Chemical Energetics

The reacting chemicals are the system; the solution, vessel and room can act as surroundings. At constant pressure, heat absorbed by the system corresponds to an increase in enthalpy. An exothermic reaction releases heat and has ΔH < 0; an endothermic reaction absorbs heat and has ΔH > 0. Bond breaking absorbs energy and bond formation releases it, so the difference between these contributions determines the overall sign.

A catalyst provides a different mechanism with a lower activation barrier. It does not change the enthalpies of the initial and final states, so it leaves ΔH unchanged. In a multistep profile, minima between peaks represent intermediates; the highest absolute peak is not automatically the step with the largest barrier measured from its preceding intermediate.

Standard enthalpy changes refer to substances in their standard states at a specified temperature, commonly 298 K, with standard pressure 100 kPa. School calculations conventionally use 1 mol dm^-3 for specified aqueous standard conditions. Temperature must still be stated: the standard symbol does not itself mean 298 K. Changing a state, multiplying an equation or reversing it changes the corresponding enthalpy value.

For a strong acid and strong base in dilute aqueous solution, the main reaction is H⁺(aq) + OH^-(aq) → H₂O(l), so neutralisation enthalpies are similar. A weak acid or base also undergoes ionisation, changing the measured total. Do not assume every acid-base combination has exactly the same enthalpy.

q = mcΔT, where m is the mass being warmed, c is its specific heat capacity and ΔT is final minus initial temperature. With m in g and c in J g^-1 K^-1, q is in J. A temperature interval of 1 K equals an interval of 1 °C. For an insulated arrangement, q_reaction = -q_surroundings.

If the calorimeter heat capacity C is supplied, include its heat gain: q_surroundings = (mc + C)ΔT. In a combustion experiment, use the mass of water warmed and the amount of fuel actually burnt. In a dissolution experiment, use the appropriate solution mass and heat capacity, and divide by the amount of solute dissolved.

A temperature-time record before and after mixing can support extrapolation to estimate the temperature at mixing, reducing a systematic heat-loss error. Repeated trials reveal scatter but do not by themselves remove the same heat loss from every trial. Always identify the limiting reagent before dividing q by an amount.

Hess Law: the enthalpy change of a reaction depends on its initial and final states, not the route taken. Write the target equation first. Reverse any auxiliary equation that runs the wrong way and change its sign; multiply its enthalpy when multiplying its coefficients. Cancel identical species only when their states also match.

With formation data, ΔH_reaction = ΣνΔH_f(products) - ΣνΔH_f(reactants), where ν is the stoichiometric coefficient. Elements in their standard states have standard formation enthalpy zero by definition. This does not mean they contain no energy or have zero atomisation enthalpy.

The bond-energy route passes through separated gaseous atoms. Include extra phase-change terms if the target uses liquids or solids. Mean bond energies give estimates because bond strength depends on molecular environment; formation enthalpies for the exact substances and states can give a more specific value.

A first electron affinity is the enthalpy change when one mole of gaseous atoms each gains an electron to form gaseous singly negative ions: Cl(g) + e^- → Cl^-(g). It is often negative. Adding an electron to an already negative ion, as in O^-(g) + e^- → O^2-(g), requires energy to overcome repulsion and is positive. Ionisation energies always refer to removing electrons from gaseous species and are positive.

For MgO, include Mg atomisation, half an O=O bond dissociation, both the first and second ionisation energies of Mg, and both electron affinities of oxygen. Each coefficient follows the target formula. A highly exothermic lattice-formation step can compensate for costly ion formation; do not omit a positive second electron affinity to make a cycle look easier.

A simple electrostatic model predicts a larger lattice-energy magnitude for greater ionic charge product and smaller interionic separation. Thus MgO has a much more negative lattice energy than NaCl, while MgO is generally more negative than CaO because Mg²⁺ is smaller. Use these comparisons for similar structures; appreciable covalent character can cause departures from a purely ionic model.

Entropy, S, measures the dispersal of energy among accessible microscopic arrangements. A state with more accessible arrangements has greater entropy. Qualitative predictions work best when a large change, such as gas production or a phase change, dominates; do not infer a precise value from appearance.

Dissolution often increases positional dispersal, but strongly hydrated ions can order nearby water molecules. The net entropy of dissolution is not guaranteed positive. If gas amounts are unchanged or competing changes are similar, supplied entropy data is stronger evidence than a simple counting rule. The syllabus does not require calculating reaction entropy from tables of absolute standard entropies.

ΔG° = ΔH° - TΔS°. T is absolute temperature in K. If ΔH is in kJ mol^-1 and ΔS in J mol^-1 K^-1, divide the entropy by 1000 first. At constant temperature and pressure, a negative actual ΔG favours the forward process; zero corresponds to equilibrium; a positive value favours the reverse process.

The standard value ΔG° assesses the driving tendency from standard-state conditions. A negative standard value is often described as thermodynamically feasible or spontaneous under standard conditions. It does not promise that all reactant turns into product, that the change is fast, or that the same direction remains favoured at every composition.

If ΔH and ΔS have the same sign, solve T = ΔH/ΔS for the crossover where ΔG = 0, using consistent units. For the +36.0 kJ mol^-1, +120 J mol^-1 K^-1 example, T = 300 K. The direction of the inequality depends on the signs: positive/positive is favourable above the crossover; negative/negative below it.

Heat gained by the solution is heat lost by the reaction. Hess cycles require identical endpoint states. Bond-energy estimates use gaseous bonds broken minus gaseous bonds formed. The syllabus lattice energy is negative for lattice formation; dissolution uses its reverse plus hydration.

ΔG° = ΔH° - TΔS° with T in K and matching units. Heating favours processes with positive ΔS. A negative standard free-energy change indicates a thermodynamic tendency under standard conditions, not a guaranteed fast or complete reaction.