Energetics

## Identify the system and the sign of heat transfer - The reacting substances form the system; the solution, calorimeter and wider environment can form parts of the surroundings. Always name the boundary before assigning a heat sign. - At constant pressure, with only pressure–volume work under the usual calorimetry assumptions, the heat transferred to the system equals its enthalpy change. Exothermic changes have negative ΔH; endothermic changes have positive ΔH. - A temperature rise in an initially equilibrated surrounding solution can indicate an exothermic reaction. A controlled-temperature bath can absorb heat without warming, so warming is not the definition of exothermicity. ## Account for bond breaking and bond formation - Breaking a specified covalent bond requires energy; forming that bond from the separated gaseous atoms releases energy. Reaction enthalpy reflects the balance of all relevant changes. - Using positive bond-dissociation magnitudes, estimate ΔH as the total for bonds broken minus the total for bonds formed. A larger release on formation gives a negative overall result. - The imagined route through gaseous atoms is a bookkeeping route. It does not claim that every real reaction proceeds by first breaking all reactant bonds into isolated atoms. ## Read enthalpy changes from a reaction profile - A reaction profile compares reactant, transition-state and product levels along a reaction coordinate. The horizontal coordinate is not elapsed time. - The vertical difference products minus reactants gives ΔH. The forward activation barrier is measured from reactants to the transition-state maximum. - The supplied piecewise-linear profile is an idealised reading model, not a measured molecular trajectory. Read the endpoint levels and maximum separately; the zero on the axis is a reference level. ## Separate activation barriers from reaction enthalpy - For the same single-step profile, forward and reverse barriers reach the same maximum. The forward barrier minus the reverse barrier therefore equals ΔH in this simple energy-level model. - An exothermic reaction can still have a large activation barrier and proceed slowly. The sign of ΔH alone does not determine a rate or establish thermodynamic spontaneity. - A catalyst offers a different route with a lower barrier while leaving reactant and product enthalpies unchanged. Transition states are not isolable reaction intermediates. ## Specify standard conditions and physical states - A standard enthalpy change refers to standard pressure, 100 kPa, and a stated temperature, often 298 K. The standard-state symbol does not force every measurement to be at 298 K. - Use the appropriate reference state at that temperature and pressure. State symbols matter: forming H₂O(l) and H₂O(g) gives different enthalpy changes. - For solution problems, state the concentration or standard-solution convention when relevant. One bar is 100 kPa; one atmosphere is 101.325 kPa, a distinct pressure. ## Define standard enthalpy of formation - The standard enthalpy of formation is the enthalpy change when one mole of a substance forms from its constituent elements in their reference states at standard pressure and the stated temperature. - Fractional coefficients can be necessary to form exactly one mole. For H₂O(l), H₂(g) + ½O₂(g) → H₂O(l) is a formation equation at the stated conditions. - An element in its reference state has zero standard formation enthalpy by convention. This does not assign zero to every allotrope or to isolated gaseous atoms of that element. ## Define standard enthalpy of combustion - The standard enthalpy of combustion refers to complete burning of one mole of a substance in oxygen, with reactants and products in their specified standard states at the stated temperature. - Normalise the equation to one mole of fuel and include product states. Carbon dioxide and liquid water are the usual specified products for standard hydrocarbon combustion near 298 K. - A formation equation can also be a combustion equation, as when one mole of carbon in its reference form burns to one mole of CO₂(g). These labels need not be mutually exclusive. ## Reverse and scale thermochemical equations - Reversing a thermochemical equation reverses the sign of its enthalpy change. Multiplying every coefficient by a factor multiplies the enthalpy change for that equation by the same factor. - State whether an energy refers to a particular reacting amount or to one mole of reaction as written. Scaling an equation is different from silently changing the molar basis. - Keep state symbols unchanged while scaling. If a state changes, an additional phase-change enthalpy may be required. ## Use a solution calorimeter correctly - The cutaway shows an insulated cup with a lid, support beaker, immersed temperature probe and stirrer. In use the cup is opaque; the cutaway reveals the contents for explanation. - Mix measured quantities, stir, and record temperature against time. Keep the probe clear of the base and sides so that it samples the solution rather than the vessel. - Insulation and a lid reduce heat exchange; they do not remove all heat loss or calorimeter heat capacity. Use the total heated solution mass when the calculation assumes both liquids form one solution. ## Calculate heat from mass and temperature change - For a material with approximately constant specific heat capacity, q = mcΔT. Use units consistently: grams with J g⁻¹ K⁻¹ give q in joules. - A temperature difference has the same numerical value in kelvin and degrees Celsius. ΔT is final temperature minus initial temperature, not the final temperature alone. - The calculated q belongs to the body whose mass and temperature are used. Convert joules to kilojoules before reporting a molar result in kJ mol⁻¹. ## Obtain a molar neutralisation enthalpy - For dilute strong acid–strong base neutralisation, H⁺(aq) + OH⁻(aq) → H₂O(l) forms one mole of water per mole of reaction. The measured value is usually close to −57 kJ mol⁻¹ under familiar conditions. - In the supplied experiment, 100.0 g of solution warms by 6.8 K and 0.0500 mol of water forms. Ignoring other transfers, q_solution = 2842.4 J and ΔH = −2.8424/0.0500 = −56.848 kJ mol⁻¹. - Weak-acid or weak-base ionisation can contribute to the measured enthalpy, so do not define every possible neutralisation as having an identical value or sign. Report sensible precision from the measurements. ## Choose the correct reacting amount and molar basis - Use concentration and volume to calculate amounts, then compare them with the balanced equation. Excess reagent is not all consumed. - For reaction enthalpy per mole of reaction as written, divide the reaction heat by the reaction extent: amount consumed divided by its stoichiometric coefficient. - For a neutralisation value per mole of water, use moles of water formed instead. These molar bases differ if the written equation produces more than one mole of water. ## Include calorimeter heat capacity when supplied - A calorimeter can absorb heat as well as the solution. If its heat capacity is C_{cal} in J K⁻¹, its heat gain is C_{cal}ΔT. - When the solution and calorimeter undergo the same rise and other transfers are negligible, the reaction heat is −(mc + C_{cal})ΔT. - Omitting a positive calorimeter heat gain from an exothermic calculation makes the estimated reaction enthalpy less negative, provided the other quantities and assumptions are unchanged. ## Calculate an experimental combustion enthalpy - In simple combustion calorimetry, a spirit burner heats water in a metal can. Weigh the burner before and after burning to estimate fuel consumption, and measure the water temperature rise. - The illustrated copper can has a solid base between flame and water; a metal probe and stirrer are immersed above it. Heat lost to the air, the vessel and other surroundings limits the simple water-only estimate. - Use moles of fuel, not water, for a combustion value per mole of fuel. The result can be less negative than a reference value because the water captures only part of the released heat. ## Evaluate systematic errors in calorimetry - Heat escaping an exothermic reaction’s measured solution generally reduces its observed temperature rise. A water-only calculation then underestimates the magnitude of the released heat. - Incomplete combustion reduces energy released per measured mass of fuel. Evaporation from a burner can exaggerate the mass apparently burnt; both can make the calculated combustion value less negative. - Repeated measurements improve the estimate of random variation but do not automatically remove a systematic loss. Choose a practical improvement that targets the stated limitation. ## Estimate a temperature at the mixing time - A temperature–time trace can be extrapolated to estimate the temperature at a defined mixing time before later cooling has reduced the reading. - Apply the chosen model only over the region it describes. A linear cooling correction is an approximation, not proof that the entire experiment cools linearly. - Subtract the appropriate initial baseline from the estimated mixing-time value. Keep the temperature correction separate from the energy and molar calculations. ## Apply Hess’s law to identical endpoints - Enthalpy is a state function. The overall change between specified initial and final states is independent of the route taken. - Add equations and their signed enthalpy changes, reversing or scaling them as required. Intermediate species cancel only when both their formulae and physical states match. - A cycle’s layout is a drawing choice. Follow the defined equations and arrow directions rather than assuming that every formation arrow must point upward on the page. ## Calculate reaction enthalpy from formation data - Multiply each standard formation enthalpy by its coefficient in the balanced equation. Add the product contributions and subtract the reactant contributions. - For CO(g) + ½O₂(g) → CO₂(g), supplied formation values −393.5 and −110.5 kJ mol⁻¹ give −393.5 − (−110.5) = −283.0 kJ mol⁻¹. - O₂(g) contributes zero only because it is the reference form at the stated conditions. Preserve the states and the reaction’s molar basis. ## Use combustion data with common final products - A combustion Hess cycle takes both sides of a target equation to the same final products in the same states. Oxygen amounts must also balance across the combined route. - For formation of ethanol, compare combustion of 2C(graphite) + 3H₂(g) with combustion of one mole of ethanol to 2CO₂(g) + 3H₂O(l). Reverse the ethanol-combustion route when returning to ethanol. - With the supplied values, ΔH_f = 2(−394) + 3(−286) − (−1367) = −279 kJ mol⁻¹. Do not subtract an already negative value without tracking the reversal. ## Build a cycle by cancelling equations - Write the target equation, then manipulate the available equations so unwanted intermediates cancel. A reversed equation carries the opposite enthalpy change. - Check atoms, charge and state symbols after addition. A correct numerical sum cannot repair an unbalanced chemical route. - Hess cycles can link experimentally accessible changes to an inaccessible target. For example, hydration of an anhydrous salt can be related to dissolution of the anhydrous and hydrated forms to the same final solution. ## Relate atomisation to a molecular bond average - Standard atomisation forms one mole of gaseous atoms from an element in its reference state at standard pressure and the stated temperature. - A Hess route can compare a molecule with its separated gaseous atoms. For methane, use carbon atomisation, four hydrogen atomisations and the reverse of methane formation. - Dividing the complete methane atomisation enthalpy by four gives its average C–H dissociation value. This does not require its four successive bond dissociations to have equal energies. ## Estimate a gas-phase reaction from mean bond enthalpies - Mean bond enthalpy is a gaseous bond-dissociation enthalpy averaged across relevant molecular environments. It estimates a bond type’s behaviour rather than giving an exact value for every molecule. - Count every affected bond using the balanced equation. For N₂ + 3H₂ → 2NH₃, break one N≡N and three H–H bonds and form six N–H bonds. - Using the supplied positive dissociation values, the estimate is 945 + 3(436) − 6(391) = −93 kJ mol⁻¹ of reaction as written, which produces two moles of ammonia. ## Correct phase differences and recognise approximations - A mean-bond calculation describes the specified gaseous species. A target reaction containing liquid water needs a condensation contribution if the bond calculation first produces water vapour. - For two moles of water vapour becoming liquid, add twice the supplied molar condensation enthalpy to the gaseous reaction value. - Mean bond enthalpy calculations also use Hess’s law. Differences from compound-specific data arise from averaging and possibly different states, not from a failure of path independence. ## Report uncertainty and evaluate a comparison - Retain unrounded intermediate values and round the final result to a precision justified by the measurements. Exact stoichiometric coefficients do not limit significant figures. - A temperature rise calculated from two thermometer readings has uncertainty from both readings. For a conservative absolute-uncertainty estimate, add their absolute uncertainties. - Distinguish random spread, systematic error and model assumptions when comparing an experimental result with a reference value. Agreement to many decimal places is not meaningful if the measurements are much less precise.