Unit 1 Flashcards
Same molec
Same T
Different V
VP?
Same VP
Same molec
Diff Temp
VP?
Higher temp -> higher KE -> more molec in gas phase -> higher VP
Which molecule would you expect to have a lower H(l) at room temperature?
A. H2CO
B. CH3OH
Lower H(l) -> most stable -> the one with H bonding
Strong IMF -> most stable -> lowest E and lowest H(l)
Answer: B
Which molecule would you expect to have a lower H(g) at room temperature?
A. H2CO
B. CH3OH
Lower H(g) -> most stable -> IMF
Ideal gas -> no IMF
Answer: same
🔺S equation
🔺S = qrev/T = 🔺H/T
Which molecule would you expect to have a higher S(g) at room temperature?
A. H2CO
B. CH3OH
Higher S(g) -> more molec in gas phase -> weaker IMF Answer: A
🔺G° equation
🔺G° = 🔺H° - T🔺S°
What is 🔺G and 🔺Suniv for a spontaneous process?
Favored
Spontaneous -> 🔺G = negative
🔺Suniv = positive
What is 🔺G and 🔺Suniv for a nonspontenous process?
“Uphill”
🔺G = positive
🔺Suniv = negative
Does breaking IMFs require energy or release energy? Endo or exo? Sign of 🔺H? Examples?
Requires energy
Endo
🔺H = positive
Ex. Vaporization (l -> g), fusion (s -> l), sublimation (s -> g)
Does forming IMFs require energy or release energy? Endo or exo? Sign of 🔺H? Examples?
Releases energy
Exo
🔺H = negative
Ex. Condensation (g -> l), freezing (l -> s), deposition (g -> s)
At equilibrium, 🔺G = ? Greactants = ?
🔺G = 0 Greactants = Gproducts
MP/FP equation
T = 🔺Hfus / 🔺Sfus
Boiling point definition
When VP = the current atmospheric pressure -> liquid boils
Normal boiling point definition
Boiling point at atmospheric pressure (1 atm)
What will be the boiling point of water in Denver? Assume the atmospheric pressure is 0.82 atm.
A. >100 °C
B. <100 °C
C. 100 °C
0.82 atm -> not as many molecules in the gas phase
Patm ⬇️, Tbp⬇️
BP ⬇️
Answer: B
What’s the difference between boiling point and evaporation?
BP specific @ P
Evaporation -> room temp -> min KE needed to break IMF
Partial pressure equation
PV = nRT
* PA = nRT / V *
BP equation
T = 🔺Hvap / 🔺S
Which of the below substances would you expect to have the largest value for 🔺Hvap?
A. Methane
B. Water
C. Acetone
B
Clausius-Clapeyron Equation
ln(P2/P1) = 🔺Hvap/R (1/T1 - 1/T2) 1 molec. 2 temps 🔺Hvap -> J/mol R -> 8.314 J/mol•K T -> K
Warming solid/liquid/gas equation
q = mC🔺T
Comes out in J or kJ
Phase change equation
Melting solid -> q = n🔺Hfus
Boiling liquid -> q = n🔺Hvap
Comes out in J or kJ
For a phase change, we assume that 🔺T=? 🔺G=?
🔺T=0
🔺G=0
Equilibrium
Triple point definition
All 3 phases exist at this T & P
Critical point definition
The T & P limits for gas/liquid phases
Cannot distinguish gas/liquid above critical
If there is a positive slope, can you melt dry ice by applying a higher pressure? Is the solid or liquid more dense?
No
The solid
If there is a negative slope, can you melt ice by applying a higher pressure? Is the solid or liquid more dense?
Yes
The liquid
🔺Hsolution equation
🔺Hsolution = 🔺HLE + 🔺Hsolvation
Is dissolution for a solid solute endothermic or exothermic? Why?
Endothermic 🔺Hsolution = 🔺HLE + 🔺Hsolvation Positive = positive + negative 🔺HLE > 🔺Hsolvation Solute bonds > solvent bonds Ionic/ ion-ion bonds > H bonds
What is the sign of 🔺Ssolution when we dissolve a solid solute in a liquid solvent?
🔺Ssolution = positive
What will happen to a supersaturated aqueous sodium chloride solution when the temperature is increased?
A. No change.
B. The solubility will increase
C. The solubility will decrease
Solid solute ⬆️ T, ⬆️ KE ⬆️ solubility Breaking the solute-solute bonds in the lattice Answer: B
Is dissolution for a gaseous solute endothermic or exothermic? Why?
Exothermic
🔺Hsolution = 🔺HLE + 🔺Hsolvation
Negative = 0 + negative
No LE for gas solute
What is the sign of 🔺Ssolution when we dissolve a gaseous solute in a liquid solvent?
🔺Ssolution = negative
What will happen to a carbonated soda when the temperature is increased?
A. No change
B. The solubility will increase
C. The solubility will decrease
Gas solute ⬆️ T, ⬆️ KE ⬇️ solubility Break solute-solvent bonds (IMFs) Answer: C
Henry’s Law equations
Cgas = kH * Pgas
kH -> Henry’s constant [M/atm]
Pgas = K * Xgas
Xgas = # mol A / tot mol
Strong electrolyte definition
100% dissociation
AB -> A + B
Ionic = strong
Weak electrolyte definition
<100% dissociation
AB -> A + B, AB
Nonelectrolyte definition
0% dissociation
AB -> AB
Covalent, i = 1
How to calculate van’t Hoff factor
i = moles of particles in solution / moles of formula units dissolved
Molarity equation
M = moles solute / L solution
What does “concentration of particles” mean?
Molarity with the van’t Hoff factor -> iM = i(moles solute) / L solution
Molality equation
♍️ = moles of solute / kg solvent
Molality of the particles of a solution equation
♍️ = (i) mol solute / kg solvent
What is the effect of ⬆️ mixing?
⬆️ mixing, ⬆️ S, ⬇️ G, more stable, ⬇️ VP, ⬆️ BP
Gsolvent ? Gsolution
Ssolvent ? Ssolution
Gsolvent > Gsolution
Ssolvent < Ssolution
Raoult’s Law for VPsolution
VPsolution = Xsolvent * VPsolvent
Xsolvent = # moles solvent / tot moles
VPsolvent -> p°
Raoult’s Law for 🔺VP
🔺VP = -Xsolute * VPsolvent Xsolute = mol solute / tot moles Xsolute -> watch for i! VPsolvent-> VP° * VPsolution = VP°solvent + 🔺VP *
What is Xsolute for 1 mole of sodium chloride in 55.55 mol H2O?
X = mol solute / tot mol = 1 mol Na+ + 1 mol Cl- / (55.55 + 1 + 1) = 0.035
Raoult’s Law (for liquids)
VPsolution = VPsolvent A + VPsolvent B
Dalton’s Law (for gases)
Ptotal = PGas A + PGas B
Sample A: 250 mL of water
Sample B: 1 mole of sugar in 250 mL of water
Which sample has the highest boiling point? Why?
B
IMFs
solute-solvent
Boiling Point Elevation equation
🔺Tb = iKb♍️ 🔺BP = iKb♍️ 🔺BP -> °C Kb = BP elevation constant [°C/♍️] ♍️ -> molal Tbp’ = Tbp° + 🔺Tbp BP’ = BP° + 🔺BP
Nonvolatile solute van’t hoff factor
1
Sample A: 250 mL of water
Sample B: 1 mole of sugar in 250 mL of water
Which sample has the lowest freezing point? Why?
B
Solution is more stable over a wider Temp range
Freezing Point Depression equation
🔺Tf = -iKf♍️ FP = -iKf♍️ FP -> °C Kf = FP depression constant [°C/♍️] FP’ = FP° + 🔺Tf
If there are 0.0719 mol NaCl, how many mol solute are there?
mol solute = 0.0719 mol Na+ + 0.0719 mol Cl- = 0.1438 mol
You find a U tube that has allowed for the solvent to pass through the barrier after a period of time. A is lower and B is higher.
- Is A the solvent or the solution?
- Is B the solvent or the solution?
- Why?
- Solvent
- Solution
- Solvent moves to solution to try to establish equilibrium
Osmotic Pressure equation
pi = iMRT pi -> atm M -> molarity ionic -> iM R = 0.08206 L•atm / mol•K T -> Kelvin
Magnitude of Q or K Very Small
Reaction Progression?
Spontaneity?
Reactant Favored
nonspon
Magnitude of Q or K 1-ish
Reaction Progression?
[Products] = [Reactants]
Magnitude of Q or K Very Large
Reaction Progression?
Spontaneity?
Product Favored
spon
KC vs. KP equation
KP = KC(RT)^🔺n R -> 0.08206 L•atm / mol•K T -> Kelvin 🔺n = products - reactants gases only!
Product Favored Reaction
Product Favored -> Exo
G° products is lower than G° reactants
Reactant Favored Reaction
Reactant Favored -> Endo
G° reactants is lower than G° products
How to find 🔺G° on free energy diagram
Right side of curve - left side of curve
Low right side - high left side = negative (spon)
High right side - low left side = positive (nonspon)
Which side of the curve is Q < K?
Left side is always Q < K
🔺G° vs. 🔺G equation
🔺G = 🔺G° + RTlnQ RTlnQ -> mixing -> S 🔺G = standard + mixing (real) expt
🔺G° equation at equilibrium
🔺G = 0 Q = K 🔺G° = -RTlnK R = 8.314 J / mol•K T = 298 K
When K is small, what is the sign of 🔺G°?
Positive
When K is big, what is the sign of 🔺G°?
Negative
What happens when I increase the reactants?
⬆️ Reactant, ⬇️ Q, shift to make more products
Reactant —> Product
want: Q=K
“Add Away”
What happens when I decrease the reactants?
⬇️ Reactants, ⬆️ Q
Q > K
React ⬅️⬅️ Prod
“Take Toward”
Where does the reaction shift when heat energy is added to an exothermic reaction?
Reactants -> Products + Heat ⬆️ Prod, ⬆️ Q Q > K ⬅️⬅️ The reaction shifts to “add” some reactants.
Where does the reaction shift when heat energy is added to an endothermic reaction?
Reactants + Heat -> Products ⬆️ React, ⬇️ Q Q < K ➡️➡️ The reaction shifts to “add” some products.
van’t Hoff Equation
ln(K2/K1) = (🔺H°/R) * (1/T1 - 1/T2) K -> eq constants 🔺H° -> J/mol R -> 8.314 J/mol•K T -> Kelvin
Where does the reaction shift when we decrease the volume of the container of gases?
⬇️ V, ⬆️ P, P is too high
The reaction will shift in such a way to decrease the pressure.
It will move towards the side of the reaction with fewer gas molecules. -> wants to remove gas particles in order to ⬇️ P
Where does the reaction shift when we increase the volume of the container of gases?
⬆️ V, ⬇️ P, P is too low
The reaction will shift in such a way to increase the pressure.
It will move towards the side of the reaction with the greater number of gas molecules.
want more gas particles
How does decreasing the temperature of an exothermic reaction affect the value of K?
K will increase
How does increasing the temperature of an endothermic reaction affect the value of K?
K will increase
How does increasing the temperature of an exothermic reaction affect the value of K?
K will decrease
How does decreasing the temperature of an endothermic reaction affect the value of K?
K will decrease
