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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 . Phase Changes. note the scale: energy (heat). also called fusion. Phase Diagram.
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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.
Phase Changes note the scale: energy (heat) also called fusion
Phase Diagram another way to see phase changes
Working Phase Change Heat Problems • Putting heat into a system is putting energy into a system. This energy appears mostly 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. • 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 = CΔT (often, C = sp. heat * mass) • 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) (Δphase changeHA)
When a solid is heated but is not at its melting point • For a solid, adding heat causes larger vibrations of the atoms or molecules from their resting positions. The temperature of the solid increases. • The solid remains a solid because the particles of the solid are held in place by intermolecular forces.
When a solid is heated but is not at its melting point, • heat flow, T, or specific heat may be found using q = CΔT • A half-pound block of iron (sp. heat = 0.46J/g*K) at 25°C is given 25 kJ of heat. What is the resulting temperature of the block of iron? • ΔT = _q_ (where here C = sp. heat * mass) • C • = _______25000 J________ • 0.500 lb x 453.6 g x 0.46 J • lb g*K • = 240 K …so the final T = 25 + 240 = 260°C
Phase Change When a Solid is Heated at its Melting Point • At the melting point, the particles in the solid are moving so much that additional heat will the overcome intermolecular forces holding them in place and the solid will undergo a phase changeto a liquid. • The intermolecular forces are still present, but are weaker because the average distance between particles has increased. The heat in effect changed (reduced) potential energy, not kinetic.
Phase Change When a Solid is Heated at its Melting Point • The stronger the intermolecular forces, the more heat must be added before the particles can break free. The higher the melting point of a solid, the stronger its intermolecular forces.
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 = mass* ΔfusH= (mol of solid) (ΔfusH) • molar mass • where ΔfusH is the heat of fusion of the solid.
Heat of Fusion Depends on the Strength of the Intermolecular Attractions Present • The stronger the attractions, the larger the heat of fusion. • 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
Heat of Fusion • The equation that shows the heat flow in a phase change (here, melting or fusion) is • q = (moles of solid) ΔfusH • ΔfusH is expressed as energy/mol or energy/gram. • The heat of fusion of iron is 15.19 kJ/mol (or 272 J/g) at its normal melting point of 1535 ºC.
Heat of Fusion Heat of fusion, ΔfusH: solid liquid Why can we not use q = CΔT to calculate how much heat is necessary to melt a certain mass of a solid? During a phase change, the phase changes but the temperature does not (ΔT=0).
When a Solid is Heated at its Melting Point • How much heat is needed to melt a half-pound block 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) • mol of Fe: 0.500 lb x 453.6 g x mol of Fe = 4.061 • lb 55.845 g • q = 4.061 mol (15.19 kJ/mol) = 61.69 kJ • The final T is 1535°C, because T doesn’t change during a phase change.
When a liquid is heated, but is not at its boiling point, • adding heat increases the average speed of the liquid particles. • 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, but weaker due to greater distance between particles).
When a liquid is heated but is not at its boiling point, • heat flow, T, or specific heat may be found using q = CΔT • A half-pound 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 = sp. heat * mass) • C • = _______25000 J________ • 0.500 lb x 453.6 g x 4.184 J • lb g*K • = 26 K …so the final T = 25 + 26 = 51°C
Phase Change When a Liquid is Heated at its Boiling Point • Eventually, the speeds will be so large that the liquid particles will no longer be significantly attracted by the intermolecular forces of the other liquid particles and will undergo a phase change to a gas.
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 = mass* ΔvapH= (mol of liquid) (ΔvapH) • molar mass • where ΔvapH is the heat of vaporization of the liquid.
Heat of Vaporization • Heat of vaporization, ΔvapH: liquid gas • We can not use q = CΔT, because during a phase change, the phase changes but the temperature does not (ΔT=0). • Use: q = (moles of liquid) ΔvapH • This has the same form as for fusion. • The heat of vaporization of water is 40.67 kJ/mol (or 2258 J/g) at its normal boiling point of 100ºC.
When a Liquid is Heated at its Boiling Point • How much heat is needed to vaporize a half-pound 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) • mol of H2O : 0.500 lb x 453.6 g x mol of H2O = 12.59 • lb 18.015 g • q = 12.59 mol (40.67 kJ/mol) = 512.0 kJ • The final T is 100°C, because T doesn’t change during a phase change.
Heating a Gas Does Not Result in a Phase Change • For a gas, adding heat causes the average speed of the gas particles to increase (Kinetic Theory of Gases). This does not result in a phase change, but in an increase in the temperature of the gas. • NOTE: Unless a solid is at its melting point (or sublimation point) or a liquid is at its boiling point, adding heat to a substance will always cause its temperature to rise.
Phase Changes: Heating Curvefor 1 mole of H2O at 1 bar water and water vapor water vapor The heat required to boil water is called the heatofvaporizationΔvapH. Temperature (°C) ice and water water The heat required to melt ice is called the heatoffusionΔfusH. 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 temperatures, and amounts. 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 Heat of vaporization of water ∆vapH = 40.67 kJ/mol at 100°C
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 temperatures, and amounts. 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 q = (25.0 g)(2.05 J/g-°C)(12°C) = 615 J There are 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 We cannot use q = C∆T because ∆T=0. We have the molar heat of fusion, so first we find the moles of ice: mol of H2O = 25.0 g / 18.015 g/mol = 1.3877 q = ∆fusH = (6.008 kJ/mol )(1.3877 mol) q = 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 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 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. Heat used to this point=0.615+8.337+10.46=19.412 kJ There are 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 We cannot use q = C∆T because ∆T=0. We have the molar heat of vaporization, and we already know we have 1.3877 mol H2O. q = ∆vapH = (40.67 kJ/mol )(1.3877 mol) q = 56.438 kJ…but 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 50.588 kJ = (40.67 kJ/mol ) (mol of water converted to steam) mol of water converted to steam = 50.588/40.67 = 1.244 1.244 mol H2O is converted to steam. 22.4 g H2O is converted to steam. 2.6 g H2O remains liquid. What is the temperature of the steam? What is the temperature of the water?
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 Curvefor 1 mole of H2O at 1 bar water vapor water and water vapor The heat released when water vapor condenses is called the heatofcondensationΔcondH. Temperature (°C) water ice and water The heat given off when water freezes is called the heatoffreezingΔfreezeH. ice
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?
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?