Q: What is Hess's Law and how do you use it to calculate ΔH for a reaction?
A: Hess's Law states that because enthalpy (H) is a state function, the total enthalpy change for a reaction is independent of the pathway — only the initial and final states matter. To apply it: (1) Arrange given equations so reactants and products match your target. (2) Reverse any equation that needs it — reversing changes the sign of ΔH. (3) Multiply equations by coefficients as needed — multiplying scales ΔH by the same factor. (4) Add all equations — intermediates that appear on both sides cancel. (5) Sum all ΔH values. Also applies using standard enthalpies of formation: ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants). Elements in their standard state have ΔH°f = 0 by definition.
Q: Explain the four combinations of ΔH and ΔS signs and when each reaction is spontaneous.
A: Gibbs equation: ΔG = ΔH − TΔS. ΔG negative = spontaneous. (1) ΔH negative, ΔS positive: always spontaneous at all temperatures — both enthalpy and entropy favor it. Example: combustion. (2) ΔH positive, ΔS negative: never spontaneous at any temperature — both factors oppose it. (3) ΔH negative, ΔS negative: spontaneous at low temperatures where ΔH dominates. TΔS term is small at low T. Example: water freezing below 0°C. (4) ΔH positive, ΔS positive: spontaneous at high temperatures where TΔS dominates. Example: ice melting above 0°C. The crossover temperature where ΔG = 0: T = ΔH/ΔS. Below this temperature one direction is favored; above it, the other.
Q: Describe the three laws of thermodynamics and their practical implications.
A: First Law (Conservation of Energy): The total energy of the universe is constant. Energy can be converted between forms but cannot be created or destroyed. ΔE = q + w (heat absorbed plus work done on system). Implication: you can never get more energy out of a process than you put in. Second Law (Entropy): The total entropy of the universe always increases in any spontaneous process. Heat flows spontaneously from hot to cold, never the reverse. ΔS_universe ≥ 0. Implication: no process is 100% efficient — some energy always becomes unavailable as waste heat. Third Law (Absolute Zero): The entropy of a perfect crystal at absolute zero (0 K) equals zero. Absolute zero is unattainable in practice. Implication: establishes the reference point for all absolute entropy values; all substances have positive entropy at any temperature above 0 K.
Q: How do you calculate the total heat required to convert ice at −20°C to steam at 120°C?
A: Five separate calculations using q = mcΔT for temperature changes and q = n·ΔH for phase changes. Step 1: Heat ice from −20°C to 0°C. q = m × c_ice × 20. (c_ice = 2.09 J/g·°C.) Step 2: Melt ice at 0°C. q = n × ΔH_fus. (ΔH_fus = 6.01 kJ/mol.) Step 3: Heat liquid water from 0°C to 100°C. q = m × c_water × 100. (c_water = 4.184 J/g·°C.) Step 4: Boil water at 100°C. q = n × ΔH_vap. (ΔH_vap = 40.7 kJ/mol.) Step 5: Heat steam from 100°C to 120°C. q = m × c_steam × 20. (c_steam = 2.01 J/g·°C.) Total = sum of all five steps. The boiling step (Step 4) requires by far the most energy because all intermolecular hydrogen bonds must be broken.
Q: What is the relationship between ΔG° and the equilibrium constant K?
A: ΔG° = −RT ln(K), where R = 8.314 J/(mol·K) and T is temperature in Kelvin. This equation links thermodynamics (ΔG°) to equilibrium (K). Large negative ΔG°: K >> 1, products strongly favored at equilibrium. Large positive ΔG°: K << 1, reactants strongly favored. ΔG° = 0: K = 1, roughly equal concentrations of products and reactants. Note: ΔG° is the free energy change under standard conditions (1 M, 1 atm, 25°C). ΔG (without degree symbol) applies under non-standard conditions: ΔG = ΔG° + RT ln(Q), where Q is the reaction quotient. When Q = K, ΔG = 0 (system is at equilibrium). When Q < K, ΔG negative (reaction proceeds forward). When Q > K, ΔG positive (reaction proceeds in reverse).