Physical Chemistry Problem Solving: Integrated Review

Introduction

This unit has covered gases, solutions, thermochemistry, equilibria, electrochemistry and kinetics. The topics look different, but almost every calculation rests on three conservation laws: mass is conserved, charge is conserved and energy is conserved. Seeing these laws as the common framework makes unfamiliar problems less intimidating, because each one can be approached by asking which quantity must balance. This review shows the three laws at work and how they can check each other.

Core explanation

Conservation of mass. In a chemical reaction atoms are rearranged, never created or destroyed, so each element balances and the total mass of reactants equals the total mass of products. In calculations this appears as balanced equations, mole ratios, limiting-reagent reasoning, combustion analysis and empirical formulae. A quick check on any stoichiometry problem is to add up the masses of everything entering and leaving; the totals must agree.

Conservation of charge. Every solution is electrically neutral overall. For ions, the sum of (concentration × charge magnitude) for cations equals the same sum for anions. This rule finds an unknown ion concentration, checks analytical data and underpins pH calculations with several equilibria. In electrochemistry, charge is conserved as electrons: the electrons lost at the anode equal those gained at the cathode, and the charge passed Q = It links current to amount of electrons through n(e⁻) = Q ÷ F.

Conservation of energy. In a calorimeter, heat released by a reaction equals heat absorbed by the solution and container, q = mcΔT. Because enthalpy is a state function, Hess's law allows an unknown enthalpy change to be found from any route between the same start and end points — formation data, combustion data or combined equations. The same law connects the electrical work of a cell to Gibbs energy, ΔG = −nFE.

Using laws to check each other. A complete problem often uses all three. Mass balance fixes amounts, charge balance fixes ion or electron amounts, and energy balance gives heat or work. An error in one step often shows up as an imbalance elsewhere, so a second law acts as an independent check.

Formulae

Mass: Σ m(reactants) = Σ m(products)

Charge: Σ (c × charge) for cations = Σ (c × charge ) for anions; n(e⁻) = It ÷ F

Energy: q = mcΔT; ΔH(reaction) = −q ÷ n; ΔrH = ΣΔfH(products) − ΣΔfH(reactants)

Step-by-step reasoning

1. Identify which conserved quantity links the givens to the target. 2. Write the balance as an equation (mass, charge or energy). 3. Convert everything to amounts in moles where possible. 4. Solve for the unknown. 5. Use a second conservation law, if available, as a cross-check.

Visual explanation

Picture three accounting ledgers side by side, headed "mass", "charge" and "energy". Each has an "in" column and an "out" column. A correct solution shows the columns of every ledger adding to the same totals; an error shows up as a mismatch in one ledger even if the others balance.

Real-world analogy

A shop's accounts must balance money, stock and hours worked. If the till shows more money than the goods sold could explain, something is wrong, even before you know what. Chemical conservation laws are the laboratory's balance sheets.

Real-world example

Electroplating plants monitor current and time to predict how much metal is deposited, weigh the plated parts to confirm the prediction, and track bath temperature to manage heat from resistive losses. Mass, charge and energy balances run side by side, and a mismatch between predicted and measured mass flags problems such as side reactions producing hydrogen.

Why?

Why do these three laws apply everywhere in chemistry? Chemical change rearranges atoms and electrons without destroying them, and energy is conserved in every physical process. Any quantitative model that broke one of these laws would predict something that never happens.

Common misconception

"Conservation of mass means the number of moles is conserved." Mass and atoms are conserved, but the total amount of molecules can change: 2H₂ + O₂ → 2H₂O turns 3 mol of molecules into 2 mol.

Worked example

Question: 0.160 g of methane burns completely in oxygen. Find the masses of CO₂ and H₂O formed, confirm mass balance, and calculate the heat released (ΔcH = −890 kJ mol⁻¹).

Reasoning: CH₄ + 2O₂ → CO₂ + 2H₂O. n(CH₄) = 0.160 ÷ 16.04 = 9.98 × 10⁻³ mol.

O₂ used = 2 × 9.98 × 10⁻³ = 0.01995 mol = 0.638 g.

CO₂ = 9.98 × 10⁻³ × 44.01 = 0.439 g; H₂O = 0.01995 × 18.02 = 0.359 g.

Mass check: 0.160 + 0.638 = 0.798 g in; 0.439 + 0.359 = 0.798 g out.

Energy: 9.98 × 10⁻³ × 890 = 8.88 kJ.

Answer: 0.439 g CO₂ and 0.359 g H₂O; masses balance at 0.798 g; 8.88 kJ of heat is released.

Quick check

1. A current of 0.500 A flows for 30.0 minutes through copper(II) sulfate solution. What mass of copper is deposited? Answer: Q = 900 C; n(e⁻) = 900 ÷ 96 485 = 9.33 × 10⁻³ mol; n(Cu) = 4.66 × 10⁻³ mol, so 0.296 g of copper.

Exam focus

Name the conservation law you are applying; it makes the method clear to an examiner. For charge balance, remember to multiply by the charge: 0.10 mol dm⁻³ sulfate contributes 0.20 mol dm⁻³ of negative charge. In calorimetry, the sign of ΔH is opposite to the heat gained by the surroundings.

Advanced insight

In nuclear processes mass and energy are interconverted according to E = mc², but chemical energy changes are so small that the associated mass change is undetectable: the 890 kJ from burning a mole of methane corresponds to about 10⁻⁸ g. For all chemical calculations, mass and energy can be conserved separately.

Summary

Mass, charge and energy are conserved in every chemical process. Mass balance underlies stoichiometry and formula calculations; charge balance governs ionic solutions and electrolysis through Q = It and n(e⁻) = Q ÷ F; energy balance governs calorimetry and Hess cycles. Using one law to cross-check another is one of the most reliable ways to catch errors in multi-step problems.

Practice questions

1. A solution contains 0.10 mol dm⁻³ K⁺, 0.050 mol dm⁻³ Mg²⁺ and chloride as the only anion. Find [Cl⁻]. Answer: Positive charge = 0.10 + 2 × 0.050 = 0.20 mol dm⁻³, so [Cl⁻] = 0.20 mol dm⁻³. 2. What mass of silver is deposited by 0.0100 mol of electrons? Answer: Ag⁺ + e⁻ → Ag, so 0.0100 mol Ag = 0.0100 × 107.9 = 1.08 g. 3. 50.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH and the temperature rises by 6.8 K. Calculate ΔH per mole of water formed (assume 100 g of solution, c = 4.18 J g⁻¹ K⁻¹). Answer: q = 100 × 4.18 × 6.8 = 2.84 kJ; n = 0.0500 mol; ΔH = −2.84 ÷ 0.0500 = −57 kJ mol⁻¹. 4. Given ΔfH(CO) = −110.5 kJ mol⁻¹ and ΔfH(CO₂) = −393.5 kJ mol⁻¹, find ΔH for CO + ½O₂ → CO₂. Answer: ΔH = −393.5 − (−110.5) = −283.0 kJ mol⁻¹, since oxygen as an element has ΔfH = 0.