1. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson The change of state from liquid to gas for chloroform, formerly used as an anesthetic, has and a. (a) Is the change of state from liquid to gas favored or unfavored by  H? by  S? (b) Is the change of state from liquid to gas favored or unfavored at 35 °C? (c) Is this change of state spontaneous at 65 °C? (a) The  H does NOT favor this change of state (  H = positive), but the  S does favor the process. Since the two factors are not in agreement, we must use the equation for free-energy change to determine if the process is favored at a given temperature. Worked Example 8.1 Change of State: Enthalpy, Entropy, and Free Energy A process will be favored if energy is released (  H = negative) and if there is a decrease in disorder (  S = positive). In cases in which one factor is favorable and the other is unfavorable, then we can calculate the free-energy change to determine if the process is favored: When  G is negative, the process is favored. Analysis Solution

2. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Because  G is negative in this case, the change of state is favored at this temperature. Since the  G = positive, this change of state is not favored at 35 °C. (c) Repeating the calculation using the equation for free-energy change at 65 °C (338 K): (b) Substituting the values for  H and  S into the equation for free-energy change we can determine if  G is positive or negative at 35 °C (308 K). Note that we must first convert degrees celsius to kelvins and convert the  S from cal to kcal so the units can be added together. Continued Worked Example 8.1 Change of State: Enthalpy, Entropy, and Free Energy

3. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 8.2 Identifying Intermolecular Forces: Polar versus Nonpolar The intermolecular forces will depend on the molecular structure, what type of bonds are in the molecule (polar or non-polar), and how the bonds are arranged. Analysis (a) Since methane contains only bonds, it is a nonpolar molecule; it has only London dispersion forces. (b) The bond is polar, so this is a polar molecule; it has both dipole–dipole forces and London dispersion forces. (c) Acetic acid is a polar molecule with an bond. Thus, it has dipole–dipole forces, London dispersion forces, and hydrogen bonds. Solution Identify the intermolecular forces that influence the properties of the following compounds: (a) (b) (c)

4. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 8.3 Unit Conversions (Pressure): psi, Atmospheres, and Pascals Using the starting pressure in psi, the pressure in atm and pascals can be calculated using the equivalent values in appropriate units as conversion factors. Analysis STEP 1: Identify known information. STEP 2: Identify answer and units. STEP 3: Identify conversion factors. Using equivalent values in appropriate units, we can obtain conversion factors to convert to atm and pascals. STEP 4: Solve. Use the appropriate conversion factors to set up an equation in which unwanted units cancel. Solution A typical bicycle tire is inflated with air to a pressure of 55 psi. How many atmospheres is this? How many pascals?

5. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 8.4 Unit Conversions (Pressure): mmHg to Atmospheres Since the mercury level is higher in the arm open to the flask, the gas pressure in the flask is lower than atmospheric pressure (1 atm = 760 mmHg). We can convert the difference in the level of mercury in the two arms of the manometer from mmHg to atmospheres to determine the difference in pressure. Analysis The pressure in a closed flask is measured using a manometer. If the mercury level in the arm open to the sealed vessel is 23.6 cm higher than the level of mercury in the arm open to the atmosphere, what is the gas pressure (in atm) in the closed flask? The height difference (23.6 cm) is about one-third the height of a column of Hg that is equal to 1 atm (or 76 cm Hg). Therefore, the pressure in the flask should be about 0.33 atm lower than atmospheric pressure, or about 0.67 atm. Ballpark Estimate

6. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson The pressure in the flask is calculated by subtracting this difference from 1 atm: This result agrees well with our estimate of 0.67 atm. Ballpark Check Since the height difference is given in cm Hg, we must first convert to mmHg, and then to atm. The result is the difference in gas pressure between the flask and the open atmosphere (1 atm). Solution Continued Worked Example 8.4 Unit Conversions (Pressure): mmHg to Atmospheres

7. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 8.5 Using Boyle’s Law: Finding Volume at a Given Pressure This is a Boyle’s law problem because the volume and pressure in the cylinder change but the amount of gas and the temperature remain constant. According to Boyle’s law, the pressure of the gas times its volume is constant: Knowing three of the four variables in this equation, we can solve for the unknown. Analysis In a typical automobile engine, the fuel/air mixture in a cylinder is compressed from 1.0 atm to 9.5 atm. If the uncompressed volume of the cylinder is 750 mL, what is the volume when fully compressed? Since the pressure increases approximately 10-fold (from 1.0 atm to 9.5 atm), the volume must decrease to approximately 1/10, from 750 mL to about 75 mL. Ballpark Estimate

8. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson STEP 1: Identify known information. Of the four variables in Boyle’s law, we know P 1, V 1, and P 2. STEP 2: Identify answer and units. Our estimate was 75 mL. Ballpark Check STEP 3: Identify equation. In this case, we simply substitute the known variables into Boyle’s law and rearrange to isolate the unknown. STEP 4: Solve. Substitute the known information into the equation. Make sure units cancel so that the answer is given in the units of the unknown variable. Worked Example 8.5 Using Boyle’s Law: Finding Volume at a Given Pressure Continued Solution

9. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson STEP 1: Identify known information. Of the four variables in Charles’s law, we know T 1, V 1, and T 2. STEP 2: Identify answer and units. Worked Example 8.6 Using Charles’s Law: Finding Volume at a Given Temperature This is a Charles’s law problem because the volume and temperature of the air change while the amount and pressure remain constant. Knowing three of the four variables, we can rearrange Charles’s law to solve for the unknown. Analysis Solution An average adult inhales a volume of 0.50 L of air with each breath. If the air is warmed from room temperature (20 °C = 293 K) to body temperature (37 °C = 310 K) while in the lungs, what is the volume of the air exhaled? Charles’s law predicts an increase in volume directly proportional to the increase in temperature from 273 K to 310 K. The increase of less than 20 K represents a relatively small change compared to the initial temperature of 273 K. A 10% increase, for example, would be equal to a temperature change of 27 K; so a 20-K change would be less than 10%. We would therefore expect the volume to increase by less than 10%, from 0.50 L to a little less than 0.55 L. Ballpark Estimate

10. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson STEP 3: Identify equation. Substitute the known variables into Charles’s law and rearrange to isolate the unknown. STEP 4: Solve. Substitute the known information into Charles’s law; check to make sure units cancel. This is consistent with our estimate! Ballpark Check Continued Worked Example 8.6 Using Charles’s Law: Finding Volume at a Given Temperature

11. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 8.7 Using Gay-Lussac’s Law: Finding Pressure at a Given Temperature This is a Gay-Lussac’s law problem because the pressure and temperature of the gas inside the can change while its amount and volume remain constant. We know three of the four variables in the equation for Gay-Lussac’s law, and can find the unknown by substitution and rearrangement. Analysis STEP 1: Identify known information. Of the four variables in Gay-Lussac’s law, we know P 1, T 1 and T 2. (Note that T must be in kelvins.) STEP 2: Identify answer and units. Solution What does the inside pressure become if an aerosol can with an initial pressure of 4.5 atm is heated in a fire from room temperature (20 °C) to 600 °C? Gay-Lussac’s law states that pressure is directly proportional to temperature. Since the Kelvin temperature increases approximately threefold (from about 300 K to about 900 K), we expect the pressure to also increase by approximately threefold, from 4.5 atm to about 14 atm. Ballpark Estimate

12. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson STEP 3: Identify equation. Substituting the known variables into Gay-Lussac’s law, we rearrange to isolate the unknown. STEP 4: Solve. Substitute the known information into Gay-Lussac’s law; check to make sure units cancel. Our estimate was 14 atm. Ballpark Check Continued Worked Example 8.7 Using Gay-Lussac’s Law: Finding Pressure at a Given Temperature

13. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 8.8 Using the Combined Gas Law: Finding Temperature This is a combined gas law problem because pressure, volume, and temperature change while the amount of helium remains constant. Of the six variables in this equation, we know P 1, V 1, T 1, P 2, and V 2, and we need to find T 2. Analysis STEP 1: Identify known information. Of the six variables in combined gas law we know P 1, V 1, T 1, P 2, and V 2. (As always, T must be converted from Celsius degrees to kelvins.) Solution A 6.3 L sample of helium gas stored at 25 °C and 1.0 atm pressure is transferred to a 2.0 L tank and maintained at a pressure of 2.8 atm. What temperature is needed to maintain this pressure? Since the volume goes down by a little more than a factor of about 3 (from 6.3 L to 2.0 L) and the pressure goes up by a little less than a factor of about 3 (from 1.0 atm to 2.8 atm), the two changes roughly offset each other, and so the temperature should not change much. Since the volume-decrease factor (3.2) is slightly greater than the pressure-increase factor (2.8), the temperature will drop slightly. Ballpark Estimate

14. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson The relatively small decrease in temperature (38 °C, or 13% compared to the original temperature) is consistent with our prediction. STEP 2: Identify answer and units. STEP 3: Identify the equation. Substitute the known variables into the equation for the combined gas law and rearrange to isolate the unknown. STEP 4: Solve. Solve the combined gas law equation for T 2 ; check to make sure units cancel. Ballpark Check Continued Worked Example 8.8 Using the Combined Gas Law: Finding Temperature

15. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson STEP 1: Identify known information. We are given the room dimensions. STEP 2: Identify answer and units. STEP 3: Identify the equation. The volume of the room is the product of its three dimensions. Once we have the volume (in m 3 ), we can convert to liters and use the molar volume at STP as a conversion factor to obtain moles of air. STEP 4: Solve. Use the room volume and the molar volume at STP to set up an equation, making sure unwanted units cancel. Worked Example 8.9 Using Avogadro’s Law: Finding Moles in a Given Volume at STP We first find the volume of the room and then use standard molar volume as a conversion factor to find the number of moles. Analysis Solution Use the standard molar volume of a gas at STP (22.4 L) to find how many moles of air at STP are in a room measuring 4.11 m wide by 5.36 m long by 2.58 m high.

16. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson STEP 1: Identify known information. We know three of the four variables in the ideal gas law. STEP 2: Identify answer and units. STEP 3: Identify the equation. Knowing three of the four variables in the ideal gas law, we can rearrange and solve for the unknown variable, n. Note: because pressure is given in atm, we use the value of R that is expressed in atm: Worked Example 8.10 Using the Ideal Gas Law: Finding Moles This is an ideal gas law problem because it asks for a value of n when P, V, and T are known: n = PV/RT. The volume is given in the correct unit of liters, but temperature must be converted to kelvins. Analysis Solution How many moles of air are in the lungs of an average person with a total lung capacity of 3.8 L? Assume that the person is at 1.0 atm pressure and has a normal body temperature of 37 °C.

17. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Continued Worked Example 8.10 Using the Ideal Gas Law: Finding Moles STEP 4: Solve. Substitute the known information and the appropriate value of R into the ideal gas law equation and solve for n.

18. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson STEP 1: Identify known information. We know two of the four variables in the ideal gas law; V, T, and can calculate the third, n, from the information provided. STEP 2: Identify answer and units. STEP 3: Identify the equation. First, calculate the number of moles, n, of methane in the cylinder by using molar mass (16.0 g/mol) as a conversion factor. Then use the ideal gas law to calculate the pressure. STEP 4: Solve. Substitute the known information and the appropriate value of R into the ideal gas law equation and solve for P. Worked Example 8.11 Using the Ideal Gas Law: Finding Pressure This is an ideal gas law problem because it asks for a value of P when V, T, and n are given. Although not provided directly, enough information is given so that we can calculate the value of n (n = g/MW). Analysis Solution Methane gas is sold in steel cylinders with a volume of 43.8 L containing 5.54 kg. What is the pressure in atmospheres inside the cylinder at a temperature of 20.0 °C (293.15 K)? The molar mass of methane (CH 4 ) is 16.0 g/mol.

19. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 8.12 Using Dalton’s Law: Finding Partial Pressures According to Dalton’s law, the partial pressure of any gas in a mixture is equal to the percent concentration of the gas times the total gas pressure (750 mmHg). In this case, Analysis Humid air on a warm summer day is approximately 20% oxygen, 75% nitrogen, 4% water vapor, and 1% argon. What is the partial pressure of each component if the atmospheric pressure is 750 mmHg? Note that the sum of the partial pressures must equal the total pressure (within rounding error). Solution

20. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Worked Example 8.13 Heat of Fusion: Calculating Total Heat of Melting The heat of fusion tells how much heat is required to melt 1 g. To find the amount of heat needed to melt 0.300 mol, we need a mole-to-mass conversion. Analysis STEP 1: Identify known information. We know heat of fusion (cal/g), and the number of moles of naphthalene. STEP 2: Identify answer and units. Solution Naphthalene, an organic substance often used in mothballs, has a heat of fusion of 35.7 cal/g (149 J/g) and a molar mass of 128.0 g/mol. How much heat in kilocalories is required to melt 0.300 mol of naphthalene? Naphthalene has a molar mass of 128.0 g/mol, so 0.300 mol has a mass of about one-third this amount, or about 40 g. Approximately 35 cal or 150 J is required to melt 1 g, so we need about 40 times this amount of heat, or. Ballpark Estimate

21. © 2013 Pearson Education, Inc. Fundamentals of General, Organic, and Biological Chemistry, 7e John McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson Continued Worked Example 8.13 Heat of Fusion: Calculating Total Heat of Melting STEP 3: Identify the equation. First convert moles of naphthalene to grams using the molar mass (128 g/mol) as a conversion factor. Then use the heat of fusion as a conversion factor to calculate the total heat necessary to melt the mass of naphthalene. STEP 4: Solve. Multiplying the mass of naphthalene by the heat of fusion then gives the answer. The calculated result agrees with our estimate (1.4 kcal or 6.0 kJ) Ballpark Check