Exam 2 Lecture Notes

February 6th Lecture

Electrolyte: Separates into ions when it dissolves in water.

Strong electrolytes: Dissociates completely(100% of molecules dissociate to \n form ions).

Weak electrolytes: Dissociates very little (much less than 100%).

Nonelectrolyte: Does not dissociate to form ions.

Ionic Electrolytes

Ion dipole attraction: When the dipole of an ion (solute) is attracted to the polar dipole of the solvent

Dissociation: Polar solvent molecules surround and solvate the ions, reducing the strong electrostatic forces between them.

Covalent Electrolytes: Forms Ions due to a reaction with the water rather than dissociation.

Solubility

Solubility: The maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium.

Unsaturated Solution: Contains less than the equilibrium concentration of a dissolved solute.

Saturated Solution: Contains the maximum equilibrium concentration of a dissolved solute, with a little bit extra.

Supersaturated: Contains more than the equilibrium concentration of a dissolved solute.

Gas solubility

Gases become less soluble as temperature increases

Gases become more soluble as pressure increases

Henry’s Law: Cg = Kh * Pg

Cg: Molarity Mols/L | Kh: Henry’s laws constant | Pg: Partial pressure of the gas

If a diver breathes air (XN2 = 0.78) at a depth where the total pressure is 2.5 atm, calculate the concentration of nitrogen in his blood (in M). Henry’s law constant for N2 in blood at body temperature is 6.2 × 10−4 mol L−1 atm−1.

Pn2=(0.78)(2.5) = 1.95

Cg=6.2 x 10^-4 (1.95) = 1.2*10^-3

Miscible: If one liquid dissolves in another liquid with infinite mutual solubility

Immiscible: If two liquids do not mix (very low mutual solubility)

Colligative properties

Molarity (M): Amount (mols) of solute/ Volume (litters) of solution

• Useful for stoichiometric calculations and titrations
• Temperature-dependent due to volume
• Does not allow the determination of the exact amount of solvent used to prepare a solution
• A 1 M solution is called a “1 molar solution”

Molality (m): Amount (mols) of solute/ Mass (Kg) of solvent

• NOT temperature dependent (does not depend on volume!)
• Allows determination of the exact amount of solvent used to prepare a solution
• A 1 m solution is called a “1 molal solution”

\
\
\n

February 8th Lecture

^^Molality^^ (m)

molality (m) = amount (mols) of solute/mass (kg) of solvent

• NOT temperature dependent (does not depend on volume!)
• Allows determination of the exact amount of solvent used to prepare a solution
• A 1 m solution is called a “1 molal solution”

What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?

mols of CaCl2= 32.0g/110.98g = 0.2883 mols

271 g of H2O/ 1000g = 0.271kg of H20

0.2883 mols of CaCl2 / 0.271kg of H20 = 1.06 m

^^Mole Fraction^^

𝑋= 𝑎𝑚𝑜𝑢𝑛𝑡 𝑚𝑜𝑙 𝑜𝑓𝑠𝑜𝑙𝑢𝑡𝑒 / 𝑎𝑚𝑜𝑢𝑛𝑡 𝑚𝑜𝑙 𝑜𝑓𝑠𝑜𝑙𝑢𝑡𝑒+𝑎𝑚𝑜𝑢𝑛𝑡 𝑚𝑜𝑙 𝑜𝑓𝑠𝑜𝑙𝑣𝑒𝑛𝑡

^^Mass Percent^^

𝑀𝑎𝑠𝑠 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 = 𝑚𝑎𝑠𝑠 𝑜𝑓𝑠𝑜𝑙𝑢𝑡𝑒/ 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × 100

An alcohol solution contains 35.0 g of 1-propanol (CH3CH2CH2OH) and 150. g of ethanol \n (CH3CH2OH). Calculate the mass percent of each alcohol.

35g +150g = 185g

1-Propanol = 35g /185g x 100 = 18.9%

Ethanol = 150 g / 185g x 100 = 81.1%

An aqueous solution of H2O2 is 30.0% by mass and has a density of 1.11 g/mL. \n Calculate the (a) molality, (b) mole fraction of H2O2, and (c) molarity.

Given: Density -1.11 g/mL Solution - Solvent: H2O Solute: H2O2 Mass % - 30.0%

30.0 g of H2O2 / 100.0g solution

\
a) Mols of H2O2/ Kg of H2O

30.g of h2O2/ 34.02g of H2O2 = 0.8818 mols of H2O2

70g of H2O/ 1000 g = 0.07 Kg of H20

0.8818 mols of H202/ 0.07 kg of H20 = 12.6m

\
b) 70.0g / 18.02g = 3.885 mol H2O

0.8818 mols of H2O2/ 3.885 mol H2O + 0.8818 mols of H2O2 = 0.185

\
c) 100 g solution / 1.11 g /1000mL = 0.09009 L solution

0.8818 mols of H2O2 / 0.09009 L solution = 9.79 M

^^Colligative Properties^^

Colligative properties: Properties that depend on the # of solute particles, not their chemical identity.

1. Vapor pressure lowering

   1. A mole fraction is needed

2. Boiling point elevation

   1. molality will be needed

3. Freezing point depression

   1. molality will be needed

4. Osmotic pressure (we will not cover this one)

We will need to use two mole-based, temperature-independent concentration units \n for this section:mole fraction and molality.

^^Vapor Pressure Lowering^^

The vapor pressure ofa solution containing a nonvolatile solute is always \n lower than the vapor pressure of the pure solvent.

^^Raoult’s Law^^: The partial pressure (PA) exerted by any component \n of a solution is equal to the vapor pressure of the pure component \n (P°A) multiplied by its mole fraction in the solution (XA).

PA= XA × P°A

If the solute is nonvolatile (negligible vapor pressure), the vapor \n the pressure of the solution (P vap, soln) is only due to the solvent:

Pvap,soln = Xsolvent × P°solvent

\
Calculate the vapor pressure of a solution containing 92.1 g of glycerin, \n C3H5(OH)3, and 184.4 g of ethanol, C2H5OH, at 40 °C. The vapor \n pressure of pure ethanol is 0.178 atm at 40 °C.

Solvent: Ethanol (46.07 g/mol)

Solute: Glycerin (92.09 g/mol)

P° Ethanol: 0.178 atm

Moles ethanol: 184.4 g of ethanol / 46.07 g/mol = 4.002 mols of ethanol

Total moles: (92.1 g of glycerin / 92.09 g/mol) + 4.002 mols of ethanol = 5.002 total moles of solution

Mole fraction: 4.002 mols of ethanol / 5.002 mols of solution = 0.8001

Pvap, soln = 0.8001 x 0.178atm = 0.142 atm

^^Van’t Hoff Factor^^

A strong electrolyte dissociates completely to form ions in solution.

Use the compound’s formula to determine the # of particles in the solution. \n Each mole of MgCl2 is expected to give 3 moles of dissolved ions (Mg2+ and 2 Cl−).

van’t Hoff factor (i):a multiplying factor to take into account the dissociation of \n a strong electrolyte to predict the effect on the solution.

i = total moles of ions formed via dissociation / ionization 1 mol undissociated or unionized electrolyte

^^Boiling Point Elevation^^

A solution always boils at a higher temperature than the pure solvent.

Boiling point: The temperature at which the vapor pressure (Pvap) equals the external pressure, Pext.

Pvap = Pext → Tb

The boiling point elevation (ΔTb) is proportional to the molality of the solution.

ΔTb = iKbm

• ΔTb = Tb(solution) − Tb(pure solvent)
• i = van’t Hoff factor
  • If the solute is a nonelectrolyte (e.g. sugars, alcohols), i = 1.
  • If the solute is an electrolyte (e.g. NaCl, MgCl2), i > 1.
• Kb = boiling point elevation constant (°C/m)
• m = molality

You add 1.00 kg of ethylene glycol (C2H6O2; MW = 62.07 g/mol) antifreeze to 4450 g \n of water in your car's radiator. The boiling point of water is 100.00 °C, and the molal \n boiling point elevation constant for water is 0.512 °C/m. What is the boiling point of \n the solution in °C?

Moles of ethylene glycol: 1.00 kg of ethylene glycol * 1000 g / 62.07 g/mol = 16.11 mols of ethylene glycol molality of ethylene glycol: 16.11 mols of ethylene glycol / (4450 g of water /1000 g) = 3.62 m of ethylene glycol

Van’t Hoff Factor of ethylene glycol: C2H6O2 (s) → C2H6O2 (aq) i=1

Kb: 0.512 °C/m

ΔTb: 1 x 0.512 °C/m x 3.62 m of ethylene glycol = 1.85 °C

1.85 = Tb(solution) − 100.0 °C = 101.85°C

February 10th Lecture

^^Freezing point depression^^

A solution always freezes at a lower temperature than the pure solvent.

The freezing point depression (ΔTf) is proportional to the molality of the \n solution.

ΔTf = iKfm

• ΔTf = Tf(pure solvent) − Tf(solution)
  • WARNING: ΔTf is always positive
• i = van’t Hoff factor
  • If the solute is a nonelectrolyte (e.g. sugars, alcohols), i = 1.
  • If the solute is an electrolyte (e.g. NaCl, MgCl2), i > 1.
• Kf = freezing point depression constant (°C/m)
• m = molality

Applications of ΔTf: car antifreeze, de-icing sidewalks and roads

You add 1.00 kg of ethylene glycol (C2H6O2; MW = 62.07 g/mol) antifreeze to 4450 g \n of water in your car's radiator. The freezing point of water is 0.00 °C, and the molal \n freezing point depression constant for water is 1.86 °C/m. What is the freezing point \n of the solution in °C?

Moles of ethylene glycol: 1.00 kg of ethylene glycol * 1000 g / 62.07 g/mol = 16.11 mols of ethylene glycol molality of ethylene glycol: 16.11 mols of ethylene glycol / (4450 g of water /1000 g) = 3.62 m of ethylene glycol

Van’t Hoff Factor of ethylene glycol: C2H6O2 (s) → C2H6O2 (aq) i=1

Kf=1.86°C/m

ΔTf= 1 x 1.86°C/m x 3.62 m = 6.784°C

6.784°C = 0.00°C − Tf(solution) = -6.784°C

\
\
\
\

February 13th Lecture

Chemical kinetics: the study of how fast the change from reactants to \n products occurs.

Rate: the measure of how some property varies with time (ex: speed)

Reaction rate: the change in the amount of a reactant or product per unit of time.

• Any given reaction has a different rate under different conditions.

Rate = -(Δ [A]/Δ T)

In the reaction, aA + bB → cC + dD:

Relative rate: @@-(1/a)((Δ [a]/ Δ T) = -(1/b)((Δ [b]/ Δ T)@@ = ^^(1/c)((Δ [c]/ Δ T) = (1/d)((Δ [d]/ Δ T)^^

@@Orange@@ is negative because those are the reactants

^^Blue^^ is positive because those are the products.

Instantaneous rate: Rate at any one-time point ( slope of the tangent to the curve)

Initial rate: The instantaneous rate at time = 0 (always the fastest)

Average rate: Any two, time points to get the rate

Factors that Influence Reaction Rate

1. Chemical nature of the reacting substances
2. The physical state of the reactants: substances must mix for particles to collide. \n In 2 different phases, reactions only occur at the interfaces.
3. The temperature of the reactants: higher T = greater reaction rate \n With higher T, particles have more energy and therefore collide more \n often and more effectively.
4. Concentrations of the reactants: higher [reactants] = greater reaction rate
5. Presence of a catalyst: catalysts increase the rate of reaction

\

February 15th Lecture

The Rate Law

In the reaction, aA + bB → products

Rate = k[A]^m[B]^n

The term k is the rate constant, which is specific for a given reaction at a given \n temperature.

The exponents m and n are reaction ordersand are determined by experiment. \n They are NOT the coefficients a and b!

^^A → Products^^

• If the rate doubles when [A] doubles, the rate depends on [A]1 and the reaction is “first order with respect to A.”
• If the rate quadruples when [A] doubles, the rate depends on [A]^2 and the reaction is “second order with respect to A.”
• If the rate does not change when [A] doubles, the rate does not depend on [A], and the reaction is “zero order with respect to A.”

^^Units of the Rate Constant (k)^^

Overall Rxn (Order) Units of k (t in seconds)

0 mol/L·s (or mol L-1s-1)

1 1/s (or s^-1)

2 L/mol·s (or L mol^-1s^-1)

3 L^2/mol^2·s (or L^2 mol^-2s^-1)

General formula: units of K = (L/mol) ^order-1 / Units of t

^^Integrated Rate Laws^^

First-order rate equation: rate = – delta [A]/ delta t= k [A] or ln[A]t = –kt + ln[A]0

Second-order rate equation: rate = – delta [A]/ delta t = k [A]^2 or 1/[A]t= kt + 1/[A]0

Zero-order rate equation: rate = – delta [A]/ delta t= k [A]^0 or [A]t = –kt + [A]0

^^Half-Life^^

The **half-life (t½ )**for a reaction is the time taken for the concentration of a \n reactant to drop to half its initial value.

<<God Table<<

\n \n

February 17th Lecture

^^We did a lot of problems in this class, check paper notebooks for problems!^^

Collision theory: Reactants must collide in order to react!

1. The rate of a reaction is proportional to the rate of reactants
2. They must collide in the right orientation
3. They must collide with enough energy

\

February 20th Lecture

Activation energy: The energy needed to start the reaction

• The lower the activation energy the faster the reaction

When particles collide effectively they reach an activated state or transition state

Temperature and Activation energy: They are inversely proportional

Arrhenius equation: K=Ae^(-Ea/Rt)

• K= rate constant
• A= frequency factor
• R= universal gas constant 8.314 J/mol*K
• T= Temperature

Straight Line form of the Arrhenius equation:

lnk = lnA - Ea/R(1/T)

• K= rate constant
• A= frequency factor
• R= universal gas constant 8.314 J/mol*K
• T= Temperature

Differences in two temperatures:

ln (K1/K2) = -Ea/R((1/T2)-(1/T1))

• K= rate constant
• A= frequency factor
• R= universal gas constant 8.314 J/mol*K
• T= Temperature

^^Reaction mechanism^^

The mechanismof a reaction is the sequence of single reaction steps \n (elementary steps) that make up the overall equation.

Intermediates are substances that are made in a reaction and consumed completely in another part of the reaction

The rate law for an elementary step can be deduced from the reaction stoichiometry

• reaction order equals molecularity for an elementary step only.

Molecularity = the number of particles involved in the reaction

The slowest step in a reaction is the rate-determining or rate-limiting step.

\

February 22nd Lecture

The rate-limiting steps rate law becomes the rate law for the overall reaction

Energy diagrams for multi-step reactions: You will see the same number of peaks for the number of steps for the reactants.

• The rise before each peak is the activation energy required for each step
• The step with the highest peak is the rate-limiting step.
• The troth between the two peaks is where you would find intermediates.

Correlating Mechanism with Rate Law

1. The elementary steps must add up to the overall balanced equation.
2. Elementary steps should be reasonable (unimolecular or bimolecular. maybe trimolecular)
3. The mechanism must correlate with the observed rate law (rate law for RLS = overall rate law)

A mechanism is a hypothesis - we can not prove it to be correct, but we can use it to predict.

2NO2(g) + F2(g) →2NO2F(g) \n Rate = k[NO2][F2] (Observed through experiment) Problem

The accepted mechanism is: \n (1) NO2(g) + F2(g) → NO2F(g) + F(g) [slow; rate limiting] \n (2) NO2(g) + F(g) → NO2F(g) [fast]

Add up the two accepted mechanisms:

NO2(g) + F2(g) → NO2F(g) + ~~F(g)~~ \n NO2(g) +~~F(g)~~ → NO2F(g)

2NO2(g) + F2(g) →2NO2F(g) We passed step one!

Are they reasonable?

NO2(g) + F2(g) → NO2F(g) + F(g) Bimolecular \n NO2(g) + F(g) → NO2F(g) Biomolecular

Yes, we passed this step!

Rate law test:

NO2(g) + F2(g) → NO2F(g) + F(g)

Rate law= k[NO2][F2]

Rate law is the same as experimental which means we passed step three!

\

Catalyst

Substance that speeds up the reaction and does not get consumed by the reaction

• Different mechanisms with a lower activation energy
• does NOT affect delta H or the yield of products

Enzymes are Catalysts

• Lock and key, you know this you sexy beast ;)

\