CHM 101 Mock Exam 4 Answer Key

CHM 101 Mock Exam #4 1. Select the correct equilibrium constant, Keq, expression for 3 H2(g) + N2(g) ⇌ 2 NH3(g). A. B. C. D. E. [NH3]2 / ([H2][N2]) [NH3]2 / ([H2]3[N2]) [NH3]2[H2]3[N2] [NH3] / ([H2][N2]) [NH3]2 / ([H2]3[N2]2) For a A + b B ⇌ c C + d D, Keq = ([C]c[D]d)/([A]a[B]b) 2. Consider 2 HI(g) ⇌ H2(g) + I2(g) at a specific temperature. If [HI] = 1.17 M, [H2] = 1.37 M, and [I2] = 0.100 M at equilibrium, what is the value of the equilibrium constant, Keq, under these conditions? A. 0.100 B. 1.17 C. 1.26 D. 8.54 E. 10.0 For 2 HI ⇌ H2 + I2, Keq = ([H2][I2]) / [HI]2 = [(1.37)(0.100)] / (1.17)2 = 0.100 3. Given below are equilibrium constants for different chemical reactions. Which one corresponds to a reaction that least favors the formation of products (i.e., which one is the most reactants-favored reaction)? A. 0.025 B. 1.0 C. 5.2 D. 2.4 × 105 E. 1.8 × 10–6 K (or Keq) equals to 1 (not zero) means that the amounts of reactants and products at equilibrium are equal. Therefore, if K is much greater than 1 (K >> 1), it means that there are more products than reactants at equilibrium (products are favored). K << 1 means that there are more reactants than products (reactants are favored). 0.0000018 is therefore the best answer to this question because even 0.025, which is less than 1, does not indicate a K value that is as reactants-favored as 0.0000018. 240000, on the other hand, is the most productsfavored equilibrium constant as opposed to 5.2. 4. Consider the following exothermic reaction initially at equilibrium: NO(g) + Cl2(g) ⇌ NOCl2(g). What change (shifting) will be observed if the amount of NO(g) in the reaction mixture were to increase? A. B. C. D. E. Right shifting; the concentrations of NO and Cl2 will increase No shifting; the concentration of NOCl2 will remain unchanged Right shifting; the concentration of NOCl2 will increase No shifting; the concentrations of NO and Cl2 will remain unchanged Left shifting; the concentration of NOCl2 will decrease 5. Consider the following exothermic reaction initially at equilibrium: NO(g) + Cl2(g) ⇌ NOCl2(g). What change (shifting) will be observed if the temperature of the reaction mixture were to decrease? A. B. C. D. E. Right shifting; the concentrations of NO and Cl2 will increase No shifting; the concentration of NOCl2 will remain unchanged Right shifting; the concentration of NOCl2 will increase No shifting; the concentrations of NO and Cl2 will remain unchanged Left shifting; the concentration of NOCl2 will decrease Both Questions 4 and 5 are about Le Chatelier’s principle. They have the same answer because in both cases, the reaction NO + Cl2 ⇌ NOCl2 is subject to stresses that would require a “right shift” meaning “producing more products” to offset the impacts of the stresses. In Question 4, the stress is an increase in the amount of a reactant (too much reactant). In Question 5, however, the stress is a decrease in -1- temperature. To “see” how a temperature change affects a chemical equilibrium, you must pay attention to whether the reaction is endothermic (heat-absorbing) or exothermic (heat-releasing). In this question, the reaction is exothermic (which is stated explicitly in the question statement), meaning that NO + Cl2 ⇌ NOCl2 + Heat. Lowering the reaction temperature means that heat is temporarily removed from the reaction. To this effect, a “product” (heat is a product) removed from an equilibrium necessitates a “right shift” to produce more products on the right side, including NOCl2 and heat, by lowering (or reacting) NO and Cl2 on the left side. 6. Factors that influence reaction rates include all of the following except ____. A. B. C. D. E. reaction temperature presence of a catalyst size of solid reactant particles concentrations of reactants size of the equilibrium constant What are the factors that affect the rate (or speed) of a chemical reaction that is far from equilibrium? The collision theory has informed us that the frequency of collisions between reactant molecules is directly proportional to the rate of the reaction between them. Hence, the reaction temperature, the concentrations of reactants, and the surface area of solid reactants (i.e., the smaller size of broken-up solid particles as opposed to a big chunk of a solid) all contribute to the rate of a chemical reaction. The presence of catalysts in a reaction can increase its speed by lowering the activation energy of the reaction. See Question 21 for an activation energy diagram. Questions 7-11 are concerning your knowledge of intermolecular forces. Intermolecular Forces: There are three types of intermolecular forces (IMF’s): London dispersion forces, dipole-dipole forces, and hydrogen bonding. Many more IMF’s exist but won’t be discussed in this course. Also, we are only going to focus on attractive IMF’s. Therefore, the stronger is an IMF, the more energy is required to overcome the IMF’s, and the more pronounced certain properties of a substance is. These properties depend on how strongly the particles of a substance are associated with each other: melting, boiling, the pressure of a substance’s vapor at equilibrium (called the “vapor pressure”), etc. are all good examples of these properties. Similar substances can be distinguished by their properties (such as boiling points), and differences in these properties can be attributed to differences in the strength of IMF’s operating in the respective substances. In each type of IMF’s, you need to know the nature of such IMF (why it exists), its relative strength to other types of IMF’s, and how to distinguish the strength of this type of IMF among several substances possessing such IMF’s. It is also true that these predictions we are making based on strengths of IMF’s are only “best guesses.” They are generally good enough at the CHM 101 level, and as such, the questions that you are expected to answer on this exam should not go beyond this level. Dipole-dipole attraction refers to the attractive forces that exist between polar molecules, such as H2O molecules. The polarity (measured by the dipole moment) of a substance determines the strength of dipole-dipole forces. Substances with higher dipole moments have stronger dipole-dipole forces (no calculation required – we will simply “accept” this based on face value). Dipole-dipole attraction is significantly stronger London dispersion forces (LDF’s), which refer to attractive forces between instantaneous dipoles. Since any substance can become instantaneous dipoles, LDF’s are found among all substances, including substances without dipoledipole attraction. Indeed, non-polar molecules attract each other exclusively by LDF’s. 7. Which of the following is expected to have the highest magnitude of London dispersion forces? A. H2 B. O2 C. N2 D. Cl2 E. F2 The strength of LDF’s depends on the substance’s size or mass. The higher mass possessed by the substance, the stronger is its LDF. The reason is that the polarity of an instantaneous dipole is given by uneven distribution of electrons within a molecule due to the electrons’ movements. For instance, when a negative charge is placed next to a molecule, the electrons of the molecule should move away from the negative charge and thus creating the instantaneous dipole. This instantaneous dipole can produce other instantaneous dipoles by induction. Since the polarity in an instantaneous dipole is relatively small, LDF is a weak IMF. Yet, the mass of a substance indicates the number of atoms (and the number of electrons) in the substance. The more electrons there are, the higher is the polarizability of the molecule, and the stronger the LDF’s are in the substance. 8. Which of the molecules shown below is/are polar? That is, which of the molecules will exhibit dipole-dipole forces of attraction in the liquid state? A. I and II B. I, II, and III C. I, III, and IV D. II only E. I and III See explanation before Question 7. 9. Which of the following is not an attractive force that acts between the individual molecules of H2O(l)? A. B. C. D. E. London dispersion forces Hydrogen bonding Dipole-dipole forces Covalent bonds All of the above are attractive forces that act between molecules of H2O(l) Hydrogen bonding is a special case of dipole-dipole attraction. Not all substances with dipoledipole attraction can have hydrogen bonding, but all hydrogen-bonding substances must have dipole-dipole attraction in the first place. We recognize hydrogen bonding by identifying within the structure of a molecule the presence of a hydrogen atom (electronegativity = 2.1) which is bonded directly to an element with an extreme electronegativity: N (3.0), O (3.5), or F (4.0). The difference in electronegativity of 0.9 or more is considered “pretty large,” thus creating unusually high polarity within the molecule. The hydrogen bond itself can be represented by “O—H ∙∙∙ O” where the O—H bond is a covalent bond in one molecule and O ∙∙∙ H is hydrogen bonding (a strong attractive force) between the hydrogen atom of one molecule and the oxygen atom of a different molecule (also bonded to its own hydrogen atom). Hydrogen bonding brings about unusually higher melting and boiling points of water relative to similar substances such as CO2 and CH4, and hence, hydrogen bonding is recognized as one of the strongest IMF’s. Small alcohol molecules can also undergo hydrogen bonding, but the strength of alcohol’s hydrogen bonds is generally not as high as water’s due water having two O—H bonds within a single H2O molecule: water can do twice as many hydrogen bonds as an alcohol molecule such as methanol or ethanol. See Questions 9, 10, and 11. 10. Which of the following can participate in hydrogen bonding? A. CH4 B. CHCl3 C. H2S D. H2O E. CH3OCH3 11. Which of the following statements is incorrect regarding intermolecular forces? A. B. C. D. E. Dipole-dipole forces are present in all atoms and molecules. Hydrogen bonding is considered one of the strongest intermolecular forces. London dispersion forces increase with increasing molecular size. For a substance to participate in hydrogen bonding, it must have hydrogen bonded to O, N, and F. Intermolecular forces are much weaker than intramolecular (bonding) forces. Questions 12-15 are concerning solid-liquid, liquid-liquid, and gas-liquid solubility and forces in a solid. 12. Which of the following statements is incorrect regarding the solution process for the dissolving of potassium iodide, KI, in water? A. B. C. D. E. In a KI solution, the ions interact with the water molecules through ion-dipole forces. Hydration is the process whereby water molecules surround solute particles. When an ionic compound dissolves, the ionic bonds break. Some of the hydrogen bonds among the water molecules must break. A hydrated iodide ion is surrounded by the partially negative end of the water molecules. Ionic compounds can dissolve in water because of the strong ion-dipole attraction. This question tests you on your understanding of how (in addition to why) ionic compounds can dissolve in water. When an ionic compound dissolves, how many ions of what kinds does it produce? Also, what is the step-by-step process that describes the solvation process applied to an ionic compound? For example, how do water molecules pull apart the ions in the compound: By surrounding each ion and pulling each away one at a time! The negative (or oxygen) ends of water molecules surround a cation; the positive (or hydrogen) ends of water molecules surround an anion. In answer choice E of this question, iodide is an anion (a negative ion); it can’t be surrounded by the negative ends of water molecules! 13. The bonding forces that hold copper, Cu, atoms together in the solid state are? A. Metallic B. Network covalent C. Dipole-dipole D. Covalent E. Ionic There are generally four different types of solids. You are probably most familiar with ionic solids. In a metallic solid, the atoms of a metal (or an alloy) are held together by an attractive force given by an array of metal atoms (without their valence electrons) surrounded by a “sea” of delocalized electrons. This is called metallic bonding. In a molecular solid, such as ice, molecules of a substance are held together by intermolecular forces. Since intermolecular forces are much, much weaker than intramolecular forces (chemical bonds), the melting and boiling points of molecular solids are expected to be much, much lower than that of metallic and ionic solids. Finally, in a network solid, such as diamond, the atoms are held together by covalent bonds (not IMF’s). Since covalent bonds are intramolecular forces (strong forces), the melting point of a network solid would be comparable to a metallic or ionic solid. 14. Why do we need a frequent and regular source of vitamin C in our diets but not of vitamin A? -4- CHM 101 Mock Exam #3 A. Vitamin A is polar and can be stored in our fat; vitamin C is nonpolar so it is soluble in body fluids. B. Vitamin A is nonpolar and can be stored in our fat; vitamin C is polar so it is soluble in body fluids and released in our urine. C. The –OH group on vitamin C make it very nonpolar so it dissolves in vitamin A. D. The long chain of carbon atoms makes vitamin A very polar so it is water-soluble. E. Vitamins A and C are both water-soluble. Enzymes are protein-based, biological catalysts. Vitamins, on the other hand, can be thought of as coenzymes (or compounds significant for the synthesis of coenzymes) that become vital for the functions of various enzymes. An enzyme can lose its ability to catalyze a reaction unless the coenzyme is present. So, “Why do we need a frequent and regular source of vitamin C in our diets but not of vitamin A?” The answer is in the solubility of vitamin C, which is polar and can readily dissolve in water and is excreted in the urine. Vitamin A, which is non-polar, can dissolve and be stored in our body fats. 15. Which of the following should be most soluble in ethanol, CH3CH2OH(aq)? A. Olive oil B. CH3CH3(g) C. CH3NH2(l) D. C6H6(l) E. I2(s) You must be able to determine the polarity of various substances to identify those that are polar/non-polar. Polar substances can dissolve only in a polar solvent (such as water). Nonpolar substances can dissolve only in a non-polar solvent (such as carbon tetrachloride or acetone). Questions 16-18 are concerning your knowledge of colligative properties of solutions. 16. What happens to the vapor pressure, boiling point, and freezing point of a pure liquid solvent when a solute is dissolved in the liquid to form a solution? A. B. C. D. E. Vapor pressure increases, boiling point decreases, and freezing point decreases Vapor pressure decreases, boiling point increases, and freezing point decreases Vapor pressure decreases, boiling point increases, and freezing point increases Vapor pressure increases, boiling point decreases, and freezing point increases Vapor pressure increases, boiling point increases, and freezing point increases Salt is applied to icy roads to melt the ice by lowering the freezing point of water. This is freezing point depression. Boiling point elevation and vapor pressure lowering can be understood together. When salt is added to boiling water, the temperature of boiling water increases (this is boiling point elevation). It means that at the original boiling point of water, salt water does not boil because it has a lower vapor pressure (this is vapor pressure lowering). 17. Which of the following aqueous solutions is expected to have the highest boiling point? A. 1.0 M MgCl2 B. 1.0 M FeCl3 C. 1.0 M NaCl -5- D. 1.0 M CH3OH E. All the same This question is about the van’t Hoff factor. When an ionic compound dissolves to make a solution, the compound dissociates into multiple dissolved ions. NaCl gives two ions, but MgCl2 gives three ions, and so on. The effect of dissolving MgCl2 is therefore greater than dissolving NaCl because MgCl2 produces more significant colligative properties. FeCl3 gives four ions, and hence, FeCl3 is the correct answer to this question. The number of ions produced by dissolving an ionic compound is known as the van’t Hoff factor. The more ions there are, the greater is the colligative property effect (it applies to all four colligative properties – boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure (described in the next question)). 18. Extracellular (outside the cell) fluid with a lower concentration of solutes relative to the cellular concentration of solutes is considered hypotonic to red blood cells. What will happen to red blood cells that are placed in a hypotonic solution? A. B. C. D. E. The cells will shrink because there will be a net flow of water to the outside of the cell. The cells will expand because there will be a net flow of water to the inside of the cell. The cells will shrink because there will be a net flow of water to the inside of the cell. The cells will expand because there will be a net flow of water to the outside of the cell. Nothing will happen because the cells are impermeable. Imagine two solutions separated by a semi-permeable membrane (or imagine one solution inside a sac enclosed by a semi-permeable membrane while another solution is outside). Any difference in concentrations between the two solutions may result in the solute from the solution with higher concentration of the solute moving across the membrane to the solution at a lower concentration of the solute. The “pressure” that must be applied to “stop” the solute from moving across the membrane is called “osmotic pressure.” Osmotic pressure can be measured. Alternatively, the solvent (water) can also move instead of the solute. In this case, the solvent moves from the solution at the lower concentration to the solution with the higher concentration (to dilute the more concentrated solution). For example, an extracellular fluid that is hypotonic (lower in concentration) to a red blood cell means that water should flow into the red blood cell. The red blood cell will expand, as a result. Perform the calculations in Questions 19-20. 19. What is the percent-by-mass concentration of KCl in a solution that is prepared by adding 13.0 g of KCl to 85.0 g of water? A. 13.3% B. 15.3% C. 17.2% D. 65.4% E. 72.0% The percentage by mass is defined to be the mass of a solute divided by the mass of the entire solution (expressed as a percentage). Hence, 13.0 / (13.0 + 85.0) = 0.133 = 13.3%. 20. The solubility of an unknown compound at 20°C is 31.0 g per 100.0 g of water. What is the maximum amount of this compound that will dissolve in 200.0 g of water at the same temperature? A. 3.23 g B. 0.150 g C. 31.0 g D. 62.0 g E. 310 g Using the solubility of a substance (31.0/100.0) as a conversion factor, the question asks for the maximum amount of that substance that can dissolve in 200.0 g of solvent (i.e., twice as much solvent). Hence, 200.0 x (31.0/100.0) = 62.0 or 62.0 g of the unknown compound. Answer Questions 21-25 based on the provided figure in each question. 21. Which Roman numeral in the figure below corresponds to a reaction’s activation energy with a catalyst? -6- A. I only B. II only C. IV only D. III only E. None of the above In an energy diagram, the difference in energy between the reactants and products would tell you whether the reaction is endothermic (products energy > reactants energy) or exothermic (reactants energy > products energy), but the activation energy is always there as an addition to the reactants energy, which represents a “barrier” in energy for reactants to become products. Lowering of activation energy occurs to allow the reaction an “easier” pathway to turn reactants to products. Hence, the reaction speed increases. See Question 6. 22. According to the figure below, the normal (1.0 atm pressure) boiling point of propane (C3H8) is about ____. A. –110ºC B. –28ºC C. –42ºC D. –65ºC E. 42ºC 23. Refer again to the figure in Question 22. At lower temperatures, the vapor pressure of propane is ____. A. B. C. D. E. higher lower the same at all temperatures lower than butane (C4H10) lower than pentane (C5H12) Propane always has higher (not lower) vapor pressure than butane and pentane regardless of temperature because of lower LDF’s (London dispersion forces). See Question 7 and its explanations of LDF’s. On the other hand, at lower temperatures, vapor pressures are lower regardless of which substance. 24. A heating curve can be used to show what happens to a particular substance as heat is added. The curve usually starts at a low temperature (Point A), and the substance is a solid. As heat is added continually, -7- warming and phase changes occur until the substance eventually becomes a gas (Point F). What process occurs in the B-C region in the figure? A. B. C. D. E. Converting the solid state of the substance to its liquid state Converting the liquid state of the substance to its gaseous state Warming of the gaseous state as heat is added Warming of the liquid state as heat is added Cooling of the solid state as heat is removed It is a Heat vs. Temperature diagram (or a heating curve) for a typical substance. There are two phase transitions: melting and boiling. The main idea is that as energy (heat) is added, the temperature of a substance increases. During melting, however, a solid gains heat but does not have an increase in temperature. The added energy is used instead to overcome the IMF’s among the particles of the solid. Once sufficient energy is provided, the entire solid is melted and becomes a liquid. The same phenomenon is observed at boiling, where the added energy is used to turn the liquid- into gas-phase of the substance. At other incidences where no melting or boiling occurs, the added energy simply raises the temperature of a solid/liquid/gas. You can understand the cooling curve (same as heating curve) by reversing the direction of change from right to left on “Heat Added” to make it “Heat Removed” instead. 25. What phase change is occurring in the figure, and is it endothermic or exothermic? A. B. C. D. E. Freezing; exothermic Melting; exothermic Freezing; endothermic Melting; endothermic Boiling; endothermic During melting (see the heating curve for diagrams depicting such a molecular-level change), the process is endothermic. During boiling, the process is also endothermic. However, during freezing and condensing, energy must be removed from a substance, and hence, either of these processes must be exothermic. -8-