CHEM 123 SAPLING LEARNING CHAPTER 12

Calculate the heat energy released when 12.7 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point. Constants for mercury at 1 atm heat capacity of Hg(l) 28.0 J/(mol⋅K) melting point 234.32 K enthalpy of fusion 2.29 kJ/mol 𝑞p=

0.258 kJ Solution The number of moles, 𝑛, of mercury is 𝑛=(12.7 g)(1 mol Hg/200.6 g Hg)=0.0633 mol The amount of energy as heat, 𝑞P, released when 0.0633 mol Hg(l) is cooled from 298.15 K to 234.32 K is 𝑞P=𝑛𝐶P(𝑇2−𝑇1) =(0.0633 mol)(28.0 Jmol⋅K)(−63.83 K) =−113 J =−0.113 kJ The energy as heat released when 0.0633 mol of mercury freezes at its melting point is 𝑞P=−𝑛Δ𝐻fus=−(0.0633 mol)(2.29 kJmol)=−0.145 kJ The total heat released is (−0.113 kJ)+(−0.145 kJ)=−0.258 kJ The negative sign indicates that 0.258 kJ of heat is released by the mercury during this process.

A 37.02 g sample of a substance is initially at 28.2 °C. After absorbing 2727 J of heat, the temperature of the substance is 161.0 °C. What is the specific heat (𝑐) of the substance? 𝑐=

0.555 Jg⋅°C Solution The specific heat of a substance, 𝑐, is the amount of heat required to increase the temperature of 1 g of that substance by 1 °C. It is calculated using the equation 𝑐=𝑞/(𝑚×Δ𝑇) where 𝑞 is heat, 𝑚 is mass, and Δ𝑇 is the change in temperature. The change in temperature is equal to the final temperature minus the initial temperature. 𝑐=𝑞/(𝑚(𝑇final−𝑇initial)) Insert the values from the question into the equation and solve for the specific heat. 𝑐=2727 J/((37.02 g)(161.0 °C−28.2° C))=0.5547Jg⋅°C

How many atoms are in a simple cubic (primitive cubic) unit cell? number of atoms: How many atoms are in a body‑centered cubic unit cell? number of atoms: How many atoms are in a face‑centered cubic unit cell? number of atoms:

1 2 4 Solution Each unit cell is a cube. A cube has eight corners, each of which is occupied by an atom. If several of these cubes were stacked side‑by‑side, front‑to‑back, and top‑to‑bottom, each corner would be shared by a total of eight cubes. Thus, only 1/8 of a particular corner atom is inside any given cube. The total number of atoms in a simple cubic (primitive cubic) unit cell is 8×(1/8)=1 atom . a six sided cube with a sphere on each corner and a line along each edge of the cube connecting the center of each sphere with the centers of three other spheres In a body‑centered cubic unit cell, there is one body atom in the very center in addition to the eight partial atoms at the corners of the cube. The body atom is completely inside the cube and is not shared with neighboring cubes. The total number of atoms in a body‑centered cubic unit cell is therefore 8×(1/8)+1=2 atoms . a six sided cube with a sphere in the center of the cube, a sphere on each corner, and a line along each edge of the cube connecting the center of each corner sphere with the centers of three other corner spheres In a face‑centered cubic unit cell, there is one atom located on each of the six faces in addition to the eight partial atoms in the corners of the cube. If several face‑centered cubic unit cells were stacked in all directions, each face would be shared by a total of two cubes. Thus, only 1/2 of a particular face atom is inside any given cube. The total number of atoms in the face‑centered cubic unit cell is 8×(1/8)+6×(1/2)=4 atoms . A six sided cube with a sphere in the center of each face, a sphere on each corner, and a line along each edge of the cube connecting the center of each corner sphere with the centers of three other corner spheres. Dashed lines connect each face sphere with the sphere on the opposite face.

The phase diagram for carbon is shown. PICTURED: A carbon phase diagram with four phases. The x axis is labeled T in Kelvins and starts at zero and ends at six thousand. The y axis is labeled P in atmospheres and starts at zero then increases to one, then increases again until ending at ten to the sixth power. A right leaning slightly curved vertical line splits the graph at four thousand K. The phases on the left of the vertical line are solid 1 diamond in the upper left and solid 2 graphite in the lower left. The two solid phases are split by a horizontal line between ten to the fourth power atmospheres and ten to the fifth power atmospheres, and runs to the vertical line. The phases on the right of the vertical line are liquid in the upper right and gas in the lower right. They are seperated by a horizontal line between ten to the second power atmospheres and ten to the third power atmospheres. Which phases are present at the upper triple point? P= T= Which phase is stable at 105 atm and 1000 K?

1 * 10^5 4000 Solution A triple point is a temperature and pressure at which three phases exist in equilibrium. The triple point for diamond, graphite, and liquid carbon occurs at roughly 105 atm and 4300 K. More dense materials usually tend to be more stable at higher pressures. Therefore, considering that increasing the pressure will cause graphite to become diamond, diamond is the more dense form of carbon.

The vapor pressure of ethanol, CH3CH2OH, at 35.0 °C is 13.67 kPa. If 2.28 g of ethanol is enclosed in a 2.50 L container, how much liquid will be present? mass of liquid:

1.67g Solution Vapor pressure is the pressure of a gas in dynamic equilibrium with its liquid. To determine how much of the ethanol is in the liquid phase, you must determine how much of the ethanol is in the gas phase. Use the ideal gas equation to determine the moles of ethanol in the gas phase 𝑃𝑉=𝑛𝑅𝑇 where 𝑃 is pressure, 𝑉 is volume, 𝑛 is number of moles, 𝑅 is the ideal gas constant (0.08206 L⋅atmmol⋅K), and 𝑇 is temperature. Convert the pressure from kilopascals to atmospheres so that the units match the ideal gas constant. To convert from kilopascals to atmospheres, use the conversion factor 1 atm=101.325 kPa. 13.67 kPa×1 atm/101.325 kPa=0.1349 atm Convert the temperature from degress Celsius to Kelvin so that the units match the units of the ideal gas constant. 𝑇(K)=𝑇(°C)+273.15=35.0 °C+273.15=308.15 K Next, use the ideal gas law to calculate the number of moles of ethanol in the gas phase. 𝑛=𝑃𝑉/𝑅𝑇 𝑛=((0.1349 atm)(2.50 L))/(0.0821 L⋅atmmol⋅K)(308.15 K)=0.0133 mol Determine the mass of ethanol gas using the molar mass of ethanol. 0.0133 mol×46.07 g/1 mol=0.614 g The amount of liquid present is equal to the total amount of ethanol minus the amount in the gas phase. 2.28 g ethanol−0.614 g gas=1.67 g liquid ethanol

Consider a hexagonal close‑packed unit cell, as shown in the image. An equilateral hexagonal prism has a sphere on each corner and in the center of both the top and the bottom six sided face. Lines connect each corner sphere to two adjacent corner spheres and the center sphere on the same face and a corner sphere on the opposite face. Three spheres form an equilateral triangle inside the prism, centered between the hexagonal faces. How many corner atoms are shown in the image? number of corner atoms: 12 What fraction of each corner atom is inside the boundaries of the cell? fraction inside the cell: 1/6 How many face atoms are shown in the image? number of face atoms: 2 What fraction of each face atom is inside the boundaries of the cell? fraction inside the cell: 1/6 Additionally, there are three atoms that are 100% inside the boundaries of the unit cell. If you sum all the whole and fractions of atoms, how many atoms are actually inside a hexagonal close‑packed unit cell? number of atoms inside the cell: 6

12 1/6 2 1/2 Solution A hexagonal prism has 12 corners. Each corner is occupied by an atom. If several of these hexagonal prisms were stacked side to side, front to back, and top to bottom, each corner would be shared by a total of six prisms. Thus, only one‑sixth of a particular corner atom is inside any given hexagonal prism. The hexagonal prism has eight faces, but only two faces are occupied by an atom. Even when several hexagonal prisms are stacked together in all directions, each face atom is only shared with one other prism. Thus, one‑half of a face atom is inside any given hexagonal prism. The three atoms in the middle are completely inside the unit cell. Summing the atoms and fractions of atoms, the total number inside the unit cell is 12×(1/6)+2×(1/2)+3=6 atoms.

How much heat energy is required to boil 96.6 g of ammonia, NH3? The molar heat of vaporization of ammonia is 23.4 kJ/mol. 𝑞=

133 kJ Solution Heat energy must be supplied to boil ammonia. This heat energy is used to overcome the intermolecular forces that maintain liquid ammonia. This overcoming of the intermolecular forces leads to a phase change, the conversion from liquid to gaseous ammonia in this case. The amount of heat necessary to boil ammonia depends on the amount of ammonia and the molar heat of vaporization. The molar heat of vaporization of ammonia is the amount of heat necessary to convert 1 mol of liquid ammonia to gaseous ammonia. In other words, the molar heat of vaporization is a quantitative measure of how readily ammonia is boiled. The amount of heat (𝑞) necessary to boil the ammonia is calculated by multiplying the moles (𝑛) of ammonia by the molar heat of vaporization of ammonia. 𝑞=𝑛×molar heat of vaporization Since the amount of ammonia is given in grams, it must first be converted to moles of ammonia using the molar mass of ammonia. 96.6 g NH3×1 mol NH3/17.03 g NH3=5.67 mol NH3 Now, calculate the heat energy necessary to boil 96.6g of ammonia. 𝑞=5.67 mol×23.4 kJ/mol=133 kJ

The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 22.45 g of ethanol. heat energy:

18540 J Solution The heat of vaporization for ethanol is the amount of heat energy required to boil one gram of ethanol. Use the heat of vaporization as a conversion factor to convert the mass of ethanol to the amount of energy needed to boil 22.45 g of ethanol. 22.45 g×0.826 kJ/1 g=18.5 kJ Then, convert the energy from kilojoules to joules. Recall that 1 kJ=1000 J. 18.5 kJ×1000 J/1 kJ=18500 J

A metal crystallizes in the face‑centered cubic (FCC) lattice. The density of the metal is 19320 kg/m3, and the length of a unit cell edge, 𝑎, is 407.83 pm. Calculate the mass of one metal atom. mass: Identify the metal.

3.2763×10−22 g gold Solution Density is 𝑑=𝑚/𝑎3 where 𝑚 is the mass of one unit cell and 𝑎 is the length of the unit cell edge. There are four atoms in a face‑centered cubic unit cell, so the mass of one unit cell is equal to four times the mass of one atom. Therefore, density is 𝑑=4𝑚atom/𝑎3 The unit cell edge is given in picometers and the density has units of kg/m3. Convert the unit cell length to meters so that the units cancel properly. 407.83 pm×1×10−12 m/1 pm=4.0783×10−10 m Rearrange to solve for the mass of one metal atom. 𝑚=𝑑𝑎3/4=((19320 kg/m3)×(4.0783×10−10 m)^3)/4=3.276×10−25 kg Convert the mass to grams. 3.276×10−25 kg×1000 g/1 kg=3.276×10−22 g Multiply the mass of the metal atom by Avogadro's number to calculate the molar mass of the element. 3.276×10−22 g/1 atom×(6.022×1023 atoms/1 mol)=197.3 g/mol Use the periodic table to determine the identity of the metal. Of the metals listed, gold, with a molar mass of 196.97 g/mol, is closest to the calculated molar mass. So, the mass of one gold atom is 3.276×10−22 g.

The atomic radius of metal X is 1.40×102 picometers (pm) and a crystal of metal X has a unit cell that is face-centered cubic. Calculate the density of metal X (atomic weight = 42.3 g/mol). density:

4.525 g/cm3 Solution Density (𝑑) is mass (𝑚) divided by volume (𝑉). Therefore, the mass and the volume of the unit cell need to be determined. Start by calculating the mass of the unit cell by converting the number of atoms in the unit cell to grams in the unit cell. Recall that there are four atoms in a face-centered cubic unit cell. 𝑚=4 atoms1 unit cell×1 mol6.022×1023 atoms×42.3 g1 mol=2.810×10−22gunit cell Next, the volume (in cm3) of the unit cell needs to be determined. Since the cell is a cube, the volume is calculated by taking the edge length (𝑎) to the third power as shown in the equation 𝑉=𝑎3 For a face-centered cubic unit cell, the edge length can be calculated using the equation 𝑎=√(8)×𝑟 where 𝑟 is the atomic radius. Using the given atomic radius, the edge length for this problem is calculated as 𝑎=√(8)×1.40×10^2 pm=396 pm Next, convert the edge length from picometers (pm) to centimeters (cm). 𝑎=396 pm× (1×10−10 cm)/1 pm=3.96×10−8 cm Then, calculate the volume. 𝑉=𝑎3=(3.96×10−8 cm)^3 = 6.21×10−23 cm3 Finally, now that the mass and volume of the unit cell have been determined, calculate the density. 𝑑=𝑚/𝑉=(2.810×10−22 g)/6.21×10−23 cm3 = 4.53 g/cm3

A certain substance has a heat of vaporization of 38.87 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 times higher than it was at 353 K? 𝑇=

402 K Solution The Clausius‑Clapeyron equation relates vapor pressure (𝑃), enthalpy of vaporization (Δ𝐻vap), and temperature (𝑇). ln(𝑃2/𝑃1)=Δ𝐻vap/𝑅(1/𝑇1−1/𝑇2) If the vapor pressure increases by a factor of 5.00, then 𝑃2/𝑃1 is equal to 5.00. ln(5.00)=Δ𝐻vap/𝑅·(1/𝑇1−1/𝑇2) Since 𝑅=8.3145 J/(mol·K), the enthalpy of vaporization should be expressed in joules per mole. Δ𝐻vap=38.87 kJ/mol×1000 J/1 kJ=38870 J/mol Finally, plug the remaining values into the equation and solve for 𝑇2. In (5.00) = 38870 J/mol/8.3145 J/(mol·K)·(1/353K - 1/T2) 1.609/4675K + 1/T2=1/353 K 1/T2 = 1/353 K−1.609/4675 K =0.00249 K−1 T2=1/0.00249 K−1 =402 K

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 ∘C and 371.0 Torr at 45.0 ∘C. Calculate the value of Δ𝐻∘vap for this liquid. Δ𝐻∘vap= Calculate the normal boiling point of this liquid. boiling point:

49.4 kJ/mol 56.1∘C Solution First, use the Clausius-Clapeyron equation to find Δ𝐻∘vap. ln(𝑃2/𝑃1)=Δ𝐻∘vap/𝑅(1/𝑇1−1/𝑇2) where 𝑃 is the pressure, 𝑇 is the temperature, and 𝑅 is the gas constatnt equal to 8.3145 J/mol·K. The temperature must be converted from degrees Celsius to kelvin. Δ𝐻∘vap=(ln(𝑃2/𝑃1)𝑅)/(1/𝑇1−1/𝑇2)=(ln(371.0 Torr/92.0 Torr)(8.3145Jmol⋅K))/(1/(23.0+273.15)K−1/(45.0+273.15))K=49700 J/mol=49.7 kJ/mol The boiling point of a liquid occurs when the vapor pressure is equal to the ambient pressure. Normal boiling point is the temperature at which the vapor pressure is exactly 1 atm=760 Torr. Knowing that Δ𝐻∘vap=49700 J/mol, use the Clausius-Clapeyron equation again, where 𝑃2=760 Torr and 𝑇2 is the normal boiling point (the unknown). The values of 𝑃1 and 𝑇1 can be either set of pressure and temperature values given in the problem. ln(𝑃2/𝑃1)=Δ𝐻∘vap/𝑅(1/𝑇1−1/𝑇2) ln(𝑃2/𝑃1)×𝑅/Δ𝐻∘vap=1/𝑇1−1/𝑇2 1/𝑇2=1/𝑇1−ln(𝑃2/𝑃1)×𝑅/Δ𝐻∘vap =1/(23.0+273.15)K−ln(760 Torr/92.0 Torr)×(8.3145 J/(mol⋅K)/49700 J/mol)=0.003023 K−1 𝑇2=1/0.003023 K−1=330.8 K 330.8 K−273.15=57.6∘C

How many grams of ethanol, C2H5OH, can be boiled with 558.3 kJ of heat energy? The molar heat of vaporization of ethanol is 38.6 kJ/mol. mass of ethanol:

667 g Solution Heat energy must be supplied to boil ethanol. This heat energy is used to overcome the intermolecular forces that maintain liquid ethanol. This overcoming of the intermolecular forces leads to a phase change, the conversion from liquid to gaseous ethanol in this case. The amount of ethanol that can be boiled depends on the amount of heat energy supplied and the molar heat of vaporization. The molar heat of vaporization of ethanol is the amount of heat necessary to convert 1 mol of liquid ethanol to gaseous ethanol. In other words, the molar heat of vaporization is a quantitative measure of how readily ethanol is boiled. The amount of ethanol boiled, in moles (𝑛), is related to the heat energy (𝑞) and the molar heat of vaporization of ethanol by the equation 𝑞=𝑛×molar heat of vaporization Rearrange this equation to solve for the moles of ethanol. 𝑛=𝑞/molar heat of vaporization=558.3 kJ38.6 kJ/mol=14.46 mol C2H5OH Then, convert from moles of ethanol to grams of ethanol. 14.46 mol C2H5OH×46.07 C2H5OH/1 mol C2H5OH=666.3 g C2H5OH

A 0.548 g sample of steam at 106.4 °C is condensed into a container with 4.90 g of water at 16.9 °C . What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18 J g⋅ °C , the specific heat of steam is 2.01 J g⋅ °C , and Δ𝐻vap=−40.7 kJ/mol. 𝑇f=

79.9°C Solution The heat given off by the steam is absorbed by the liquid water. −𝑞steam=𝑞water The liquid water is warmed from its initial temperature to the final temperature of the mixture. The equation for the heat gained by the water is 𝑞water=𝑚water𝐶s, waterΔ𝑇 where 𝑞water is the heat gained by the water, 𝑚 is the mass of the water, 𝐶𝑠 is the specific heat of liquid water, and Δ𝑇 is the change in temperature. The steam is first cooled to 100 °C. Then, it undergoes a phase change from gas to liquid. Finally, it is cooled from 100 °C to the final temperature of the mixture. The heat required to cool the steam from 106.4 °C to 100 °C is 𝑞steam=𝑚steam𝐶s, steamΔ𝑇 where 𝑚 is the mass of steam, 𝐶s, steam is the specific heat of steam, and Δ𝑇 is the change in temperature. The amount of heat required to condense the steam to liquid is 𝑞steam=𝑛Δ𝐻vap where 𝑛 is the number of moles of steam and Δ𝐻vap is the enthalpy of vaporization. The amount of heat required to cool the liquified steam to the final temperature of the mixture is 𝑞steam=𝑚s𝐶s, waterΔ𝑇 where 𝑚 is the mass of steam, 𝐶s, water is the specific heat of water, and Δ𝑇 is the change in temperature. All together, the equation for the heat given off by the steam is 𝑞steam=𝑚steam𝐶s, steamΔ𝑇+𝑛Δ𝐻vap+𝑚steam𝐶s,waterΔ𝑇 Calculate the amount of heat given off when the steam is cooled from 106.4 °C to 100 °C. 𝑞steam=(0.548 g)(2.01 J/g·°C)(100 °C−106.4 °C)=−7.0 J Then, calculate the amount of heat given off when the steam condenses to liquid water. 𝑞steam=(0.548 g×1mol/18.02 g)(−40.7 kJ/mol)=−1.24 kJ×1000 J/1 kJ=−1240 J Insert these values into the equation for the total heat given off by the steam. 𝑞steam=−7.0 J−1240 J+(0.548 g)(4.18 J/g·°C)(𝑇f−100 °C) Insert the values into the equation for heat gained by the water. 𝑞water=(4.90 g)(4.18 J/g·°C)(𝑇f−16.9°C) Then, set the equation for the heat gained by the water equal to the heat given off by the steam. −𝑞steam=𝑞water −[−7.0 J−1240 J+(0.548 g)(4.18 J/g·°C)(𝑇f−100 °C)]=(4.90 g)(4.18 J/g·°C)(𝑇f−16.9°C) −[−1470 J+(2.29 J/°C)𝑇f]=−346 J+(20.5 J/°C)𝑇f Now, solve for 𝑇f. 1470 J−(2.29 J/°C)𝑇f=−346 J+(20.5 J/°C)𝑇f 1820 J=(22.8 J/°C)𝑇f 79.9 °C=𝑇f The final temperature of the water mixture is 79.9 °C.

Consider the simple cubic (primitive cubic) unit cell shown in this image. PICTURED: A six sided cube with a sphere on each corner and a line along each edge of the cube connecting the center of each sphere with the centers of three other spheres, labeled simple (primitive) cubic How many atoms are shown in the image? atoms in image: atoms What fraction of each atom is inside the boundaries of the cube? If you sum all the fractions of atoms, how many atoms are actually inside a simple cubic unit cell? atoms in unit cell: atoms

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Two 20.0 g ice cubes at −13.0 ∘C are placed into 255 g of water at 25.0 ∘C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, 𝑇f, of the water after all the ice melts. heat capacity of H2O(s) 37.7 J/(mol⋅K) heat capacity of H2O(l) 75.3 J/(mol⋅K) enthalpy of fusion of H2O 6.01 kJ/mol 𝑇f=

9.91∘C Solution Begin by calculating the heat needed to bring 40.0 g of ice at -13.0 ∘C to its melting point (0 ∘C), 𝑞ice, using the equation 𝑞ice=𝑛ice𝐶iceΔ𝑇ice where 𝑛ice is the number of moles of ice, Δ𝑇ice is the temperature change in kelvin for the ice, and 𝐶ice is the heat capacity of H2O(s), 37.7 J/(mol·K). Recall that a change in temperature will have the same value in kelvins as it does in degrees Celsius. Thus, Δ𝑇=0−(−13.0 ∘C)=13.0 K Using the molar mass of H2O (18.02 g/mol), 𝑞ice is 𝑞ice=𝑛ice𝐶iceΔ𝑇ice =(40.0 g×1 mol/18.02 g)(37.7J/mol⋅K)(13.0 K)=1090 J Next, calculate the amount of energy required to melt the ice, 𝑞melt, using the formula 𝑞melt=𝑛iceΔ𝐻fus where Δ𝐻fus is the enthalpy of fusion of H2O, 6.01 kJ/mol. Recall that 1 kJ=1000 J. Therefore, 𝑞melt=𝑛iceΔ𝐻fus=(40.0 g×1 mol/18.02 g)(6.01kJmol)(1000 J/1 kJ)=13340 J Summing 𝑞ice and 𝑞melt gives the total amount of energy needed to fully convert the solid ice from -13.0 ∘C to liquid water at 0 ∘C, 𝑞total. 𝑞total=1090+13340=14430 J The energy required to warm and melt the ice is transferred from the water. Therefore, the energy can be used to calculate the change in temperature of the water that results from the ice melting. Calculate the temperature change of the 255 g water sample, Δ𝑇water, using the rearranged form of the heat energy equation Δ𝑇water=𝑞total𝑛water𝐶water where 𝐶water is the heat capacity of H2O(l), 75.3 J/(mol·K), and 𝑛water is the moles of water in the 255 g sample. Again, use the molar mass of H2O (18.02 g/mol) to calculate 𝑛water and Δ𝑇water. Δ𝑇water=𝑞total/𝑛water𝐶water =14430 J/255 g ×(1 mol18.02 g)×(75.3 JK⋅mol) =13.5 K Subtracting Δ𝑇water from the original temperature of the water (25.0 ∘C) gives the new cooler temperature of the water, 𝑇cool. Recall that a change in temperature will have the same value in kelvins as it does in degrees Celsius. Thus, 𝑇cool=(25.0−13.5) ∘C=11.5 ∘C At this stage, there are 40.0 g H2O(l) at 0 ∘C and 255 g H2O(l) at 11.5 ∘C. Therefore, to find the final temperature of the water after all of the ice melts, 𝑇f, use the equation 𝑛ice𝐶Δ𝑇warm=𝑛water𝐶Δ𝑇cool where 𝐶 is the heat capacity of H2O(l), 75.3 J/(K·mol), Δ𝑇warm is the temperature difference between 𝑇f and the temperature of the water formed from the melted ice, 0 ∘C, and Δ𝑇cool is the temperature difference between 𝑇f and Δ𝑇cool, 11.5 ∘C. Mathematically, the 𝐶 values cancel out, as do the conversions of grams to moles of H2O. Subsequently, the equation simplifies to (𝑇f−0) ∘C×40.0 g=(11.5−𝑇f) ∘C×255 g Rearrange to solve for 𝑇f. 𝑇f(40.0 g)=(11.5 ∘C−𝑇f)(255 g) 𝑇f(40.0 g)=(255 g)(11.5 ∘C)−(255 g)𝑇f 𝑇f(40.0 g+255 g)=(255 g)(11.5 ∘C) 𝑇f=(255 g)(11.5 ∘C)/(40.0 g+255 g)=9.91 ∘C

Consider the table. Solid Triple point Normal melting point A 0.91 atm, 124 °C 108 °C B 0.35 atm, -17 °C 50 °C C 0.0072 atm, 88 °C 89 °C Based on the data in the table, which solids will melt under applied pressure?

A Solution The normal melting point of a substance occurs at a pressure of 1 atm. Thus, the normal melting point of solid A is at a pressure and temperature above and to the left of its triple point. The normal melting points of solids B and C are at a pressure and temperature above and to the right of their triple points. There is not enough data to sketch the entire phase diagram, but in this case the slope of the solid-liquid line is the only feature of interest. PICTURED: Phase diagram for solid A (general shape) A phase diagram of pressure versus temperature. A line starts at the origin and has a positive slope that spans the diagram. The phase gas lies underneath the line. Above the line lies the phase solid on the left and liquid on the right. They are separated by a negatively sloped line. PICTURE: Phase diagram for solids B and C (general shape) A phase diagram of pressure versus temperature. A line starts at the origin and has a positive slope that spans the diagram. The phase gas lies underneath the line. Above the line lies the phase solid on the left and liquid on the right. They are separated by a positively sloped line. Increasing pressure (i.e., moving directly upwards in the phase diagram) will eventually cause solid A to melt (i.e., cross the line into the liquid portion of the phase diagram). The only way to melt solid B or C is to increase the temperature (i.e., move to the right in the phase diagram).

Acetone, CH3COCH3, has a boiling point of 56°C, and ethanol, CH3CH2OH, has a boiling point of 78°C. Predict which compound, acetone or ethanol, would have the greater vapor pressure at 25°C.

ACETONE Solution Both the vapor pressure and boiling point of a substance are dependent on the strength of the intermolecular forces of the substance. A substance with weak intermolecular forces has a lower boiling point than a substance with stronger intermolecular forces forces because the intermolecular forces are easily overcome by the thermal energy. For the same reason, a substance with weak intermolecular forces has a higher vapor pressure than a substance with stronger intermolecular forces. Therefore, the boiling points of two substances can be compared to predict which substance will have the higher vapor pressure. The boiling point of acetone, 56°C, is lower than the boiling point of ethanol, 78°C. This indicates that the intermolecular forces in acetone are weaker than those in ethanol. Therefore, acetone will have a higher vapor pressure than ethanol.

The properties of several unknown solids were measured. Solid Melting point Other properties A >1000 °C does not conduct electricity B 850 °C conducts electricity in the liquid state, but not in the solid state C 750 °C conducts electricity in the solid state D 150 °C does not conduct electricity Classify the solids as ionic, molecular, metallic, or covalent. Note that covalent compounds are also known as covalent network solids or macromolecular solids. Ionic Molecular Metallic Covalent

B D C A Solution An ionic solid is held together with ionic bonds throughout the substance, making it difficult to melt. The ions in an ionic solid are not free to move and so ionic solids do not conduct electricity. If the solid is melted, the ions become mobile; consequently, ionic liquids conduct electricity. A molecular substance has covalent bonds only within molecules. One molecule is held to another molecule by relatively weak attractive forces, making the solid relatively easy to melt (low melting point). Molecular substances do not contain ions and thus do not conduct electricity. A metallic substance is held together by electrostatic attraction between cations and mobile electrons. Metals conduct electricity as solids, whereas other substances generally do not. A covalent solid has covalent bonds throughout the substance, making it very difficult to melt. Covalent substances do not contain ions and generally do not conduct electricity. The fact that solid B conducts electricity as a liquid but not as a solid immediately identifies it as an ionic substance. The high melting point of solid B confirms this classification. The fact that solid C conducts electricity as a solid means it is most likely a metal. Metals have a wide range of melting points, so the melting point of solid C is not really a factor in this classification. Since solids A and D do not conduct electricity at all, they must be molecular or covalent. The very high melting point of solid A suggests that it is covalent. The low melting point of solid D suggests that it is molecular.

Arrange the compounds from highest boiling point to lowest boiling point. Highest boiling point Lowest boiling point

CH3CH2CH2CH2CH2OH CH3CH2CH2CH2CH2Br CH3CH2CH2CH2CH3 CH3CH2CH3 Solution The boiling point of a substance is related to the strength of the intermolecular forces between the molecules in the substance. The greater the strength of the intermolecular forces between the molecules in the substance, the higher the boiling point of the substance. Among the four types of intermolecular forces, ion-dipole interactions are the strongest and occur between an ion and the dipole moment of a polar molecule. Hydrogen bonds are the second strongest intermolecular interaction and form between a lone pair on a nitrogen, N, oxygen, O, or fluorine, F, atom, and a hydrogen H, atom that is directly bonded to an N, O, or F atom. Dipole-dipole interactions are weaker than hydrogen bonds, and occur between the dipole moments of polar molecules. The weakest intermolecular forces, London dispersion forces, are caused by temporary distortions of the electron clouds surrounding all molecules, and become stronger with increasing molecule size and polarizability. To order the compounds from highest boiling point to lowest boiling point, consider the relative strengths of the intermolecular forces for each substance. None of the substances contain an ion and a polar molecule; thus, none can participate in ion-dipole interactions. One substance, 1-pentanol, CH3CH2CH2CH2CH2OH, contains lone pairs on an O atom and a highly polar bond with an H atom directly bonded to an O atom; therefore, 1-pentanol can participate in hydrogen bond formation. Due to its ability to form hydrogen bonds, 1-pentanol has the highest boiling point of the compounds. 1-bromopentane, CH3CH2CH2CH2CH2Br, is similar in size to 1-pentanol; however, it cannot participate in hydrogen bonds. Instead, the inclusion of a bromine atom forms a small dipole moment that can participate in dipole-dipole interactions, and adds a large polarizable electron cloud that provides the molecule with more London disperson forces than are available for pentane, which also contains a five-carbon chain. Collectively, the intermolecular forces for 1-bromopentane are weaker than those for 1-pentanol, but greater than those of pentane; thus, 1-bromopentane has the second highest boiling point. Pentane, CH3CH2CH2CH2CH3, and propane, CH3CH2CH3, are both nonpolar hydrocarbons and only have London dispersion forces. As pentane is larger than propane, and London dispersion forces increase with increasing molecule size, pentane has stronger London dispersion forces and a higher boiling point than propane.

Arrange these compounds by their expected boiling point. Highest boiling point Lowest boiling point

CH3OH CH3Cl CH4

Which molecules exhibit only London (dispersion) forces?

Cl2 BCl3 CH4. Solution London dispersion forces are present within all molecular substances. However, they are the only intermolecular force that exists between nonpolar molecules or atoms. To determine whether a molecule is polar or nonpolar, consider the polarity of the bonds as well as the geometry. Polar bonds have a difference in electronegativity between atoms, whereas nonpolar bonds do not. The shape of a molecule also affects polarity. Polar bonds can be treated as vectors. If the vectors are equal in magnitude, and opposite in direction, they cancel each other out. So if bonds have equal polarity and are evenly distributed in space, the molecule as a whole is nonpolar. In Cl2, there is not a difference in electronegativity between the atoms because there is only one type of atom in the molecule. This means the bond and molecule as a whole are nonpolar and it exhibits only London dispersion forces. In CH4, there is not a difference in electronegativity between C and H. This means the bonds and molecule as a whole are nonpolar and it exhibits only London dispersion forces. In BCl3, there is a difference in electronegativity between B and Cl, so the bonds are polar. However, BCl3 has trigonal planar geometry. The bonds are equal in polarity and evenly distributed in space. This means the molecule as a whole is nonpolar and exhibits only London dispersion forces. In NH3, there is a difference in electronegativity between N and H, so the bonds are polar. NH3 has trigonal pyramidal geometry, so the bonds are not evenly distributed in space and the molecule is polar. Since NH3 is polar, it is capable of additional intermolecular forces. In CH3Cl, there is not a difference in electronegativity between C and H. However, there is a difference in electronegativity between C and Cl. CH3Cl has tetrahedral geometry. There is only one polar bond in the molecule, so the molecule as a whole is polar. Since CH3Cl is polar, it is capable of additional intermolecular forces.

Which of the substances have polar interactions (dipole-dipole forces) between molecules?

ClF NF3 Solution Cl2 is a homonuclear diatomic molecule, and so it is nonpolar. The Lewis formula for ClF is A chlorine atom is bonded to a fluorine atom by a single bond. There are three lone pairs of electrons around both the chlorine and fluorine atoms. Fluorine is more electronegative than chlorine; therefore, the fluorine atom has a small negative charge and the chlorine atom has a small positive charge. The molecule is polar. The Lewis formula for NF3 is A nitrogen atom is the center atom. It is bonded to the three fluorine atoms by single bonds. There is one lone pair of electrons on the nitrogen atom. Each fluorine atom contains three lone pairs of electrons. Because fluorine is more electronegative than nitrogen, there is a small negative charge on each fluorine atom and a small positive charge on the nitrogen atom. The molecule has a trigonal pyramidal shape, and so the molecule is polar. The molecule F2 is nonpolar because it is a homonuclear molecule.

Identify the true statements about surface tension.

Cohesive forces attract the molecules of the liquid to one another. Molecules along the surface of a liquid behave differently than those in the bulk liquid. Water forming a droplet as it falls from a faucet is a primary example of surface tension. Solution Surface tension is caused by the unbalanced net inward attraction of the molecules along the surface of a liquid (red sphere) towards the molecules in the bulk liquid (orange sphere), as shown in the diagram. A beaker filled with a liquid. A molecular view of the surface shows a molecule on the surface that is attracted to molecules to the left, right and below it. A molecular view of the bulk liquid shows a molecule in the bulk that is attracted to molecules all around it. The black arrows are added to help visualize the intermolecular forces present in the liquid. Notice that the molecule in the bulk liquid (orange sphere) is pulled in all directions by attractive forces, known as cohesive forces, whereas the molecule on the surface (red sphere) is only pulled sideways and downwards. The net difference between these forces causes the molecules in the bulk liquid to behave differently than the molecules along the surface. The strength of the cohesive forces is caused by the different types of intermolecular forces present in the liquid. For example, mercury has a surface tension roughly six times higher than that of water because the covalent network the individual mercury molecules can create with each other is much stronger than the hydrogen bonding occurring between the water molecules. One of the most common examples of surface tension is water dropping from a faucet. As gravity pulls the water down, the cohesive forces between the water molecules pull the outer molecules into a spherical shape until the cohesive forces overtake the adhesive forces between the water and faucet. At that point, the water molecules shift to form a sphere, the shape with the lowest surface area to volume ratio, as it falls into the sink. As the temperature of a liquid increases, the molecules move faster and the attractive forces between the molecules decrease. The decrease in cohesive forces causes the surface tension of the liquid to decrease.

Arrange the binary hydrogen-containing compounds of group 6A in order from lowest boiling point to highest boiling point. Lowest Highest

H2S H2Se H2Te H2O Solution The boiling point of a substance is related to the strength of the intermolecular forces between the molecules in the substance. The greater the strength of the intermolecular forces between the molecules in the substance, the higher the boiling point of the substance. Among the four types of intermolecular forces, ion-dipole interactions are the strongest and occur between an ion and the partially positive or negative end of a dipole in a polar molecule. Hydrogen bonds are the second strongest intermolecular interaction and form between a lone pair on a nitrogen, N, oxygen, O, or fluorine, F, atom, and a hydrogen, H, atom that is directly bonded to an N, O, or F atom. Dipole-dipole interactions are weaker than hydrogen bonds, and occur between the partially positive and negative ends of polar molecules. The weakest intermolecular forces, London dispersion forces, are caused by temporary distortions of the electron clouds surrounding all molecules, and become stronger with increasing molecule size and polarizability. None of the compounds contain an ion and a polar substance; therefore, ion-dipole interactions do not influence the boiling points of these substances. H2O can participate in hydrogen bonds because it contains a hydrogen atom, H, directly bonded to a nitrogen, N, oxygen, O, or fluorine, F, atom. Substances that can participate in hydrogen bonding have relatively high boiling points; thus, H2O has the highest boiling point. Comparing the relative strengths of the dipoles for the remaining three polar compounds, the dipoles and thus the strength of the intermolecular dipole-dipole interactions decrease moving down the group. At the same time, the atomic radius of the nonhydrogen atom increases moving down the group, resulting in increased London dispersion forces. For these compounds, the change in the strength of the dipole-dipole interactions is minimal, so the change in the strength of the London dispersion forces has the greatest affect on the boiling points. Thus, H2Te has the second highest boiling point, followed by H2Se and H2S.

If a solid line represents a covalent bond and a dotted line represents intermolecular attraction, which of the choices shows a hydrogen bond?

H3N⋅⋅⋅⋅⋅⋅H−O−H Solution One side of a hydrogen bond consists of a highly electronegative atom with a lone pair, N, O, or F. The other side of a hydrogen bond consists of a hydrogen atom that is directly bonded to a highly electronegative atom. Only H3N⋅⋅⋅⋅⋅⋅H−O−H shows a hydrogen bond.

Arrange the compounds by boiling point. Highest boiling point Lowest boiling point

HEXANE PENTANE PROPANE Solution All these compounds are nonpolar, so the relative strengths of the dispersion forces (London forces) will determine the relative boiling points. The strength of the dispersion forces tends to increase with the size and mass of the molecules. Thus, hexane has the strongest intermolecular forces and the highest boiling point of these three compounds. Propane has the weakest intermolecular forces and the lowest boiling point of these three compounds.

Match each property of a liquid to what it indicates about the relative strength of the intermolecular forces in that liquid. Strong intermolecular forces Weak intermolecular forces

HIGH BOILING POINT AND HIGH SURFACE TENSION AND HIGH VISCOSITY HIGH VAPOR PRESSURE Solution Vapor pressure is the pressure of the vapor phase in equilibrium with a liquid. If a large number of molecules are able to escape the attractive forces of the other liquid molecules into the vapor phase, then the vapor pressure of that liquid will be high. Boiling point is the temperature at which a liquid becomes a gas. If liquid molecules are strongly attracted to other liquid molecules, they cannot easily enter the vapor phase, and the temperature needed to boil the liquid will be high. Surface tension and viscosity both increase with the strength of intermolecular forces. Surface tension is the tendency of a liquid to bead up rather than spread out. Viscosity is a liquid's resistance to flow.

Ice floats on water. For most other substances, however, the solid sinks in the liquid. Classify each of these statements based on whether they describe water or most other substances. Water Most other substances

HIGH PRESSURE WILL CAUSE THE SOLID TO BECOME LIQUID AND IN A PHASE DIAGRAM, THE SOLID-LIQUID COEXISTENCE LINE HAS A NEGATIVE SLOPE. AND THE SOLID IS LESS DENSE THAN THE LIQUID AND THE MOLECULES ARE CLOSEST IN THE LIQUID PHASE HIGH PRESSURE WILL CAUSE THE LIQUID TO BECOME SOLID AND THE SOLID IS MORE DENSE THAN THE LIQUID AND THE MOLECULES ARE CLOSEST IN THE SOLID PHASE AND IN A PHASE DIAGRAM, THE SOLID-LIQUID COEXISTENCE LINE HAS A POSITIVE SLOPE Solution Since ice floats on water, the ice weighs less than an equal volume of water. Thus, ice is less dense than water. For other substances, the opposite is true; the solid is more dense than the liquid. Using what you know about density, you can infer that the molecules must be further apart in ice, and closer together in water. For most substances, however, the solid phase is the most condensed. Applying pressure will cause a substance to transition to its most condensed phase. For water, that is the liquid state. For most other substances, that is the solid state. Finally, compare the phase diagrams. Negative slope: Moving upward in this diagram, that is increasing the pressure at a constant temperature, the substance would cross from solid to liquid. Since applying pressure causes a substance to transition to its most condensed state, and the most condensed state of water is liquid, this phase diagram is representative of water's phase diagram. PICTURED: A phase diagram of pressure versus temperature. A line starts at the origin and has a positive slope that spans the diagram. The phase gas lies underneath the line. Above the line lies the phase solid on the left and liquid on the right. They are separated by a negatively sloped line. Positive slope: Moving upward in this diagram, that is increasing the pressure at a constant temperature, the substance would would cross from liquid to solid. Since applying pressure causes a substance to transition to its most condensed state, and the most condensed state of most substances is solid, this phase diagram is representative of most substances. PICTURED: A phase diagram of pressure versus temperature. A line starts at the origin and has a positive slope that spans the diagram. The phase gas lies underneath the line. Above the line lies the phase solid on the left and liquid on the right. They are separated by a positively sloped line.

Arrange the molecules by the strength of the London (dispersion) force interactions between molecules. Strongest London dispersion forces Weakest London dispersion forces

I2 Br2 Cl2 Solution London forces are stronger in larger, heavier molecules. I2 has a greater molar mass (253.80 g/mol) than Br2 (159.81 g/mol), and therefore has stronger London (dispersion) forces between molecules. Br2 has a greater molar mass than Cl2, so Br2 has stronger London (dispersion) forces between molecules.

Classify each of these solids as ionic, molecular, metallic, or covalent (also known as covalent‑network solids or macromolecular solids). Ionic Molecular Metallic Covalent

KCl FCl Ca C(diamond) Solution The formula of an ionic compound contains both a metal and a nonmetal. KCl fits this category. The formula of a molecular substance contains only nonmetal atoms. FCl fits this category. The formula of a metallic substance contains only metal atoms. Ca fits this category. The formula of a covalent substance will look similar to that of a molecular substance, so to distinguish them, you'll need to memorize a few key examples of covalent compounds. Typical examples of covalent compounds include diamond, graphite, and quartz. Thus, C(diamond) should be classified as a covalent solid.

Determine the solid with the highest melting point.

LiF(s) Solution Ionic solids are held to together by strong coulombic forces (ionic bonds), whereas molecular solids are held together by intermolecular forces. The coulombic forces in ionic solids are much stronger than the intermolecular forces in molecular solids, so ionic solids tend to have melting points that are much higher than molecular solids. Molecular solids tend to have low to moderately low melting points. LiF(s) is an ionic solid, whereas the other three compounds are molecular solids. Therefore, LiF(s) will have the highest melting point. The melting points of the molecular solids will depend on the strength of the intermolecular forces holding the molecules together. The stronger the intermolecular forces, the higher the boiling point. The strongest intermolecular force is hydrogen bonding, followed by dipole-dipole interactions. The weakest intermolecular force is London dispersion forces. CH3OH(s) molecules are held together by hydrogen bonding, so CH3OH(s) will have the highest melting point of the molecular solids. CCl4(s) molecules are held together by London dispersion forces, so CCl4(s) will have the lowest melting point of the molecular solids. CH2Cl2(s) molecules are held together by dipole-dipole interactions, so its melting point will be between the melting points of CH3OH(s) and CCl4(s).

Classify each substance based on the intermolecular forces present in that substance. Hydrogen bonding, dipole-dipole, and dispersion Dipole-dipole and dispersion only Dispersion only

NH3 CH3CL AND CO HE

Rank the shown compounds by boiling point. pentane: The molecule C H 3 C H 2 C H 2 C H 2 C H 3 neopentane: The structure shows a C atom with four C H 3 groups bonded to it. isopentane: The structure C H 3 C H 2 C H (C H 3) 2 Highest boiling point Lowest boiling point

PENTANE ISOPENTANE NEOPENTANE Solution All of the given compounds are nonpolar, so the relative strengths of the dispersion forces (also known as van der Waals forces) will determine the relative boiling points. For molecules of the same mass, the shape of the molecule can affect its ability to interact with other molecules. Branched molecules (such as neopentane) tend to have weaker intermolecular attraction than straight‑chain molecules of similar mass (pentane). Thus, neopentane has the lowest boiling point of these three substances and pentane has the highest boiling point.

The vapor pressure, 𝑃, of a certain liquid was measured at two temperatures, 𝑇. The data is shown in the table. T (K) P (kPa) 225 3.53 775 5.24 Keep the pressure units in kilopascals. If you were going to graphically determine the enthalpy of vaporizaton, Δ𝐻vap, for this liquid, what points would you plot? To avoid rounding errors, use three significant figures in the 𝑥‑values and four significant figures in the 𝑦‑values. point 1: 𝑥= 0.00444 point 1: 𝑦= 1.261 point 2: 𝑥= 0.00129 point 2: 𝑦= 1.656 Determine the rise, run, and slope of the line formed by these points. rise= 0.395 run= −0.00315 slope= −125 What is the enthalpy of vaporization of this liquid? Δ𝐻vap= 1040 J/mol

Solution The Clausius-Clapeyron equation can be arranged to resemble the formula for a straight line, 𝑦=𝑚𝑥+𝑏, where 𝑥 is 1/𝑇, 𝑦 is ln(𝑃), and the slope is −Δ𝐻vap/𝑅. ln(𝑃)=(−Δ𝐻vap𝑅)(1𝑇)+𝐶 Thus, to find the 𝑥‑coordinate of each point, take the reciprocal of the temperature, 𝑇. To find the 𝑦‑coordinate of each point, take the natural log of the vapor pressure, 𝑃. 𝑥1=1/𝑇1 =1/225 K =0.00444 K−1 𝑦1=ln(𝑃1) =ln(3.53 kPa) =1.261 𝑥2=1/𝑇2 =1/775 K =0.00129 K−1 𝑦2=ln(𝑃2) =ln(5.24 kPa) =1.656 The rise of a line is the distance between points in the vertical (𝑦) direction. Subtracting the two 𝑦‑values gives the rise. The run of a line is the distance between points in the horizontal (𝑥) direction. Subtracting the two 𝑥‑values gives the run. The slope of the line is the rise divided by the run. rise=𝑦2−𝑦1=0.395 run=𝑥2−𝑥1=−0.00315 K−1 slope=rise/run =0.395/−0.00315 K−1 =−125 K Since slope=−Δ𝐻vap/𝑅, you can calculate the enthalpy of vaporization, Δ𝐻vap, by plugging in values for 𝑅 and the slope. 𝑅 is equal to 8.3145 J/(mol⋅K). slope=−Δ𝐻vap/𝑅 Δ𝐻vap=−slope×𝑅=−(−125 K)×(8.3145Jmol⋅K) =1040 J/mol

In a glass tube, the meniscus of water is concave, whereas the meniscus of mercury is convex. Select the statement that explains the difference in the shape of the menisci.

The meniscus of water is concave because the adhesive forces are greater than the cohesive forces. The meniscus of mercury is convex because the cohesive forces are greater than the adhesive forces. Solution The meniscus is the curved shape of a liquid surface within a tube. Cohesive forces are attractive forces between molecules of the liquid holding them together. Adhesive forces are attractive forces between molecules of the liquid and the surface of the container. The meniscus of water is concave (rounded inward) because the adhesive forces are greater than the cohesive forces. This causes the water to creep up the side of the glass tube. The meniscus of mercury is convex (rounded outward) because the cohesive forces are greater than the adhesive forces. The mercury atoms crowd toward the center of the tube to maximize their interactions with other mercury atoms and minimize their interaction with the glass tube.

A substance has a triple point of −16.8 ∘C and 235 mmHg. What will happen to a solid sample of the substance if it is warmed from −31.7 ∘C to 0 ∘C at a constant pressure of 202 mmHg?

The solid will sublime into a gas. Solution A triple point is a point on a phase diagram that represents a set of conditions where solid, liquid, and gas are equally stable and in equilibrium. The solid is warmed from a temperature below the triple point temperature to a temperature above the triple point temperature, so a phase change will occur. The pressure of the sample is lower than the pressure at the triple point, so the solid will sublime into gas. If the pressure was greater than the triple point pressure, than the solid would melt into a liquid.

Use the atom legend to determine the intermolecular force depicted in each image. Five types of atoms are shown. A small gray atom is labeled H, a small green atom is labeled F, a medium blue atom is labeled N A plus, a slightly smaller red atom is labeled O, and a large yellow atom is labeled S. Identify the intermolecular force shown. Two molecules are shown. In the first molecule on the left side of the image, a large yellow atom is bonded to two small red atoms on its left side. The second molecule on the right of the image has a large yellow atom bonded to two small red atoms on the bottom. A dashed line connects the yellow atom of the first molecule with a red atom of the second molecule. Identify the intermolecular force shown. Two substances are shown. On the left side of the image is a medium blue atom. On the right side of the image is a red atom bonded to two gray atoms on its right side. A dashed line connects the blue atom to the red atom. Identify the intermolecular force shown. Two molecules are shown, each made of two green atoms. A dashed line connects an atom of one molecule to an atom from the other molecule. Identify the intermolecular force shown. Two molecules are shown, each made from one green atom and one gray atom. A dashed line connects the gray atom of one molecule to the green atom of the other molecule.

dipole-dipole ion-dipole London dispersion hydrogen bond Solution Four types of intermolecular interactions occur between atoms and molecules: dipole-dipole interactions, hydrogen bond interactions, ion-dipole interactions, and London dispersion forces. Polar molecules have a net dipole (regions of partially positive and negative charges) as a result of polar bonds being arranged assymetrically in the molecule. Dipole-dipole interactions occur between the partially positive region of one molecule and the partially negative region of another molecule. Hydrogen bonds are a particularly strong form of dipole-dipole interaction that occurs between a hydrogen atom, H, that is directly bonded to a nitrogen, N, oxygen, O, or fluorine, F, atom and a N, O, or F atom in another molecule. In contrast, ion-dipole interactions, the strongest intermolecular interactions, occur when the partial positive or negative region of a polar molecule is attracted to an ion of the opposite sign. The weakest intermolecular interactions, London dispersion forces, occur between instantaneous dipole moments on an atom or molecule and the resulting induced dipole moment of a neighboring atom or molecule. London dispersion forces are present in all molecules, but are the predominant intermolecular force in nonpolar molecules. Hydrogen sulfide, H2S, is a polar molecule. The two hydrogen atoms, H, each have a partial positve charge, and the sulfur atom, S, has a partial negative charge. The interaction between the partial negative charge on the S atom and the partial positive charge of an H atom is an example of a dipole-dipole interaction. Water, H2O, is a polar molecule with a net dipole formed by the partial positive charges on each of the hydrogen atoms, and the partial negative charge on the oxygen atom. The interaction between a sodium ion, Na+, and the partial negative charge on the oxygen atom of water is an example of an ion-dipole interaction. Fluorine molecules, F2, are nonpolar and do not have overall dipole moments. Therefore, they cannot participate in dipole-dipole interactions, hydrogen bonding interactions, or ion-dipole interactions. Instead, the molecules interact through London dispersion forces formed by temporary distortions in the electron clouds around each molecule. Hydrogen fluoride, HF, contains an H atom directly bonded to an F atom. The interaction between the hydrogen atom of one molecule of HF with the fluorine atom from another molecule of HF is an example of a hydrogen bond.

For liquids, which of the factors affect vapor pressure?

intermolecular forces temperature Solution Vapor pressure describes the pressure of the vapor phase in equilibrium with the liquid phase of a substance. Intermolecular forces and temperature are the only factors that affect vapor pressure. Increasing temperature increases the average kinetic energy of the molecules, which increases the force of the collisions and the probability that a molecule will be able to overcome the intermolecular forces and escape into the vapor phase. The strength of the intermolecular forces also affects that probability. The weaker the forces, the easier it is for a molecule to break away. Surface area is not a factor because pressure is force per unit area. In other words, the area cancels out. Volume is not a factor because vaporization only occurs at the surface. Humidity can affect the pressure above water at a given instant, but it does not change the pressure that would result if equilibrium were established. Remember that vapor pressure can only be measured at equilibrium when there is no net evaporation or condensation.

Consider the phase diagram for carbon dioxide. PICTURED: A phase diagram, where the x axis is labeled temperature in degrees Celsius and ranges from about -100 to 31.1 degrees. The y axis is labeled pressure in atmospheres and extends from about 0 to 73.0 atmospheres. A line extends from the origin and curves to the upper right. The gas phase is below this line, and both the solid phase and the liquid phase lay on top of this line. There are two points on this line, (-78.5, 1.0) and (-56.4, 5.11). At the triple point (-56.4, 5.11), another line extends upward from the first line, separating the solid and liquid phases. The solid phase is to the left of the second line, and the liquid phase is to the right. In what phase is CO2 at 72 atm and 0 °C? Starting from the same point, 72 atm and 0 °C, what phase change would eventually result from a decrease in pressure?

liquid vaporization Solution Locate the approximate pressure on the vertical axis and the approximate temperature on the horizontal axis. The point at which these values meet is in the liquid portion of the graph and so CO2 is a liquid at 72 atm and 0 °C. A decrease in pressure means we must move downward on the graph, so eventually you cross over into the gas portion of the graph. The phase change from liquid to gas is called vaporization.

Use the changes of state interactive to answer the question. Find the melting point and boiling point of each substance. melting point of water: 0°C boiling point of water: 100°C melting point of bromine: −5°C boiling point of bromine: 60°C melting point of benzene: 5°C boiling point of benzene: 80°C

melting point of water: 0°C boiling point of water: 100°C melting point of bromine: −5°C boiling point of bromine: 60°C melting point of benzene: 5°C boiling point of benzene: 80°C Solution In the changes of state interactive, heat each of the substances on the hot plate. The melting point and boiling point temperatures correspond to the two regions of the graph where the temperature remains constant as the amount of heat added increases. The melting point is the temperature at which a substance changes from a solid to a liquid. The boiling point is the temperature at which a substance changes from a liquid to a gas. The melting point of water is 0 ∘C, and its boiling point is 100 ∘C. The melting point of bromine is −7.2 ∘C, and its boiling point is 58.8 ∘C. The melting point of benzene is 5.5 ∘C, and its boiling point is 80.1 ∘C.

Solid matter held in place by electrostatic attraction forces between positive ions and a "sea" of freely moving electrons describes what type of bonding?

metallic Solution The statement describes a metallic bond. The electron-sea model of metallic bonding describes how the electrons are the "sea" and the positive ions are "floating" in this sea. Negative electrons and positive ions create the attractive force that holds a metal together. Ionic bonds occur when electrons are transferred, forming positively and negatively charged ions, which create an attractive force that holds them together. A covalent bond involves two atoms sharing electrons in order to form octets of electrons. Covalent bonds do not involve ions. Hydrogen bonding is a type of dipole-dipole force. Dipole-dipole forces are the attractive forces between oppositely charged parts of polar molecules.

Identify the type of bonding within each substance. Co(s) CoCl2(s) CCl4(l)

metallic ionic covalent Solution A compound made up of only nonmetal atoms will have molecules held together with covalent bonds. A substance made up of only metal atoms is held together by metallic bonding. A compound containing both metal and nonmetal atoms is held together with ionic bonds. In a solid chunk of cobalt metal, there are only metal atoms. Thus, those atoms are held to each other through metallic bonding. CoCl2(s) contains both metal and nonmetal atoms and is, therefore, an ionic compound. CCl4(l) contains only nonmetal atoms and is, therefore, a molecular compound. Each molecule in a molecular compound is held together by covalent bonds.

Identify the packing in the unit cell of the ionic crystals shown. PICTURED: A six sided cube has a gray sphere in the center of the cube, a green sphere on each corner, and a line along each edge of the cube connecting the center of each corner sphere with the centers of the three adjacent corner spheres. PICTURED:A lattice has a large cube composed of eight smaller cubes, with a large green sphere on each outer corner and in the center of each of six outer faces, and a small gray sphere on the midpoint of each outer edge and in the center of the lattice. Lines connect the center of each sphere with the centers of adjacent spheres.

simple (primitive) cubic face-centered cubic Solution First, determine the angles in the unit cell of the ionic crystal. An equilateral hexagonal prism has a sphere on each of the 12 corners and a sphere in the center of the two hexagonal faces, with lines connecting the center of each corner sphere to the centers of the three adjacent corner spheres and the closest face sphere. The spheres on each face are also connected by a dotted line. Three spheres form an equilateral triangle inside the prism, centered between the hexagonal faces. These spheres are connected by a dotted line. This is labeled hexagonal close-packed. A six sided cube with a sphere on each corner and a line along each edge of the cube connecting the center of each sphere with the centers of the three adjacent spheres, labeled simple (primitive) cubic. A six sided cube with a sphere in the center of the cube, a sphere on each corner, and a line along each edge of the cube connecting the center of each corner sphere with the centers of the three adjacent corner spheres, labeled body centered cubic. A six sided cube with a sphere in the center of each face, a sphere on each corner, and a line along each edge of the cube connecting the center of each corner sphere with the centers of the three adjacent corner spheres. Dashed lines connect each face sphere with the sphere on the opposite face. This is labeled face centered cubic. hexagonal close-packed simple (primitive) cubic body-centered cubic face-centered cubic Neither ionic crystal has angles at any vertices that are 60° or 120°, so you can rule out hexagonal close‑packed unit cell packing. Having established that the ionic crystal has a cubic unit cell (all unit cell angles are 90°), note the positions of the atoms in the unit cell. Each cubic unit cell shown has atoms occupying the corners. The atom positions that allow you to distinguish the packing of cubic unit cells are the center of the cube and the faces of the cube. The packing is determined using only ions of the same type. A six sided cube has a gray sphere in the center of the cube, a green sphere on each corner, and a line along each edge of the cube connecting the center of each corner sphere with the centers of the three adjacent corner spheres. CsCl(s) has chloride anions (green) at the corners of its cubic unit cell, and a cesium cation (grey) in the center of the cube. Looking at only the green chloride ions, you can rule out face‑centered cubic unit cell packing because there are no atoms occupying the faces of the CsCl(s) unit cell. CsCl(s) is not body‑centered cubic because the ion species in the center of the unit cell (cesium) is different than those occupying the corners (chloride). Cesium chloride has a simple (primitive) cubic unit cell packing because the unit cell has chloride ions only at the corners of the cube. If the unit cell were drawn shifted by 1/2 plane, the cesium cations would occupy the corners of the primitive unit cell with a chloride anion at the center. A lattice has a large cube composed of eight smaller cubes, with a large green sphere on each outer corner and in the center of each of six outer faces, and a small gray sphere on the midpoint of each outer edge and in the center of the lattice. Lines connect the center of each sphere with the centers of adjacent spheres. NaCl(s) has chloride anions (green) at the corners and faces of its cubic unit cell. If the view of the unit cell were drawn shifted by one plane of atoms, the sodium cations (grey) would occupy the corners and faces of the unit cell. As drawn, the sodium cations occupy the octahedral holes in the unit cell. You can rule out simple (primitive) cubic and body‑centered unit cell packing because of the chloride anions at the faces of the unit cell. Face positions are empty in simple cubic and body‑centered cubic unit cells. NaCl(s) has face‑centered unit cell packing, with atoms occupying the corners and faces of the cubic unit cell.