7. The coffee cup allows for pv work because it allows for a change in volume.

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1 1. A black body radiator is a theoretically perfect body that absorbs all energy incident upon it (or produced within it) and then emits 100% of this energy as electromagnetic radiation. 2. First, it is important to note that this question is not as straightforward as it may seem if we do not know the temperature of the pan of water. Most students will assume the pan of water is cold. However, this is a good reminder that more careful thinking is usually rewarded on the MCAT. If the pot and water happened to be at the same temperature as the heating element then no heat exchange would occur. If the pan is hotter than the element (say the element isn t even turned on yet), then heat flow will occur in the opposite direction to that most students will propose. Assuming the pot is cold and the heating element is hot: Heat will be transferred from the element to the pan via conduction because they are in contact with one another. Heat will be transferred from the pan to the water via conduction because they are also in contact with one another. The water in the bottom of the pan will heat up first and that will cause this hotter part of the liquid to rise while the cooler liquid above sinks which is convection. Finally, any part of the system that is hotter than the environment will radiate heat into the air including the heating element itself, the pan, and the water. (You can feel the heat from a red hot burner without touching it.) 3. This is our first foray into pv work. PV work is the work necessary to produce an increase in volume. For example, when a sealed balloon is heated, the gases inside the balloon will expand and must do work on the rubber walls of the balloon and the air around it to accomplish this expansion. Because some of the heat energy added to the balloon was used for pv work, only the remaining portion of the heat will go toward increasing the average kinetic energy of the molecules (i.e., temperature). So, when heat enters a system, if the system is capable of volume change, heat can go to pv work, increased temperature, or both. For this reason, the addition of a certain amount of heat will NOT necessarily be exactly proportional to the resultant increase in temperature. If the system is not capable of changing volume then no pv work can be done, so all of the added heat will go toward an increase in temperature. 4. If the volume is held constant, then 100% of the energy added will go toward an increase in temperature. If the pressure is held constant the volume can still change and therefore some of the added heat will go toward pv work. If we think of heat capacity as the amount of energy we can add before the system increases by one temperature unit, it is fairly easy to see that the system capable of pv work will be able to absorb more heat before increasing by one degree Celsius or Kelvin. It is much like asking how many gallons of water can be added to Tank A vs. Tank B? Tank A and Tank B are both 5-gallon tanks, but Tank B is connected via a hose to a reserve tank that holds 2 gallons. So, you can add 5 gallons to Tank A before it is full (analogous to a one unit increase in temperature). However, you can add 7 gallons to Tank B before it is full (the reserve tank being analogous to pv work). We would therefore say that Tank B has the higher water capacity in terms of our analogy. This indicates that the constant

2 pressure heat capacity (allows for pv work; i.e., includes the 2-gallon reserve tank) will be more than the constant volume heat capacity (does not allow for pv work; i.e., no reserve tank) for the same system. 5. The heat capacity of Beaker 2 will be greater than that of Beaker 1 because there is more water available to absorb heat in Beaker 2. However, the specific heat capacity of water in both beakers will be identical specific heat capacity is an intensive property. 6. The bomb calorimeter is a sealed steel container, meaning volume cannot change. However, because volume is constant any gases produced or consumed during the reaction will change the pressure. The coffee cup calorimeter allows for an increase in volume of the solution inside the coffee cup, but remains at atmospheric pressure throughout. The bomb is constant volume, the coffee cup constant pressure. 7. The coffee cup allows for pv work because it allows for a change in volume. 8. In physics we defined the sign of work by stating that when the force and the displacement vectors are oriented in the same direction (i.e., lifting a box) then work is positive; when the force and displacement vectors are oriented in the opposite direction (i.e., lowering a box) work is negative. 9. An isolated system is a system for which neither mass nor energy can be exchanged with the surroundings. A closed system is a system that can exchange energy with its surroundings but not mass. Definition 2 treats the system and its surroundings separately. It is overtly stated that energy transfer is occurring between the system and the surroundings (i.e., work is being done on the system by the surroundings and/or heat is being added to the system from its surroundings). The total energy change of the system in case 2 would therefore be the sum of the energy absorbed and the work done on the system ( E = q + w). However, if we defined the system as BOTH the system receiving the heat/work AND its surroundings, we would then have an isolated system. The entire isolated system would not change in energy which is exactly what definition 1 implies. What were formerly considered the system and the surroundings in definition 2 are now just parts of the system as defined for definition 1. Therefore, heat or work can be transferred back and forth between parts of the system and the total energy of the entire isolated system would not change. To further illustrate, let s say that 10J of heat were transferred to the system and 10J of work were also done on the system by the surroundings. According to definition 2, E = 10J + 10J = 20J. According to definition 1, E system = 10J + 10J = 20J and E surroundings = -10J + (-10J) = -20J. Therefore E isolated system = 20J 20J = This is a very important definition, so make sure your students get this one down. We are aware of many, many MCAT questions that rely upon the examinee remembering that temperature is a measure of the average kinetic energy of the molecules in a substance. This is exactly what students need to think of when they think temperature. In fact, it won t hurt to simply equate them: Temperature = The average kinetic energy of the molecules. Many questions, for example, substitute a phrase such as an increase in the average kinetic energy of the molecules for the phrase increase in temperature. Most students who read that phrase will not immediately recognize it as another way of saying temperature.

3 11. Isobaric means constant pressure. Isothermal means no heat exchange (i.e., constant temperature). 12. We believe it is always more effective to use a conceptual understanding of something to answer a question rather than a mathematical or memorization-based approach. In this case, if enthalpy is positive, that indicates an unfavorable change. The reaction is endothermic and energy will need to be added to move from reactants to products. At the same time we are told that entropy is negative. A negative entropy means more order which is also an unfavorable change (things spontaneously move toward more disorder, NOT more order). So, is it likely that a process will be spontaneous if both the enthalpy change and the entropy change are unfavorable? Hardly. Therefore the sign of G will be positive and the reaction will not proceed spontaneously under standard state conditions. An alternative mathematical solution would plug the signs of H and S into the fundamental thermodynamic relation: G = H - T S. 13. This is the opposite of the case examined in the previous question. This time both the enthalpy change and the entropy change are favorable, so we would definitely expect a negative sign for Gibbs Free Energy, or Answer a). Once again, this could be proven by plugging the signs into the fundamental thermodynamic relation. 14. When a solute is dissolved into solution, entropy will always be positive. A solid is far more ordered than are ions in a solution. The increase in temperature is assumed to be the result of heat evolved from the reaction, so enthalpy change must be negative (i.e., exothermic). Again, these are two favorable state changes, so according to the fundamental thermodynamic relationship, Gibbs Free Energy must be negative and the reaction must be spontaneous. 15. If the value of K is greater than one then the reaction will be spontaneous. To be more precise, if we are starting at the standard state conditions, then we know the reaction will proceed spontaneously. We know this because the natural log of a number greater than one is positive, which will result in a negative Gibbs Free Energy per the equation given. If the value of K is less than one then the reaction will not proceed spontaneously from standard state conditions. We know this because the natural log of a number less than one is negative. This negative will counteract the existing negative in front of the equation and therefore the sign of Gibbs Free Energy will be positive. If K = 1, then G = 0 because the natural log of one is zero. Again, a key principle in this section is that kinetics is walled off from thermodynamics: the value of K will not tell us anything at all about the rate of the reaction. 16. A solution with a ph of 2 has [H + ] = 10-2, while a solution of ph 4 has [H + ] = 10-4, thus the ph 2 solution has 10 2 or 100 times the [H + ] of a solution with a ph of Starting with K w = [H 3 O + ][OH - ] = 10-14, we take the negative log of all terms, yielding: -logk w = -log[h 3 O + ] + -log[oh - ] = -log(10-14 ). The middle two terms come from the log rule that states: logab = loga + logb. The first term can be replaced with pk a by definition, the second term with ph by definition, the third term with poh by definition, and the fourth term by 14 because 14 is the log of This leaves: pk w = ph + poh = Because the acid almost fully dissociates, we know that the ratio of products over reactants would have to be greater than one. In general terms, an acid with K a greater than one or a pk a less than zero is considered strong, so this acid would clearly qualify as a strong acid.

4 19. An aqueous solution with a ph of 8 is basic, but that does NOT mean that it does not contain any hydrogen ions. In fact, the presence of hydrogen ions is easily verified by solving the formula ph = -log[h + ] for [H + ]. There are 1.0 x 10-8 moles of hydrogen ions per liter of this solution. It is classified as basic because it has fewer hydrogen ions than are found in neutral water and more hydroxide ions than are found in neutral water. 20. Acid B will give the largest drop in ph. The largest decrease in ph will be caused by addition of the most acidic of the two species. You could simply recognize from the equation K w = K a *K b that K b and K a are inversely related, and therefore recognize that the smaller K b represents the stronger acid (because K a and acid strength are directly related). You could also use the above equation, plugging in 1 x for K w, and solve for K a in both cases. Either method leads to the conclusion that the acid with a K b of 1 x is the stronger acid and will therefore lower ph to the greater extent. 21. We calculate ph and poh by taking the negative log of the concentration of H + ions and OH - ions respectively. Therefore, if we take the log of the concentration of the acid or base directly we are assuming that the molar concentration of the acid or base is equal to the molar concentration of hydrogen ions or hydroxide ions, respectively. This cannot be exactly true because some of both of these ions (10-7 M) are already present in water before addition of the acid or base. The assumption is usually safe because the molar concentration of the strong acid or base is usually many magnitudes larger than If the difference were smaller this would weaken the validity of our assumption. Finally, we are also assuming that the acid and base dissociate 100%. If not, even if the first assumption were true, we could not take the log of the concentration of the acid or base directly. Say, for example, that only 75% of the acid dissociated in solution this would be a significant difference in concentration compared to 100% dissociation. Assuming that very strong acids dissociate 100% is usually a safe assumption, although ion-pairing and other factors due reduce the effective concentration of ions. This seems like a very likely MCAT topic because they tend to ask many questions that require a comparison between how we assume things to be for calculation purposes and what they are actually like in reality (real vs. ideal fluids, real vs. ideal gases, ignoring air resistance, etc.). 22. The reason HF is not a strong acid and HCl is a strong acid is a matter of structure. Looking at the conjugate bases, F - is far less stable than is Cl - due to its smaller size. When a smaller molecule has to bear a full formal charge it experiences a greater charge density and therefore more instability. A larger atom such as chlorine can spread out this charge over a greater area. 23. Both observations are easily explained by Le Chatelier s principle. In case one, sodium benzoate dissociates to release benzoate ions, which shift the acid dissociation equilibrium for benzoic acid to the left. Similarly, ammonium chloride dissociates to release ammonium ions, which shift the base dissociation equilibrium for ammonium hydroxide to the left. 24. a) Hydrolysis of NaNO 2 will result in the reaction of a nitrite ion with water to form HNO 2 and hydroxide ion, INCREASING ph; b) hydrolysis of NH 4 Cl will result in reaction of NH 4 + with water to form NH 3 and H 3 O +, DECREASING ph; c) hydrolysis of NaF will result in fluoride ion reacting with water to form HF and OH -, INCREASING ph; d) hydrolysis of NaClO 2 will result in reaction of ClO 2 - with water to form HClO 2 and OH-, INCREASING ph; e) hydrolysis of CH 3 COONa will result

5 in reaction of acetate with water to form CH 3 COOH and OH -, INCREASING ph; f) hydrolysis of NaCl will result in Cl - reacting with water to form HCl and OH -, INCREASING ph. In summary, all of the options will increase ph except for option b). 25. The terminology used in the question infers that the strong base is in the beaker, which makes it the analyte. The base is titrated with the strong acid, meaning the acid is being added dropwise and is therefore the titrant. 26. Two equivalents of base can be neutralized by one equivalent of H 2 SO 4 because each sulfuric acid produces two equivalents of hydrogen ions in solution. 27. For all of these plots, the x-axis is volume of titrant (usually ml) and the y-axis is ph. Strong Acid with Strong base: A strong acid in the flask will result in the ph being low, around 1 to 2 depending on the [SA]. As strong base is added, the ph will stay low and slowly rise until the equivalence point is reached and then it will go straight vertical for around 6 ph units. The middle of the equivalence region will be ph 7. After the equivalence point, the plot will slowly go to higher ph as more strong base is added. Strong Base with Strong Acid: With strong base present, the ph will start high. As strong acid is added, it will slowly go to lower ph and then sharply drop when the equivalence point is reached. Again, the middle of the ph region will be ph 7. It will then slowly go to lower ph as more strong acid is added. Weak Acid with Strong Base: As with the strong acid titration, this one will start at a ph below 7, but since the weak acid only partially dissociates, the ph will be higher than it is for a strong acid, or usually around 3 to 5. As strong base is added, the ph will increase slowly through the buffer region and then go vertical at the equivalence point. The equivalence point won t cover as many ph units as it did for the strong acid titration and leading up to it and after it, the ph will more quickly rise than it did for the strong acid. The half way ph of the equivalence point will not be 7 as it is for a strong acid, it will be higher than 7. This is because the conjugate base of the weak acid is now present, creating a basic solution. After the equivalence point, the ph will slowly rise as more base is added. Weak Base with Strong Acid: This will look like the strong base/strong acid titration except the ph will not start as high, the equivalence region won t last for as many ph units, and the equivalence point will be at a ph below 7. The reasons for all of these are for like reasons given in the weak acid/strong base titration, except that they need to be applied to the weak base. 28. a) At the equivalence point of the titration of a strong acid with a strong base the [OH - ] will equal the [H + ] AND the [analyte] will equal the [titrant] in the flask; b) At the equivalence point of the titration of a strong acid with a weak base the [OH - ] does NOT equal the [H + ], but the [analyte] will equal the [titrant]; c) At the equivalence point of a titration between a weak acid and a strong base the [OH - ] will NOT equal [H + ], but the [analyte] will equal the [titrant]; d) At the equivalence point of a titration between a weak acid and a weak base the [OH - ] will NOT equal [H + ], but the [analyte] will equal the [titrant] (remember WA/WB titrations are rarely attempted or useful). The pattern is that the hydroxide and hydrogen ions will be equal at the equivalence point for any strong/strong titration, but NOT for any other titrations. The concentration of the analyte will equal the concentration of the titrant at the equivalence point for all titrations.

6 29. An indicator should be chosen that will change color at a ph as close to the equivalence point as possible. A titration of a weak acid with a strong base will have an equivalence point greater than seven on the ph scale. The first two indicators would change color before the equivalence point was reached, so phenylphthalein would be the appropriate choice. 30. At the half-equivalence point the concentrations of HA and A - are equal. Therefore, the ratio of [A - ]/[HA] must be one. When we plug this into the H-H equation we get: ph = pk a + log(1). The log of one is zero, so this term falls out, demonstrating that ph = pk a at the half-equivalence point. 31. The ph would change drastically when acid or base was added to it. There would be no weak acid or base to consume the added base or acid.