This preview shows page 1. Sign up to view the full content.
Unformatted text preview: S117 Homework 9 Out of 75 Points Chapter 9 (to turn in) 22. (out of 5 points) PV work is given by , where P is the external pressure. In this case, external pressure is constant, so . Since all of the energy released from the combustion is used as work, and since the external pressure is 650.torr, where w is negative since energy is released, so . Discovering the units of this volume can be done in several ways, but since we're given the initial volume in cm3, let's try to convert this volume into cm3 (or mL). One way of finding the unit relationship between J and torr is to compare gas constants written in the relevant units. , so . Using that conversion factor, . Now to significant figures. , so , so 24. (out of 5 points) . We are given what E and q are, so Now the same method can be used as in problem 22. , so problem 22, . . since P is constant, so . Now, similar to 32. (out of 10 points) In general, it is true that at constant pressure. Therefore, , so if , then . If , then , and if , then . Recall by Avogadro's law that at constant temperature and pressure , where n is the number of moles of gas. Assuming that temperature is constant in all these reactions, we can write , so if , then . If , then , and if , then . a. There are 2 moles of gaseous product and 2 moles of gaseous reactant, so , so we can predict that . b. Here, , so we can predict that . c. Here, , so we can predict that . 34. (out of 5 points) . . At constant pressure, Now , so 36. (out of 10 points) For a monatomic ideal gas, Step 1: , and . . . Finally, . (Since this process was at constant pressure, we also could have invoked rather than having to calculate q.) Step 2: By similar calculations, , , , and . (Since this process was at constant volume, we could have invoked rather than having to calculate q.) Step 3: By similar calculations, , , , and . Overall: We can add all of the individual quantities to get the overall quantities. . , , and . 40. (out of 5 points) The heat which leaves the metal all goes into the water; that is, gained by water is , and . . , so . . . The heat , where C is the specific heat capacity. Recognize that heat flows until the temperature is the same everywhere, so if the final temperature of the water is 18.3C, that is also the final temperature of the metal. Therefore, , so . 44. (out of 5 points) This is a constant pressure process, so . The heat which goes into the NH4NO3 is coming from the solution because we're assuming no heat loss from the calorimeter, so . Since the heat capacity of the solution is known, and since the mass of the solution is , so , . On a per mole basis, 48. (out of 5 points) a. The heat given off by the reaction is gained by the calorimeter, so , so capacity. Now . . Now . Note that the mass of the bomb is unknown, so the calculated heat capacity is not the specific heat 58. (out of 5 points) We need to combine the given reactions (by adding, subtracting, and multiplying them) until we have the reaction of interest. The reaction of interest has P4O10(s) as a reactant, so we need some reaction where that compound is a reactant. Let's reverse the second reaction so that where the sign of H was switched because the reaction was reversed. We also want 6PCl5(g) to be a reactant, so we can reverse the third reaction and multiply it by 6. We want our product to be 10Cl3PO(g), so we can multiply the last reaction by 10. Adding all of these together gives The 5O2 cancels from both sides, and some PCl3 cancels as well. Now we have We still need to get rid of 4PCl3, P4(s), and 6Cl2(g). We can add the first reaction to our current overall reaction. The first reaction is Adding that to our previous overall reaction gives We have cancellations so that which is exactly the reaction we wanted! H for this reaction is the sum of all the changes in enthalpies from the individual reaction, so 62. (out of 5 points) a. By Hess's law, of a reaction is , keeping in mind that the individual s must be multiplied by the stoichiometric coefficient of the compound. For example, of the first given reaction is . Using the values given in Appendix 4, . For the second reaction, the third reaction, b. Adding these reactions gives . For . Cancelling and combining some things, This is the overall Ostwald process. The overall reaction enthalpy is the sum of the individual changes in enthalpy, so adding together the Hs from part a., , so the reaction is quite exothermic. In reality, the reactants of the Ostwald process are just NH3 and O2, but your silly book doesn't tell you that. With that information, you could combine the equations in the following way to get the correct equation for the Ostwald process. Summing those equations gives All of the individual reactions used to make this overall reaction were exothermic, so we still conclude that the Ostwald process is exothermic. 70. (out of 5 points) The balanced combustion of ethene gas is given by By Hess's law, we know that the overall enthalpy of this reaction is given by , so . Extra: (out of 10 points) Calculating the standard enthalpy of combustion using bond dissociation energies is given by . As seen by the reaction given in problem 70, it is given in this case by . This is different from the combustion enthalpy given in problem 70. The differences can come from many sources. First of all, the combustion reaction in problem 70 was written with liquid water as a product, but the bond energies used for H2O were gaseous bond energies, and the sublimation enthalpy of water was not taken into account. Secondly, bond dissociation energies are averages over many different molecules, so, for example, the OH bond may have a bond dissociation energy of 460 in some alcohol and 470 in a different molecule, so applying those averages to a particular molecule will almost always introduce some error. ...
View
Full
Document
 Fall '11
 StephenJacobson

Click to edit the document details