Thermodynamics & Energetics
Atomic & Molecular Structure
Chemical Bonding & Intermolecular Forces
Kinetics & Reaction Mechanisms
Equilibrium & Acids/Bases
100

Define enthalpy change, what two terms describe the enthalpy change?

Enthalpy is the heat gained or lost, exothermic and endothermic

100

A mixture contains 3.00 g of Mg and 5.00 g of Al. What is the mass percent of each element in the mixture?

Mg = (3.00/8.00)×100=37.50%;

Al = 62.50%

100

Define electronegativity and predict which bond (H–F, H–C, H–O) is most polar.

The pull on electrons in a bond by an atom, H–F most polar as it has the largest electronegativity difference

100

Write a generic rate law expression

Rate law: rate = k [reactants]^order

100

State Le Chatelier's principle and predict the shift when pressure is increased for the equilibrium: 

N2(g) + 3 H2(g) ⇌ 2 NH3(g)

Increasing pressure favors side with fewer gas moles: shifts to right (NH3)

200

Calculate ΔH° for the reaction using Hess's Law given these two reactions and their ΔH° values: 

A → B (–150 kJ) 

B → C (+60 kJ)

find A → C

A → C: 

ΔH = (A→B) + (B→C) 

–150 + 60 = –90 kJ.

200

Predict which atom has a larger radius: Br or Se, and justify

Se is left of Br in same period, so both atoms have 4 shells, Se is slightly larger than Br due to Br having more protons and therefore a larger effective nuclear charge

200

Rank these intermolecular forces from weakest to strongest: London dispersion, hydrogen bonding, dipole–dipole

London dispersion < dipole–dipole < hydrogen bonding

200

For a reaction with rate law rate = k[A]2[B], how does doubling [A] affect the initial rate? How about doubling [B]?

Doubling [A] (order 2) → rate ×4;

Doubling [B] (order 1) → rate ×2

200

Define Ksp and calculate the molar solubility of Ag2SO4 in pure water given Ksp = 1.2×10–5

For Ag2SO4 ⇌ 2 Ag+ + SO42–

Ksp = [Ag+]2[SO42–

(2s)2(s) = 4s3 

s = (Ksp/4)1/3

s = 0.0144

300

Explain the difference between Gibbs free energy, ΔG, and the entropy term TΔS. Give the condition(s) for a spontaneous (favorable) process at constant T and P.

ΔG = ΔH – TΔS; spontaneous if ΔG < 0

300

Given a compound whose percent composition is 40.00% C, 6.71% H, 53.29% O, determine its empirical formula

Assume 100 g: 

C = 40.00 g/12.01=3.331 mol

H = 6.71 g/1.008=6.657 mol

O = 53.29 g/16.00=3.3306 mol

Divide by smallest (~3.3306): C ≈1.000, H ≈2.000, O ≈1.000 → empirical formula CH2O

300

For a molecule with formula AB3 and one lone pair on B predict the electron geometry and molecular geometry. What would the bondf angle be

Tetrahdral

Trigonal pyramidal

<109.5

300

Describe the method of initial rates and how it can be used to determine reaction orders experimentally

Use initial-rate experiments varying concentrations; solve for reaction orders by comparing rates

300

For a weak acid HA with Ka = 1.0×10–5 and initial concentration 0.10 M, calculate pH

Ka = [H+][A-]/[HA]

1.0×10–5 = x2/(0.1-x)    assume x <<0.1

x = sqrt (1.0×10–5 * 0.1)

x = 1 x 10-3 =  [H+]

pH = log (1.0×10–3) pH ≈ 3.00

400

For a reaction, ΔH° = +40 kJ and ΔS° = +120 J/K. Determine the temperature above which the reaction becomes spontaneous (favorable). Show work.

Set ΔG = 0: 

ΔH – TΔS = 0 → T = ΔH / ΔS 

= (40,000 J)/(120 J/K) = 333.3 K

400

Given the electron configuration, identify whether the atom is in ground state 

Fe2+: [Ar],3d6 — is this correct for the ground state? Explain.

Yes, this is correct. An Fe atom has 26 electrons. This would be the Ar electron configuration plus 4s23d6. However the Fe2+ ion loses its two valence electrons (4s2).

400

Explain the hybridization and geometry for the central atom in SF4

sp3d

See-saw

400

The decomposition of N₂O₅ follows the rate law:

Rate=k[N2O5]

In an experiment, the half-life of N₂O₅ at 25°C is 2.6 hours. A 0.80 M solution of N₂O₅ is prepared. How long will it take for the concentration to decrease to 0.050 M?

For a first-order reaction: k =0.693/t1/2

0.693/2.6 hrs = 0.267 hr−1

Using the integrated rate law: ln⁡[A]=ln⁡[A]0−kt

ln⁡(0.050)=ln⁡(0.80)−(0.267)(t)

−2.996=−0.223−0.267t

0.267t=2.773     t=10.4 hours

400

A 2.0 L flask is charged with 0.50 mol of N₂O₄ at 25°C.

N2O4(g)⇌2NO2(g)

At equilibrium, the partial pressure of NO₂ is 1.2 atm. What is the value of Kp?

First, find initial pressure of N₂O₄ using PV = nRT: 

Pinitial N2O4 = 6.1 atm

Using an ICE table with the equilibrium pressure of NO₂ = 1.2 atm:

  • At equilibrium: PN2O4=6.1−(1.2/2)=5.5 atm
  • At equilibrium: PNO2=1.2 atm

Kp=(PNO2)2/PN2O4 = (1.2)2/5.5 =0.26 atm

500

A student performs a calorimetry experiment to determine the enthalpy of combustion of ethanol. A 2.50 g sample of ethanol (C₂H₅OH, molar mass = 46.07 g/mol) is completely burned in a bomb calorimeter. The temperature of the calorimeter increases from 22.5°C to 28.7°C. 

The calorimeter contains 1000 g of water (specific heat capacity of water = 4.18 J/g·°C). The bomb calorimeter itself has a heat capacity of 850 J/°C.

Calculate the heat released by the combustion of ethanol (in kJ).

Calculate the temperature change: ΔT=28.7°C−22.5°C=6.2°C

Calculate heat absorbed by the water and bomb calorimeter: 

qwater=m×c×ΔT=1000 g×4.18 J/g°C×6.2°C

qwater=25,916 J

qbomb=Cbomb×ΔT=850 J/°C×6.2°C 

qbomb=5,270 J

Calculate total heat released (by the reaction): qtotal=qwater+qbomb = 25,916+5,270 = 31,186 J

qcombustion=−31,186 J = −31.2 kJ (negative becuase it is released)


500

The first ionization energy of phosphorus (1012 kJ/mol) is greater than that of sulfur (999 kJ/mol), even though sulfur has a higher nuclear charge. Which explanation accounts for this observation?

Half-filled subshells are particularly stable due to electron pairing repulsion being minimized. 

P's 3p³ configuration is more stable than S's 3p⁴, requiring more energy to remove an electron from P

500

Use resonance and formal charge to determine the most significant Lewis structure for NO3 and explain how that relates to observed bond lengths

NO3 major resonance structures have delocalized pi bonding

All N–O bonds equal length intermediate between single and double.

500

Consider the following reaction mechanism for the decomposition of ozone:

Step 1 (fast equilibrium): O⇌ O2+O

Step 2 (slow): O+O3→2O2

What is the rate law? (Use pre-equilibirum approximation)

Rate=k[O3]2

500

Design a buffer of pH 4.74 using acetic acid (Ka = 1.8×10–5). Calculate the required ratio of [A] to [HA]

Henderson–Hasselbalch: 

pH = pKa + log([A–]/[HA])

pKa = 4.74 → ratio = 1:1 for pH 4.74

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