FOR CHEM 101A

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