What is a Phase? A phase is a homogeneous, physically distinct, and mechanically separable portion of matter. It is uniform throughout, both in chemical composition and in physical state.

Why talk about phase changes in thermochemistry? Because a phase change requires either the input or release of heat (q). s  l (melting) l  g (vaporization) s  g (sublimation) are endothermic phase changes.

Why talk about phase changes in thermochemistry? Because a phase change requires either the input or release of heat (q). l  s (freezing) g  l (condensation) g  s (deposition) are exothermic phase changes.

Heating a substance that does not change phase Putting heat into a system is putting energy into a system. This energy appears as kinetic energy of the atoms or molecules of the system. Kinetic energy is energy of motion, so adding heat to a system not at a phase change increases the movement of the atoms and molecules. For a solid below its melting point, for a liquid below its boiling point, and always for a gas, putting heat into the system results in an increase in T: q = C ΔT C is the heat capacity and has units of 𝑱 𝑲 or 𝑱 𝑪 . It is often obtained by using the specific heat, which has units of 𝑱 𝒈 𝑲 or 𝑱 𝒈 °𝑪 . The units show that C = mass (sp. heat).

When a solid is heated, but is not at its melting point the amplitude of the vibrations of the atoms or molecules from their equilibrium positions increases. the temperature of the solid increases. The solid remains a solid because the particles of the solid are held in place by intermolecular forces. q = CΔT

When a liquid is heated, but is not at its boiling point, the average speed of the liquid particles increases. the temperature of the liquid increases. The liquid remains a liquid because the particles of the liquid are still loosely held by intermolecular forces (the same ones as were present in the solid). q = CΔT

When a Gas is heated the average speed of the gas particles increases. the temperature of the gas increases. q = CΔT

Heating a substance as it undergoes a phase change Putting heat into a system is putting energy into a system. At a phase change, this energy goes to increase the average separation of the atoms or molecules of the system. This can be viewed as doing work against a force, the force being the intermolecular attractions. The temperature of the substance undergoing a phase change does not change as heat is added to it. For a solid at its melting point and for a liquid at its boiling point, putting heat into the system does NOT increase T, but results in a phase change: q = (mol of A) (Δ phase changeHA)

When a Solid is Heated at its Melting Point the amount of heat needed to melt a certain mass of solid may be found using 𝒒= 𝒎𝒐𝒍 𝒔𝒐𝒍𝒊𝒅 ∆ 𝒇𝒖𝒔 𝑯 where ΔfusH is the heat of fusion of the solid. H2O(s)H2O(l) ΔfusH(0°C) = 6.008 𝒌𝑱

ΔfusH of water (hydrogen bonding) is 6.0 kJ/mol Intermolecular Attractions Influence MP, BP, and Heats of Phase Changes The stronger the intermolecular attractions, the larger the heat of fusion (and other phase changes), and the higher the mp and bp. ΔfusH of water (hydrogen bonding) is 6.0 kJ/mol ΔfusH of acetone (dipole-dipole) is 5.7 kJ/mol ΔfusH of butane (dispersion) is 4.7 kJ/mol

When a liquid is heated at its boiling point the amount of heat needed to vaporize a certain mass of liquid may be found using 𝒒= 𝒎𝒐𝒍 𝒍𝒊𝒒𝒖𝒊𝒅 ∆ 𝒗𝒂𝒑 𝑯 where ΔvapH is the heat of vaporization of the liquid. H2O(l)H2O(g) ΔvapH(100°C) = 40.67 𝒌𝑱

Heating a Solid not at its Melting Point Heat flow, T, ΔT, or specific heat may be found using q = CΔT . 0.500 lb of iron (sp. heat = 0.46 𝑱 𝒈 𝑲 ) at 25°C is given 25 kJ of heat. What is the resulting temperature of the iron? ΔT = 𝒒 𝑪 (C = sp. heat * mass) ΔT = _______25000 J_____g K 0.500 lb 453.6 g 0.46 J lb ΔT = 240 K …so the final T = 25 + 240 = 260°C

Heating a Solid at its Melting Point How much heat is needed to melt 0.500 lb of iron (ΔfusH = 15.19 𝒌𝑱 𝒎𝒐𝒍 ) at its melting point (1535°C)? What is the resulting temperature of the melted iron? q = (mol of Fe) (ΔfusH) 𝒒=𝟎.𝟓𝟎𝟎 𝒍𝒃 𝟒𝟓𝟑.𝟔 𝒈 𝒍𝒃 𝟏 𝒎𝒐𝒍 𝑭𝒆 𝟓𝟓.𝟖𝟒𝟓 𝒈 𝟏𝟓.𝟏𝟗 𝒌𝑱 𝒎𝒐𝒍 =𝟔𝟏.𝟕 𝒌𝑱 The final T is 1535°C, because T doesn’t change during a phase change.

Heating a Liquid not at its Boiling Point Heat flow, T, ΔT, or specific heat may be found using q = CΔT . 0.500 lb of water (sp. heat = 4.184 𝑱 𝒈 𝑲 ) at 25°C is given 25 kJ of heat. What is the resulting temperature of the water? (Remember how hot the Fe got?) ΔT = 𝒒 𝑪 (C = sp. heat * mass) ΔT = _______25000 J____ g K 0.500 lb 453.6 g 4.184 J lb ΔT = 26 K …so the final T = 25 + 26 = 51°C

Heating a Liquid at its Boiling Point How much heat is needed to vaporize a 0.500 lb of water (ΔvapH = 40.67 𝒌𝑱 𝒎𝒐𝒍 ) at its boiling point (100°C)? What is the resulting temperature of the water vapor? q = (mol of H2O) (ΔvapH) 𝒒=𝟎.𝟓𝟎𝟎 𝒍𝒃 𝟒𝟓𝟑.𝟔 𝒈 𝒍𝒃 𝟏 𝒎𝒐𝒍 𝑯 𝟐 𝑶 𝟏𝟖.𝟎𝟏𝟓 𝒈 𝟒𝟎.𝟔𝟕 𝒌𝑱 𝒎𝒐𝒍 =𝟓𝟏𝟐 𝒌𝑱 The final T is 100°C, because T doesn’t change during a phase change.

Heating Curves A heating curve is a quantitative way to show the effect of heat on the temperature of a substance. Two equations are used to generate the curve: q = (mol of A) (Δfus or vapHA) (for phase changes) q = mass (specific heat) ΔT (everywhere else)

“Generic” Heating Curve for Substance A bp of A A(l) + A(g) A(g) ΔvapH of A Temperature (°C) A(s) + A(l) A(l) ΔfusH of A A(s) mp of A

Phase Changes: Heating Curve for 1 mole of H2O at 1 bar bp of water water and water vapor water vapor The heat required to boil water is called the heat of vaporization ΔvapH. Temperature (°C) ice and water water The heat required to melt ice is called the heat of fusion ΔfusH. ice mp of ice

Working a Heating Curve-Type Problem 70.0 kJ of heat are added to a 25.0-g cube of ice at -12°C. Predict the final state(s) of the H2O, their temperature(s), and amount(s). Information*: Specific heat of ice Cice = 2.05 𝑱 𝒈 °𝑪 Specific heat of water Cwater= 4.184 𝑱 𝒈 °𝑪 Specific heat of steam Csteam= 1.864 𝑱 𝒈 °𝑪 Heat of fusion of ice ∆fusH = 6.008 𝒌𝑱 𝒎𝒐𝒍 at 0°C Heat of vaporization of water ∆vapH = 40.67 𝒌𝑱 𝒎𝒐𝒍 at 100°C *For other substances, you would also be given melting point and boiling point.

Working a Heating Curve-Type Problem 70.0 kJ of heat are added to a 25.0-g cube of ice at -12°C. Predict the final state(s) of the H2O, their temperature(s), and amount(s). We solve this problem in steps, according to the physical state of the H2O. Step 1. Warming the ice from -12 °C to 0°C q = C∆T = mass (sp. heat of ice) ∆T q = (25.0 g)(2.05 𝑱 𝒈 °𝑪 )(12°C) = 615 J (0.615 kJ) There are 70.0 – 0.615 = 69.385 kJ remaining, so now we go to the next step.

Working a Heating Curve-Type Problem Step 2. Melting the ice at 0°C At a phase change, in this case, fusion, 𝒒= 𝒎𝒐𝒍 𝒔𝒐𝒍𝒊𝒅 ∆ 𝒇𝒖𝒔 𝑯 𝒒=𝟐𝟓.𝟎𝒈 𝟏 𝒎𝒐𝒍 𝑯 𝟐 𝑶 𝟏𝟖.𝟎𝟏𝟓 𝒈 𝟔.𝟎𝟎𝟖 𝒌𝑱 𝒎𝒐𝒍 =𝟖.𝟑𝟑𝟕 𝒌𝑱 At the end of step 2, all of the ice is melted and at 0°C. Heat used to this point = 0.615 + 8.337 = 8.952 kJ There are 70.0 – 8.952 = 61.048 kJ remaining, so now we go to the next step.

Working a Heating Curve-Type Problem Step 3. Heating the water from 0°C to 100°C q = C∆T = mass (sp. heat of water) ∆T q = (25.0 g)(4.184 𝑱 𝒈 °𝑪 )(100°C) = 10460 J or 10.46 kJ Since there are 61.048 kJ of heat available, there is more than enough heat to bring all of the water to 100°C. At the end of step 3, the water has reached a temperature of 100°C. It is not boiling. Heat used to this point=0.615+8.337+10.46=19.412 kJ There are 70.0 – 19.412 = 50.588 kJ remaining, so now we go to the next step.

Working a Heating Curve-Type Problem Step 4. Boiling the water at 100°C At a phase change, in this case, boiling or vaporization, 𝒒= 𝒎𝒐𝒍 𝒍𝒊𝒒𝒖𝒊𝒅 ∆ 𝒗𝒂𝒑 𝑯 𝒒=𝟐𝟓.𝟎𝒈 𝟏 𝒎𝒐𝒍 𝑯 𝟐 𝑶 𝟏𝟖.𝟎𝟏𝟓 𝒈 𝟒𝟎.𝟔𝟕 𝒌𝑱 𝒎𝒐𝒍 =𝟓𝟔.𝟒𝟑𝟗 𝒌𝑱 We only have 50.588 kJ of heat, so we do not have enough heat to convert all of the water to steam. We must find how many mol of water 50.588 kJ will convert to steam.

Working a Heating Curve-Type Problem Step 5. Determining how much water is converted to steam: 𝒒= 𝒎𝒐𝒍 𝒍𝒊𝒒𝒖𝒊𝒅 ∆ 𝒗𝒂𝒑 𝑯 50.588 kJ = (mol of water converted to steam) (40.67 𝒌𝑱 𝒎𝒐𝒍 ) mol of water converted to steam = 𝟓𝟎.𝟓𝟖𝟖 𝒌𝑱 𝒎𝒐𝒍 𝟒𝟎.𝟔𝟕 𝒌𝑱 =𝟏.𝟐𝟒𝟒 𝒎𝒐𝒍 1.244 mol H2O is converted to steam. 22.4 g H2O is converted to steam. 25.0 – 22.4 = 2.6 g H2O remains liquid. What is the temperature of the steam? The water?

Working a Heating Curve-Type Problem

Phase Changes: If you know one ∆H, you know a second one. The heat required to melt a given mass of ice is equal to the amount of heat given off when water freezes: ΔfusH = - ΔfreezeH This holds for the other two phase changes as well: ΔvapH = - ΔcondH ΔsubH = - ΔdepH

Phase Changes: Cooling Curve for 1 mole of H2O at 1 bar condensation point of steam water vapor water and water vapor The heat released when water vapor condenses is called the heat of condensation ΔcondH. Temperature (°C) water ice and water freezing point of water The heat given off when water freezes is called the heat of freezing ΔfreezeH. ice

Unit 2-2: Phase Changes Vapor Pressure The vapor pressure of a liquid (or solid) is the pressure exerted by its vapor when the liquid (or solid) and vapor states are in dynamic equilibrium. System once the liquid and gaseous ethanol have reached equilibrium at 25°C. System right after ethanol is added to the flask at 25°C.

Kinetic Energy of Molecules Putting heat into a system is putting energy into a system. Away from a phase change, this energy appears as kinetic energy of the atoms or molecules of the system. Kinetic energy is energy of motion, so adding heat to a system increases the movement of the atoms and molecules.

Vapor Pressure Adding heat causes the average speed of the liquid particles to increase and, consequently, the number of liquid particles that turn into gas to increase. The vapor pressure of a liquid (or solid!) increases with increasing temperature. the liquid.

Vapor Pressure If a liquid with a high vapor pressure is in an open container, it will evaporate more quickly than a liquid with a low vapor pressure (if both are at the same temperature). Example: Acetone has a higher vapor pressure than water and evaporates much more quickly than water at room temperature. Liquids that evaporate readily are said to be volatile. The larger the intermolecular forces in the liquid, the lower its vapor pressure and the less volatile it will be.

Boiling Points and Vapor Pressure The boiling point of a liquid is the temperature at which the vapor pressure of the liquid equals the external pressure acting on the surface of a liquid. (This is why liquids boil at lower temperatures the farther up the mountain you go.) The normal boiling point of a liquid is the temperature at which the vapor pressure of the liquid is 1 atm.

Boiling Points and Vapor Pressure What are the boiling points at 200 torr?

another way to see phase changes Phase Diagram another way to see phase changes

Phase Diagram is a graphical way to summarize the conditions under which equilibria exist between the different states of matter. The black curves (AD, AB, AC) denote conditions under which two phases exist in equilibrium.

Phase Diagram For pressures and temperatures that don’t correspond to any of the lines, only one phase of the substance is stable.

Phase Diagram The line shown by AD is the melting point line, where solid and liquid coexist (are in equilibrium). It could also be called the freezing point line. The normal melting point of a substance is the temperature at which solid and liquid coexist at 1 atm.

Phase Diagram The curve AB is the boiling point line, where liquid and vapor coexist. It could also be called the condensation point line. To repeat: Along any black line, the phases on either side of the line are in equilibrium.

Phase Diagram The curve along AC is the sublimation line, where solid and vapor coexist (are in equilibrium). It could also be called the deposition line. A is the triple point, the temperature and pressure at which three phases of a substance coexist.

Phase Diagram B is the critical point, beyond which liquid and gaseous phases cannot be distinguished. At temperatures and pressures above the critical point, the substance is said to be a supercritical fluid. The critical temperature TC and the critical pressure PC are the T and P at the critical point.

Phase Diagrams What happens as we increase T for water at 1 atm? CO2? What happens as the pressure is increased for water at a constant T? Does CO2 behave the same way?

Unit 6 - Thermo 4 Phase Changes 4/15/2017 Reading a Phase Diagram What is the physical state of water at… http://en.wikipedia.org/wiki/Phase_diagram