Chem Unit7
-
Upload
uc-berkeley-extension -
Category
Documents
-
view
689 -
download
2
description
Transcript of Chem Unit7
States of Matter
Particle vibration fluid motion rapid, random motionRigid positions move past independent of each other
each otherFixed volume fixed volume volume of container
Phases and Transitions
Sublimation
Condensation
EvaporationMelting
Freezing
Intermolecular Forces
• Strongest = IONIC FORCES
• High melting points• Oppositely charged ions
Dipole-Dipole Forces
• Dipole: contains both positively and negatively charged regions:
Hydrogen Bonding
• Special case of dipole-dipole forces
• Causes water to have higher than expected boiling point
Water Properties
• Surface tension
Water Properties
• Capillary action
Boiling PointsMolecule Boiling
Point
H2O 100°C
H2S –60.7 °C
H2Se –41°C
H2Te –2°C
Hydrogen Bonding
• Weak bond between H and electronegative atom:
• Recall
Hydrogen Bonding
Hydrogen Bonding
• DNA
Dipole-Dipole Forces
London (Dispersion) Forces
• Weak attractions between non-polar
• Increases with size of molecule (number of electrons)
• Molecular Shape (less compact > compact)
> >
London (Dispersion) Forces
• Weak attractions between non-polar
• “Temporary” or instantaneous dipoles
Intermolecular Forces
IONIC >> Dipole – Dipole >> London
Intermolecular Forces
• Which intermolecular force is expected?
CH4
Intermolecular Forces
• Which intermolecular force is expected?
CH4
Non-polar covalent molecule
London (dispersion) forces
Intermolecular Forces
• Which intermolecular force is expected?
Butanol
CH3CH2CH2CH2OH
Intermolecular Forces
• Which intermolecular force is expected?
Butanol
CH3CH2CH2CH2OH
Dipole-Dipole
Energy of Phase Changes
• Energy is required for all phase changes
Phase Diagram (H2O)
Phase Diagram (CO2)
Triple Point
Triple Point: where all three phases co-exist (T, p)
Phase Diagrams (carbon)
Properties of Matter
ForcePressure = area
h = 760 mm = 1 atm
Torricelli barometer
Robert Boyle
Boyle’s Law
• For a gas at constant T and nV and p are inversely proportional
pV = constant
Charles’ Law
• At constant pressure, V/T = constant
Charles’ Law
Gay – Lussac’s Law
• In a constant volume:P/T = constant
Gay-Lussac’s Law
Combined Gas Law
• Boyle’s Law pV = constant• Charles’ Law V/T = constant• Gay-Lussac’s Law p/T = constant
Combining all three:p1V1 p2V2
So: T1 = T2
Combined Gas Law
• A sample of gas has a volume of 400 liters when its temperature is 20°C and its pressure is 300 mm Hg. What volume will the gas occupy at STP?
Combined Gas Law
p1V1 p2V2
T1 = T2 T in Kelvin
(300/760 mm Hg)(400 L) = (760 mm Hg) (V2)
(293 K) (273 K)
V2 = 147 Liters
Combined Gas Law
• A sample of He gas has a volume of 250 mL at 456 torr and 25°C. At what temperature does this gas have a volume of 150 mL and 561 torr?
p1V1 p2V2
T1 = T2
Combined Gas Law
• A sample of He gas has a volume of 250 mL at 456 torr and 25°C. At what temperature does this gas have a volume of 150 mL and 561 torr?
p1V1 p2V2 T2 = p2V2 T1
T1 = T2 p1V1
Combined Gas Law
• A sample of He gas has a volume of 250 mL at 456 torr and 25°C. At what temperature does this gas have a volume of 150 mL and 561 torr?
p1V1 p2V2 (561/760)(0.15L)(298K)
T1 = T2 T2 = (456/760)(0.25L)
= 220 K = -53°C
Ideal Gases
• Non-interacting
• Point particles
• Randomly moving with elastic collisions (no energy lost)
Ideal Gases
• Avogadro’s Law:Equal volumes of gas contain the same number of molecules at the same T & p.
n = number of moles
p1V1 = constant
n T1
Ideal Gas Law
p1V1 = constant = 0.082 L atm = R
n T1 K mole
= Universal Gas Constant
One mole of gas at STP:Volume = nRT/p = (1 mole)(0.082 Latm)
(273K)/1 atm
= 22.4 Liters
Ideal Gas Law
• pV = nRT
• How many moles of Helium are present in a balloon that has a volume of 65 L at 20° C and 705 torr?Given Needed
V, T, p, R n n = pV/RT
Ideal Gas Law
• pV = nRT
• How many moles of Helium are present in a balloon that has a volume of 65 L at 20° C and 705 torr?
n = pV/RT
= (705/760 atm)(65 L)
(0.082 Latm/K)(293)
= 2.5 moles He
Ideal Gas
6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How many moles of this gas are present?
Ideal Gas
6.2 liters of an ideal gas are contained at 3.0 atm and 37 °C. How many moles of this gas are present?
n = pV/RT
= (3 atm)(6.2 L)(0.082 L atm/mole K)(310 K)
= 0.73 moles
Ideal Gases and Density
Density
• Gas density increases with molecular mass.
Density
• What is the density of NO2 gas at 0.97 atm and 35°C?
MW = 46 g/mole Molar mass p (46 g/mole)(0.97 atm)
• Density = RT (0.082 L atm/mole K) (308 K)
= 1.767 g/L
Gas Diffusion
• Movement of particles from region of Higher density to lower density
Gas Diffusion
• Movement of particles from region of Higher density to lower density
Depends on density (molar mass)
Graham’s Law proportionality
Rate of effusion inversely to square root of molar mass
Smaller molecules escape FASTER than larger molecules
Dalton’s Law
PT = P1 + P2 + P3 + . . . .
Total pressure of a gas sample is the sun of the partial pressures.
Dalton’s Law
• A mixture of O2, CO2 and N2 has a pT of 0.97 atm; if pO2 = 0.7 atm and pN2 = 0.12 atm, what is pCO2?
Dalton’s Law
• A mixture of O2, CO2 and N2 has a pT of 0.97 atm; if pO2 = 0.7 atm and pN2 = 0.12 atm, what is pCO2?
• pT = pO2 + pN2 + pCO2 = 0.97 atm
• pCO2 = 0.97 atm - (0.7 atm + 0.12 atm)
= 0.15 atm
Dalton’s Law
• The partial pressures of CH4 and O2 are 0.175 atm and 0.25 atm.
At 65°C in a volume of 2 L, how many moles of each gas are present?
Dalton’s Law
• The partial pressures of CH4 and O2 are 0.175 atm and 0.25 atm.
At 65°C in a volume of 2 L, how many moles of each gas are present?
nCH4 = pV/RT = 0.175 atm (2L)/0.082 L atm/mole K
(338 K = 0.126 moles
nO2 = pV/RT = 0.25 atm (2L)/ 0.082 L atm/mole K (338 K)
= 0.018 moles
Unit 7 Review
• Phases of Matter and Transitions• Intermolecular Forces• Phase Diagrams• Boyle’s, Charles’, Gay Lussac’s Laws• Ideal Gas Law pV = nRT• Graham’s Law of Diffusion• Partial Pressure