CH 222 Midterm 2

IMF's

100

How do IMFs relate to boiling point and vapor pressure?

Stronger IMFs will have higher boiling points and lower vapor pressure

Boiling point: stronger forces means stronger attraction, so it’s harder to overcome those attractions in order for the liquid to become a gas

Vapor Pressure: Stronger attraction means smaller amounts of particles want to escape into the gas phase

200

What is the strongest IMF for the following molecule?

CH₃CH₃

London Dispersion

*The symmetry of the molecule would result in it being nonpolar, so it would only have london dispersion

300

What is the strongest IMF for the following molecules?

CH₂O

CH₃CH₂CH₂OH

OF₂

CH₂O ---> Dipole-Dipole

*The oxygen connected to the carbon would pull the electron density more strongly than the hydrogens, resulting in a polar molecule

CH₃CH₂CH₂OH ---> Hydrogen Bonding

*A hydrogen is bonded to an oxygen

OF₂ ---> Dipole-Dipole

*This molecule would have a tetrahedral bent structure due to the two long pairs on oxygen, making it a polar molecule

400

Rank the following compounds from lowest to highest boiling points based on their IMF's

GeCl4 , CH4 , SIH4 , GeH4

CH₃OH , CH₃CH₂CH₃ , CH₃CH₂OH

GeCl4 , CH4 , SIH4 , GeH4

CH4 < SIH4 < GeH4 < GeCl4

*All of these compounds would have london dispersion, so we go off molar mass from least to greatest to determine our ranking

CH₃OH , CH₃CH₂CH₃ , CH₃CH₂OH

CH3CH2CH3 < CH3OH < CH3CH2OH

*Two of the molecules would have hydrogen bonding due to the OH, but the longer compound would have a higher BP due to a larger molar mass

500

Define the following terms:

London-Dispersion

Dipole-Dipole

Hydrogen Bonding

London-Dispersion: temporary dipoles, something every molecule has

Dipole-Dipole: permanent dipoles that are present in polar molecules

Hydrogen Bonding: strong dipole molecules, occurs when a hydrogen is paired to a nitrogen, oxygen, or fluorine

Heating Curves

100

A

Melting starts at 50 ^oC and it melts from minute 2 to minute 5. 

200

How much energy (in kJ) is required to heat 29.0g of ice from -13 ^oC to 64 ^oC? (ΔH_fusion for water is 6.02 kJ/mol) 

(C_s,ice = 2.09 J/g C)

(C_s,liquid = 4.18 J/g C)

q = (29g)(2.09)(0+13) = 0.788 kJ
ΔH_fus = 1.609 mol x 6.02 kJ/mol = 9.69 kJ

q = (29g)(4.18)(64-0) = 7.76 kJ

0.788kJ + 9.69kJ + 7.76 kJ = 18.2 kJ

300

Identify the phase changes and whether they are endothermic or exothermic: 

1. Fusion

2. Freezing

3. Vaporization

4. Condensation

5. Sublimation

6. Deposition

1. Fusion: solid ---> liquid endothermic

2. Freezing: liquid ---> solid exothermic

3. Vaporization: liquid ---> gas endothermic 

4. Condensation: gas ---> liquid exothermic 

5. Sublimation: solid ---> gas endothermic

6. Deposition: gas ---> solid exothermic

400

Calculate the amount of heat (in kJ) that is released when 35.0g of benzene vapor(C₆H₆) is cooled from 96.0 ^oC to 26.4 ^oC

BP benzene: 80.09 ^oC    ΔHvap = 30.72 kJ/mol    

FP benzene: 5.49^oC

Cs,liq = 1.74 J/g ^oC        Cs,gas = 1.055 J/g ^oC

q = (35g)(1.055)(80.09 - 96) = -0.587 kJ

ΔHvap = -30.72 kJ/mol x 0.448 mol = -13.8 kJ

q = (35g)(1.74)(26.4 - 80.09) = -3.27 kJ

-17.7 kJ

500

How much heat (in kJ) is released when 73.0g of water is cooled from 136 ^oC to -27 ^oC? 

Cs steam = 2.09 J/g C 

Cs liquid H2O = 4.18 J/g C 

ΔHvap = 40.7 kJ/mol)

q = (73g) (2.09) (100-136) = -5.49 kJ

ΔHvap = 4.05 mol x (-40.7 kJ/mol) = -164.9 kJ

q = (73) (4.18) (0-100) = -30.51 kJ

ΔH_fus = 4.05 mol x (-6.02 kJ/mol) = -24.38 kJ

q = (73) (2.09) (-27-0) = -4.12 kJ

(-5.49 kJ) + (-164.9 kJ) + (-30.51 kJ) + (-24.38 kJ) + (-4.12 kJ) = -229.7 kJ

Vapor Pressure & Osmotic Pressure

100

Calculate the vapor pressure of a nonvolatile solution made by dissolving 50.0 g glucose, C₆H₁₂O₆, in 500 g of water. The vapor pressure of pure water is 47.1 torr at 37°C

P_soln = X_solv x P^o_solv

Water is our solvent due to there being a larger amount of it

Answer: 46.63 torr

500g H2O xx 1/18.02 = 27.75 mol

50g glu xx 1/180.16 = 0.2775 mol

27.75/(27.75 + 0.2775) = 0.99

P = 0.99 xx 47.1 = 46.63

200

Calculate the osmotic pressure for a sample of aqueous 0.10 M Na3PO4 at 20°C.

3 Na⁺ & 1 PO₄^3- → i = 4

T = 20C + 273.15 = 293.15 K

Π = (4)(0.10M)(0.08206)(293.15K) = 9.6 atm

300

A physiological saline solution is 0.92% NaCl by mass. What is the osmotic pressure of such a solution at a body temperature of 37 °C? (density = 1 g/ml)

(hint: assuming 100g of solution would help)

(2nd hint: assuming 100g of solution will give us the overall liters as well using the given density of the solution)

Assume 100g of solution, so we have 0.92g NaCl

100g solution ---> 100mL solution ---> 0.1 L

0.92g / 58.443 g/mol = 0.015742 mol / 0.100L = 0.157 M

Π = (2)(0.157)(0.08206)(310) = 7.99 atm

400

Calculate the vapor pressure of a volatile solution of 74.0 g of benzene (C₆H₆) in 48.8 g of toluene (C₇H₈) at 25.0 °C. The vapor pressure of benzene is 95.1 torr and of toluene is 28.4 torr at this temperature.

Benzene: 74g / 78.113 (g/mol) = 0.947 mol

Toluene: 48.8g / 92.14 (g/mol) = 0.5296 mol

X_ben = 0.947 / (0.947 + 0.5296) = 0.641

X_tol = 0.5296 / (0.947 + 0.5296) = 0.3586 mol

P_soln = (95.1 torr)(0.6414) + (28.4 torr)(0.3586) = 71.2 torr

500

Hemoglobin is a large molecule that carries oxygen in human blood. A water solution that contains 0.263 g of hemoglobin (Hb) in 10.0 mL of solution has an osmotic pressure of 7.51 torr at 25 ^oC 

What is the molar mass of the hemoglobin?

(Hint: Hemoglobin is considered a nonelectrolyte)

i = 1

T = 25 C + 273.15 K = 298.15 K

Π = iMRT → M = Π / (iRT)

M = (0.009882atm) / (1 x 0.08206 x 298.15K) = 0.000404M

0.000404M x 0.01 = 4.04 x 10^-6 mol  

0.263g / 4.04 x 10^-6 mol = 6.51 x 10⁴ g/mol

Freezing Point and Boiling Point

100

What is the freezing point depression when 62.2 g of toluene (C7H8) is dissolved in 481 g of naphthalene? The freezing point constant for naphthalene is 7.00 °C/m.

tf = (1) (7.00 °C/m) (0.675058 mol / 0.481 kg)

tf = 9.82 ^oC

200

A solution of 10g of sodium chloride is added to 100g of water in an attempt to elevate the boiling point. What is the new boiling point of the solution? Kb for water is 0.52 ^oC/m.

tb = (2)(0.52 C/m)(1.71 m) = 1.78

  1.78 C + 100 C = 101.78 ^oC

300

 How many grams of fucose (C₆H₁₂O₅) must be dissolved in 419 g of benzene to raise the boiling point by 3.8 °C? The boiling point constant for benzene is 2.67 °C/m.

3.8 = (1)(2.67)(m)    

m = (1.423 mol/1 kg) x 0.419 kg = 0.596 mol

(0.596 mol) x (164.16g/mol) = 97.9 g

400

2.00 g of some unknown compound reduces the freezing point of 75.00 g of benzene (C₆H₆) from 5.53 C to 4.90 C. What is the molar mass of the compound?

(4.90 - 5.53) ^oC = (1) (5.12) m

m = 0.123 mol/kg x 0.075 kg benzene = 0.00923 mol solute

2.00g / 0.00923 mol = 216.80 g/mol

500

Calculate the boiling point of a 4.81% by mass solution of magnesium chloride (MgCl₂ ; molar mass = 95.211g/mol). 

The van't Hoff factor of MgCl₂ is 2.7 

Kb of water is 0.512 °C/m

(hint: assume 100g of solution) 

Assuming 100g of solution will give us 4.81g MgCl₂ and 95.19g water 

4.81g MgCl2 x 1mol/95.211g = 0.005052 mol

Δt = (2.7)(0.512 C/m)(0.005052mol / 0.00952 kg) = 0.73 ^oC

Δt = 100 C + 0.73 = 100.73 ^oC

Rate Laws

100

Which reaction will proceed at a faster rate?

A

*the lower activation energy needed for the reaction to begin will make the reaction proceed at a quicker rate 

200

Determine the units for the following in terms of M and s

a. Units for [A]?

b. Units for the rate of reaction?

c. Units for the rate constant of first order?

d. Units for rate constant of second order?

e. Units for rate constant of zero order?

a. M

b. M/s

c. 1/s

d. 1/M s

e. M/s

300

Determine the rate law and rate constant

x = 2

y = 0 

Rate = k[NH⁺₄]

k = 2.42 1/M s

400

what is the rate law and value of the rate constant for this reaction?

500

Find the missing concentration

(hint: find the rate law, then the k value, then plug in for experiment 4 to find your missing concentration)

Rate = k[HgCl₂][C₂O₄^2-]²

k = 0.0152 1/M²s

Ex 4 [C₂O₄^2-] concentration = 0.513 M

Integrated Rate Laws

100

Write the three integrated rate law equations:

Zero

First

Second

Zero: [A]_t = -kt + [A]₀

First: ln[A]_t = -kt + ln[A]₀

Second: 1 / [A]_t = kt + 1 / [A]₀

200

A reaction of the form A→Products is second-order with a rate constant of 0.225 1/M s. 

If the initial concentration of A is 0.293 mol/L, what is the molar concentration of A after 35.4 s?

1/[A]_t = (0.225)(35.4) + 1/0.293M = 11.4 1/M

[A]_t = 0.0879 M

300

The reaction of butadiene gas (C₄H₆) with itself produces C₈H₁₂ gas as follows: 2 C₄H₆ → C₈H₁₂

The reaction is second order with a rate constant equal to 5.76×10⁻² M/min under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration remaining after 10.0 min?

1 / [A]_t = kt + 1 / [A]₀

1 / [A]_t = (5.76 x 10^-2 M/min)(10min) + 1/[0.200M]

1 / [A]_t = 5.576 M

[A]_t = 0.179 M

400

H₂O₂(l) → 2H₂O(l) + O₂(g) 

The decomposition of H₂O₂ to water and oxygen follows first order kinetics with a rate constant of 0.0410 min^-1 . Calculate the [H₂O₂] after 10 minutes if [H2O2]0 is 0.200 M.

ln[H₂O₂]_t− ln[H2O2]₀ = −kt 

ln[H₂O₂]_t − ln[0.200]₀ = (−0.0410 min⁻¹ )(10 min) 

[H2O2] = 0.132 M

500

Consider the first-order decomposition of SO₂Cl₂. The concentration of SO₂Cl₂ decreases from 0.240 M to 0.120 M after 2.16 hrs at 340°C. 

A. Calculate the time it takes for the concentration of SO₂Cl₂ to decrease from 0.100 M to 0.050 M? 

B. From 0.520 M to 0.0325 M?

ln[A]t = -kt + ln[A]0 

k = 0.321 1/h

A. from 0.100 M to 0.050 M

ln[0.050M] = -(0.321)t + ln[0.100M]

t = 2.16 hrs

B. From 0.520 M to 0.0325 M?

ln[0.0325M] = -(0.321)t + ln[0.520M]

t = 8.64 hrs

Half Lives & Activation Energy

100

Write the three half life equations:

Zero

First

Second

Zero: t_1/2 = [A]₀ / 2k

First: t_1/2 = 0.693 / k

Second: t_1/2 = 1 / k[A]₀

200

The activation energy (Ea) is 160 kJ/mol and the frequency factor(A) is 4.73×10¹⁰ 1/M s for the reaction below: 

H2 (g) + I2 (g) → 2 HI(g) 

What is the rate constant of the reaction at 300°C?

k = Ae^-Ea/RT

k = (4.73 x 1010)e^{-[(1.60x10^5)/(8.314 x 573)]}

k = 1.23 x 10^-4 1/M s

300

A. How many half lives would it take for the concentration of NOCl to reach 12.5% of its original concentration?

B. How many half lives to reach 0.39% of its original volume?

(hint: you don't need to use any half life equations for this)

A. 3 half lives

B. 8 half lives

Explanation:

100% to 50% is 1 half life

50% to 25% is 2 half lives 

25% to 12.5% is 3 half lives

12.5% to 6.25% is 4 half lives 

etc.

400

Given that the rate constant is 11 1/M s at 345 K and the frequency factor is 20 1/M s, calculate the activation energy in kJ/mol.

ln(k) = ln(a) + (-Ea/RT)

-RTln(k/a) = Ea

-(8.314)(345) ln (11/20)

Ea = 1.715 kJ/mol

500

Anthropologists can estimate the age of a bone or other sample of organic matter by its carbon-14 content. The carbon-14 in a living organism is constant until the organism dies, after which carbon-14 decays with first-order kinetics and a half-life of 5730 years. Suppose a bone from an ancient human contains 19.5% of the C-14 found in living organisms. How old is the bone?

(Hint: assume the original concentration is 100%)

(second hint: you'll need to use both the first order half life equation and first order integrated rate equation)

Use half life equation to find k value:

k = 0.693 / 5730yrs = 1.21 x 10^-4 1/yr

Now plug in everything to integrated rate law:

ln(19.5) = -(1.21 x 10^-4 1/yr)(t) + ln(100)

t = 13500 years