: : Use simple structure and bonding models to account for the following. The bond lengths in CO 3

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1 Chem 55 Problem Set #2 Spring 200 Name TA Name Lab Section # ALL work must be shown to receive full credit. Due at the beginning of lecture on Friday, February 2, 200. PS2.. Draw two resonance structures for the nitrite ion, NO 2. - N N - O O O O b) What is the hybridization around the nitrogen atom in this polyatomic ion? The hybridization on the nitrogen atom is sp 2. c) Describe the pi-bonding in the polyatomic ion. σ N O π σ O The pi bonding in this molecule extends over all three atoms. Each atom contributes a p atomic orbital to the extended pi system. d) Label the sigma and pi bonds as either localized or delocalized. The sigma bonds, as shown as lines (bonds) between the nitrogen and O oxygen atoms, are localized N between each N O bond. The bond is extended over all three O atoms through the p orbitals. PS2.2. Use simple structure and bonding models to account for the following. The bond lengths in CO 3 2 are all identical and are shorter than a carbonoxygen single bond. O O C O CO 3 2 can be drawn in three possible resonance hybrid forms. Each atom has an atomic p orbital oriented so there is an extended system for the 2 electrons. There are 2 electrons in the system spread over the three bonds so there is 0.3 of a bond for each C O or an equivalent to a bond order of.3. The larger bond order means the bond distance will be less than an C O single bond. CHEM 55 Spring 200

2 PS2.3. In the boxes below diagram the specified system as viewed at the atomic level in the space provided. Be sure to clearly label each of the substances in your diagram. Label Area A sample of oxygen at 25 C A sample of Br 2 liquid A sample of sodium at 25 C PS2.4. Explain what the terms heat of fusion and heat of vaporization mean. In your explanation include the symbol for each term and provide a chemical equation describing the fusion and vaporization process. Also explain how you could calculate the heat of fusion or heat of vaporization using the table of data in Appendix B on page A-5. The heat of fusion is the amount of heat required to convert one mol of solid at its normal melting point to one mol of liquid. The heat of vaporization is the amount of heat required to convert one mol of liquid at its normal boiling point to one mol of gas. A chemical equation which symbolizes the fusion process is; H 2 O(s) H 2 O(l) We have to be a little careful here. Fusion is the phase change from liquid to solid (freezing), but the heat of fusion is generally thought of a the heat required to convert the solid into it liquid. A chemical equation which symbolizes the vaporization process is; H 2 O(l) H 2 O(g) To calculate either the heat of fusion or the heat of vaporization we can use the relationship; H rxn = S( H f(products) - S( H f(reactants) So we can find the H f for reactants and the products in Appendix B. Finding the difference is the heat of fusion or heat of vaporization depending on the phases we use. CHEM 55 2 Spring 200

3 PS2.5.a) How much heat is produced when 75.0 g of steam at 35 C is converted to water at 20.0 C? Step ) steam at 35 C to steam at 00 C. Step 2) steam at 00 C to liquid at 00 C. Step 3) liquid at 00 C to liquid at 20 C. ).84 g C x 75.0 g x 35 C = 4827 (4.83 x 03 ) 2) 2259 g x 75.0 g = 69,425 (.69 x 05 ) 3) 4.84 g C x 75.0 g x 80 C = 25,04 (2.5 x 04 ) b) How much heat is required to convert 30.0 g of ice at -0.0 C to steam at 05.0 C? Step ) solid at -0 C to solid at 0 C. Step 2) solid at 0 C to solid at 0 C. Step 3) liquid at 0 C to liquid at 00 C. Step 4) liquid at 00 C to steam at 00 C. Step 5) steam at 00 C to steam at 05 C. ) 2.09 g C x 30.0 g x 0 C = 627 (6.27 x 02 ) 2) 334 g x 30.0 g = 0,020 (.00 x 04 ) 3) 4.84 g C x 30.0 g x 00 C = 2,552 (.26 x 04 ) 4) 2259 g x 30.0 g = 67,770 (6.78 x 04 ) 5).84 g C x 30.0 g x 5 C = 276 (3 x 02 ) = 9.2 k PS2.6. Ethyl alcohol melts at -4 C and boils at 78 C. The enthalpy of vaporization for ethyl alcohol at 78 C is 870 g and the enthalpy of fusion is 09 g. If the specific heat of solid ethyl alcohol is taken to be 0.97, and that for the g. C liquid 2.3, how much heat is required to convert 0.0 g of ethyl alcohol g. C at -20 C to the vapor phase at 78 C? Step ) solid at -20 C solid to -4 C soln Step 2) solid at -4 C to liquid at -4 C Step 3) liquid -4 C to liquid 78 C Step 4) liquid at 78 C to vapor at 78 C ) 0.97 x 0.0 g x 6 C g. C = ) 09 g x 0.0 g = 090 3) 2.3 x 0.0 g x 92 C g. C = 4,46 4) 870 g x 0.0 g = 8700 Heat required = 4 k CHEM 55 3 Spring 200

4 PS2.7. Define the term equilibrium vapor pressure. the pressure due to particles of a substance in the vapor phase above its liquid in a closed container at a given temperature. b) Use a vapor-pressure table (check the Database link on the class web site) to look up the equilibrium vapor pressure of a sample of water at 95 C and at 83 C. The vapor pressure of water at 95 C is mmhg and at 83 C the vapor pressure is mmhg. c) Consider two closed containers each partially filled with liquid water one at 95 C and the other at 83 C. Can the pressure of water vapor in the gas phase in either container ever exceed the equilibrium vapor pressure at the particular temperature? Explain why or why not. No. At a given temperature we cannot have a pressure due to the vapor above a liquid greater than the equilibrium vapor pressure. If we attempt to add additional water, in the vapor phase, to a system already at equilibrium, the rate of condensation increases until the vapor pressure re-establishes equilibrium. The net result is there is no change in the vapor pressure. PS2.8. A sample of water in the vapor phase (no liquid present) in a flask of constant volume exerts a pressure of 635 mm Hg at 00 C. The flask is slowly cooled. a) Assuming no condensation, use the Ideal Gas Law to calculate the pressure of the vapor at 95 C; at 83 C P 2 = P T C P 2 = P T 2 = = 635mmHg 368K 373K = 626 mmhg 635mmHg 356K 373K = 606 mmhg b) Will condensation occur at 95 C; 83 C? Since the calculated pressure of the sample in the vapor at 95 C is less than the equilibrium vapor pressure no condensation occurs. Condensation occurs at 83 C because the calculate pressure exerted by the vapor is greater than the equilibrium vapor pressure at that temperature. c) On the basis of your answers in a) and b), predict the pressure exerted by the water vapor at 95 C; at 83 C. d) The pressure due to the vapor at 95 C is 626 mmhg, at 83 C the vapor pressure is mmhg. Can you determine the volume of water which has condensed at 83 C? CHEM 55 4 Spring 200

5 PS2.9. ln (vapor pressure) Consider the following data for the vapor pressure Lithium Magnesium T (K) P v (mm Hg) T (K) a) Use graphing software (Microsoft Excel) to plot ln (P v ) vs. T for each metal and use your graph to determine the slope of the best line through the data. The heat of vaporization of a liquid can be obtained from such a plot. The relationship is given as, H vap slope = 8.34 mol K Calculate the heat of vaporization for lithium and magnesium. (Note Be sure to clearly label the graph.) Slope = 566 K Vapor Pressure Data for Lithium H vap slope = mol K y = -566x K 8.34 mo H vap = 96.2 k /Temperature (K) ln(vp) Linear (ln(vp)) ln (vapor pressure) 7 6 Vapor Pressure of Magnesium y = x ln(vp) 3 Linear (ln(vp)) /Temperature (K) Slope = 934 K H vap slope = 8.34 mol K 934 K 8.34 H vap = 77.4 k mol CHEM 55 5 Spring 200

6 b) In which metal is the bonding stronger? Lithium has stronger bonding than magnesium on the basis of the H vap. That lithium has the higher H vap suggests the stronger bonding. From data in another reference the H vap for lithium is 48 k and 27 k for magnesium. Either the data that I used is funky or such data for metals does not fit the ln (P v ) vs. T relationship as well. c) Determine the temperature of a sample of lithium and of magnesium when the vapor pressure is 350 mmhg. Lithium y =.6 x 0 4 x or ln(vp) =.6 x 0 4 T ln(350) =.6 x 0 4 T =.6 x 0 4 T x 0 4 T = T = x 0 4 = 8.22 x 0 4 T =.22 x 0 3 K Magnesium y = 9.3 x 0 3 x or ln(vp) = 9.3 x 0 3 T ln(350) = 9.3 x 0 3 T = 9.3 x 0 3 T x 0 3 T = T = x 0 3 =.02 x 0 3 T = 9.76 x 0 2 K CHEM 55 6 Spring 200

7 d) Determine the vapor pressure of a sample of lithium and of magnesium at 800. C. Lithium y =.6 x 0 4 x or ln(vp) =.6 x 0 4 T ln(vp) =.6 x ln(vp) = 4.59 e ln(vp) = e 4.59 vp = 98.4 mmhg Magnesium y = 9.3 x 0 3 x or ln(vp) = 9.3 x 0 3 T ln(vp) = 9.3 x ln(vp) = 6.32 e ln(vp) = e 6.32 vp = 557 mmhg PS2.0a. The normal boiling point of acetone, (CH 3 ) 2 CO is 56.2 C and its H vap = 32.0 k mol. Draw a Lewis structure for acetone and calculate the temperature at which acetone has a vapor pressure of 570. mmhg. ln P P 2 = - H ( R - ln = 8.34 T 2 ) mol ( K) mol K ln (0.750) = ( - ) = ( - ) x 0-5 = ( - ) = 3. x 0-3 = 32 K Lewis structure CHEM 55 7 Spring 200

8 b) Using data in part a of this problem, calculate the vapor pressure of acetone when the temperature is 5 C. ln P P 2 = - H ( R - VP ln 760 = 8.34 mol T 2 ) mol K ln VP 760 = (4.35 x 0-4 ) ln VP 760 = ( K) eln VP = e VP 760 = 0.88 vapor 5 C = 43 mm Hg CHEM 55 8 Spring 200