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Physical Chemistry -- Practice Problems

Physical Chemistry — Practice Problems

Worked Examples

Example 1: First-Order Kinetics — Drug Degradation

Problem: A drug degrades via first-order kinetics with k=2.5×10−4k = 2.5 \times 10^{-4} min−1^{-1} at 25°C. If the initial concentration is 100 mg/L, how long does it take for the concentration to drop to 90 mg/L?

Solution:

For first-order kinetics: [A]=[A]0e−kt[A] = [A]_0 e^{-kt}

Rearranging for time:

t=−1kln⁡[A][A]0=−12.5×10−4ln⁡90100t = -\frac{1}{k} \ln\frac{[A]}{[A]_0} = -\frac{1}{2.5 \times 10^{-4}} \ln\frac{90}{100}

t=−4000×ln⁡(0.90)=−4000×(−0.1054)=421 min≈7.0 hourst = -4000 \times \ln(0.90) = -4000 \times (-0.1054) = 421 \text{ min} \approx 7.0 \text{ hours}

Half-life check: t1/2=ln⁡2/k=0.693/2.5×10−4=2772t_{1/2} = \ln 2 / k = 0.693 / 2.5 \times 10^{-4} = 2772 min ≈46\approx 46 hours. Since we’re only losing 10% (not 50%), the time should be much less than the half-life. ✓

Key insight: First-order half-lives are independent of concentration. This is why drug shelf-life is expressed as t90t_{90} (time to 90% remaining), not as an absolute amount.

Example 2: Particle in a Box — Conjugated Systems

Problem: For β\beta-carotene (11 conjugated double bonds), model the π\pi electrons as particles in a box. The conjugation length is approximately 1.80 nm. Calculate the wavelength of the lowest-energy electronic transition.

Solution:

With 11 double bonds, there are 22 π\pi electrons. Each energy level holds 2 electrons (Pauli exclusion), so the HOMO has n=11n = 11 and the LUMO has n=12n = 12.

Energy levels: En=n2h28mL2E_n = \frac{n^2 h^2}{8mL^2}

ΔE=E12−E11=h28mL2(122−112)=h28mL2(23)\Delta E = E_{12} - E_{11} = \frac{h^2}{8mL^2}(12^2 - 11^2) = \frac{h^2}{8mL^2}(23)

ΔE=(6.626×10−34)2×238×(9.109×10−31)×(1.80×10−9)2\Delta E = \frac{(6.626 \times 10^{-34})^2 \times 23}{8 \times (9.109 \times 10^{-31}) \times (1.80 \times 10^{-9})^2}

ΔE=4.39×10−66×238×9.109×10−31×3.24×10−18\Delta E = \frac{4.39 \times 10^{-66} \times 23}{8 \times 9.109 \times 10^{-31} \times 3.24 \times 10^{-18}}

ΔE=1.01×10−642.36×10−47=4.28×10−18 J\Delta E = \frac{1.01 \times 10^{-64}}{2.36 \times 10^{-47}} = 4.28 \times 10^{-18} \text{ J}

Converting to wavelength:

λ=hcΔE=(6.626×10−34)(3.00×108)4.28×10−18=4.64×10−8 m=464 nm\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{4.28 \times 10^{-18}} = 4.64 \times 10^{-8} \text{ m} = 464 \text{ nm}

Comparison: Experimental λmax⁡=452\lambda_{\max} = 452 nm. The particle-in-a-box model gives a reasonable estimate (within 3%) for this highly conjugated system.

Example 3: Boltzmann Distribution at Different Temperatures

Problem: Calculate the ratio of molecules in the first excited rotational state (J=1J = 1) to the ground state (J=0J = 0) for HCl at 300 K and 1000 K. The rotational temperature is Θrot=15.2\Theta_{\text{rot}} = 15.2 K.

Solution:

For a linear rotor, the energy of level JJ is EJ=BJ(J+1)E_J = BJ(J+1) where B=kBΘrotB = k_B \Theta_{\text{rot}}.

The degeneracy of level JJ is gJ=2J+1g_J = 2J + 1.

n1n0=g1e−E1/kBTg0e−E0/kBT=3×e−2Θrot/T1×e0=3e−2Θrot/T\frac{n_1}{n_0} = \frac{g_1 e^{-E_1/k_BT}}{g_0 e^{-E_0/k_BT}} = \frac{3 \times e^{-2\Theta_{\text{rot}}/T}}{1 \times e^{0}} = 3e^{-2\Theta_{\text{rot}}/T}

At 300 K:

n1n0=3e−2(15.2)/300=3e−0.1013=3×0.9037=2.71\frac{n_1}{n_0} = 3e^{-2(15.2)/300} = 3e^{-0.1013} = 3 \times 0.9037 = 2.71

At 1000 K:

n1n0=3e−2(15.2)/1000=3e−0.0304=3×0.9701=2.91\frac{n_1}{n_0} = 3e^{-2(15.2)/1000} = 3e^{-0.0304} = 3 \times 0.9701 = 2.91

Key insight: At room temperature, the J=1J = 1 state is already more populated than J=0J = 0 (ratio 2.71) due to the 3-fold degeneracy advantage. At higher temperatures, the ratio approaches the degeneracy ratio (3:1) as the exponential factor approaches 1.

Thermodynamics

Chemical Kinetics

Quantum Chemistry and Statistical Mechanics

Intuition

Physical chemistry is the mathematical backbone of all chemistry: It connects abstract thermodynamic quantities (entropy, free energy) to measurable quantities (temperature, pressure, voltage). Think of it as the “physics of chemical systems” — applying energy conservation and statistical reasoning to molecules.

Why it matters: Physical chemistry principles underpin battery design, refrigeration, atmospheric chemistry, and pharmaceutical stability. Without understanding Gibbs free energy, you cannot predict whether a drug will degrade on the shelf.

The key insight: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S is the single most important equation in chemistry — it determines spontaneity at constant temperature and pressure, unifying enthalpy and entropy into one decision rule.

Cross-References

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.

Advanced Content

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Derivations and Proofs

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Extended Examples

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

Research Connections

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

Prerequisites

Ensure you have mastered the prerequisite material before attempting this advanced content.