FTCE Chemistry 6-12 study materials
FTCE Chemistry 6-12 study resources

FTCE Chemistry 6-12: Energy and Its Transformations

FTCE Chemistry 6-12 · Work through your exam, one topic at a time.

All guide sections

Energy and Its Transformations

What this domain tests

Energy and Its Transformations covers thermochemistry: heat transfer, calorimetry, enthalpy, Hess's law, bond energies, and the roles of enthalpy and entropy in spontaneity. Items ask you to calculate q = mcΔT, combine enthalpies, and interpret signs correctly.

Heat, enthalpy, and sign conventions

  • Heat q flows from a hotter object to a colder one; temperature measures average kinetic energy, not heat content.
  • Exothermic: the system releases heat, ΔH < 0, and surroundings warm. Endothermic: the system absorbs heat, ΔH > 0, and surroundings cool.
  • At constant pressure, q = ΔH. The first law is ΔE = q + w.

Calorimetry

  • q = mcΔT, where m is the mass of the substance being heated and c is its specific heat. For water, c = 4.18 J/(g·°C).
  • In a coffee-cup calorimeter, q_reaction = −q_solution when heat loss to the surroundings is negligible.
  • Phase changes use q = nΔH_fus or q = nΔH_vap; a complete heating curve is the sum of temperature-change and phase-change steps.

Enthalpy calculations

  • Standard enthalpy: ΔH°_rxn = Σ ΔH°_f(products) − Σ ΔH°_f(reactants); elements in their standard states have ΔH°_f = 0.
  • Hess's law: reversing a step changes the sign of ΔH; multiplying a step multiplies ΔH.
  • Bond energies: ΔH = Σ bonds broken − Σ bonds formed.
  • ΔG = ΔH − TΔS. Negative ΔG means the process is spontaneous under those conditions; because ΔH and ΔS can oppose each other, temperature can change the outcome.

Worked example

50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH are mixed in a coffee-cup calorimeter, and the temperature rises from 24.0 °C to 31.0 °C. Assume a density of 1.00 g/mL and c = 4.18 J/(g·°C). Total mass = 100.0 g, so q_solution = (100.0)(4.18)(7.0) = 2.93 × 10³ J = 2.93 kJ. Moles of water formed = 0.0500 L × 1.00 M = 0.0500 mol. Because q_reaction = −2.93 kJ, ΔH = −2.93 ÷ 0.0500 = −58.6 kJ/mol, reported as about −59 kJ/mol (the 7.0 °C change limits precision).

Traps that cost marks

  • Losing the negative sign for an exothermic reaction.
  • Using only the solute mass instead of the total solution mass.
  • Forgetting to multiply by moles in phase-change or molar enthalpy calculations.
  • Applying Hess's law without flipping signs or scaling coefficients.
  • Assuming ΔH alone determines spontaneity.