Thermodynamics

## Specify standard states and reaction basis - A standard enthalpy change refers to100 kPa and a stated temperature, commonly298 K. Always specify the chemical equation, physical states and amount basis. - Standard formation refers to one mole of substance formed from reference-state elements. Reference-state elements have zero standard formation enthalpy by convention; their absolute entropies need not be zero. - Changing an equation coefficient changes the enthalpy for that written reaction. Gaseous atoms are not the reference-state starting materials for forming sodium chloride from its elements. ## Define atomisation using one mole of gaseous atoms - Atomisation forms one mole of gaseous atoms from the element in its specified reference state. Its equation can require a fractional molecular coefficient. - For chlorine,½Cl₂(g)→Cl(g) is half of Cl₂(g)→2Cl(g), so the supplied bond-dissociation enthalpy 242 gives atomisation 121 kJ mol⁻¹. - Atomising ordinary bonded elements requires energy, but an already monatomic reference gas such as Ar(g) has no change for Ar(g)→Ar(g). Do not call every possible atomisation value strictly positive. ## Include successive ionisation energies - Ionisation removes an electron from a gaseous atom or ion. The first step starts with atoms; the second starts with singly positive gaseous ions. - To form Mg²⁺(g) from Mg(g), include both first and second ionisation steps. The supplied total is 738+1451=2189 kJ mol⁻¹. - Both steps are endothermic. Omitting the second ionisation would leave the wrong gaseous charge state and an incomplete Born–Haber route. ## Use a signed electron-affinity convention - An electron-affinity step adds an electron to a gaseous atom or ion. Here values are signed enthalpy changes for the equation as written. - The first oxygen step is exothermic in the supplied data, while adding another electron to O⁻ is endothermic because the incoming electron encounters an already negative ion. - The overall O(g)+2e⁻→O²⁻(g) change is the sum −141+798=+657 kJ mol⁻¹. A stable oxide lattice can compensate for the energy cost of producing its gaseous ions. ## Separate lattice formation from lattice dissociation - Lattice formation produces one mole of ionic solid from its separated gaseous ions. Lattice dissociation is the reverse process to the same gaseous states. - For the supplied NaCl model, formation is −787 kJ mol⁻¹ and dissociation is +787 kJ mol⁻¹. The reversal changes the sign, not the magnitude. - Check which definition is used before adding a lattice term to a cycle. A quoted positive magnitude must not be inserted as a negative formation value without stating that conversion. ## Connect two Born–Haber routes with the same endpoints - The supplied route diagram connects reference-state elements to solid NaCl directly and through gaseous atoms and ions. All edge values are in kJ mol⁻¹ of NaCl formed. - The indirect signed sum 229+496−349−787 equals the direct−411. Each node retains all species needed for its stage, including the electron between ionisation and electron attachment. - This is a thermochemical bookkeeping diagram, not a reaction mechanism or an energy-height plot. Its page directions do not mean every downward drawn edge is exothermic. ## Calculate a lattice term by Hess’s law - For NaCl, add sodium atomisation, chlorine atomisation, sodium ionisation and chlorine electron affinity before the lattice-formation step. - The supplied pre-lattice total is 108+121+496−349=376 kJ mol⁻¹. To reach the formation endpoint−411, lattice formation is −411−376=−787 kJ mol⁻¹. - Retain signs in brackets and check the endpoint equation. Use one internally consistent set of supplied data rather than mixing rounded values from different tables. ## Include both charge-building steps for magnesium oxide - Forming Mg²⁺ requires two magnesium ionisations; forming O²⁻ requires both oxygen electron-affinity steps. Oxygen atomisation is already per mole of atoms. - The supplied pre-lattice sum is 148+249+738+1451−141+798=3243 kJ mol⁻¹. With formation−602, lattice formation is −3845 kJ mol⁻¹. - Check gas-ion charges and the electron count before calculating. Omitting an endothermic second electron affinity creates a large sign and magnitude error. ## Scale cycle terms by the formula stoichiometry - One mole of MgCl₂ contains one mole of Mg²⁺ and two moles of Cl⁻. Each cycle term must correspond to those amounts. - If chlorine atomisation is supplied per mole of chlorine atoms, use it twice. Likewise use twice the first chlorine electron-affinity value, with magnesium IE1 andIE2 once each. - Do not double the whole magnesium contribution or insert a second chlorine electron affinity: two different chlorine atoms each accept one electron. ## Infer an unknown electron-affinity term - A Born–Haber cycle can determine any missing leg when the other terms and endpoints are known. The unknown need not be called the direct route. - For the supplied illustrative XCl data,−400=100+120+500+EA−800. Rearranging gives EA=−320 kJ mol⁻¹. - Keep the target process direction explicit. A released-energy magnitude and a signed attachment enthalpy would use different sign conventions. ## Compare ionic charge and separation under a stated model - For comparable lattice structures, greater opposite-charge product and smaller ion-centre separation usually give stronger electrostatic attraction and a larger lattice-enthalpy magnitude. - The supplied point-charge pairs isolate these factors. Increasing separation weakens attraction; changing both ion charges from unit to double increases their charge product. - Real lattice enthalpies also depend on geometry, repulsion and bonding character. Do not apply a radius-only rule across unrelated structures without stating its limits. ## Evaluate agreement with an ideal ionic model - An ideal ionic calculation can be compared with a Born–Haber value inferred from experimental thermochemical data. - Close agreement supports the model as a useful approximation; it does not prove zero covalent character. A large discrepancy may indicate polarisation and covalent contribution, alongside other model limitations. - In the supplied comparison, salt B has a larger absolute and relative discrepancy thanA. Judge the stated model and data rather than declaring a compound purely ionic from one agreement. ## Explain ion hydration and water orientation - Hydration transfers a specified gaseous ion into aqueous surroundings. Ion–dipole attractions form as polar water molecules arrange around the ion. - Water’s oxygen end is partially negative and faces a positive ion; a hydrogen end is partially positive and faces a negative ion. The illustration is a simplified orientation snapshot, not a fixed hydration-number claim. - Dissolving an ionic crystal separates its existing ions and hydrates them. The symbol(aq) means aqueous; it does not by itself certify infinite dilution or a particular hydration coordination number. ## Relate dissolution to lattice separation and hydration - Dissolution can be analysed through a route from ionic solid to gaseous ions and then to hydrated ions. It is a Hess route rather than a claim about the actual microscopic mechanism. - Using lattice dissociation as positive, ΔHsolution=ΔHlattice,dissociation+ΣΔHhydration. Multiply each ion’s hydration term by its stoichiometric coefficient. - The supplied schematic route has+700 and−680 kJ mol⁻¹, so its net solution change is +20. With formation lattice enthalpy instead, reverse that term’s sign first. ## Calculate an unknown hydration enthalpy - The supplied KCl dissolution gives 20=700+(−320)+ΔHhyd(Cl⁻), with every value in kJ mol⁻¹ on the stated reference convention. - The chloride term is therefore−360 kJ mol⁻¹. The combined hydration release−680 partially offsets the lattice-separation cost+700. - Single-ion hydration values use a convention; retain the supplied consistent set. The total solution enthalpy can be positive even though hydration releases energy. ## Include multiple ion hydration contributions - For MgCl₂ there are two chloride hydration terms per formula unit. The supplied balance is −100=2500+ΔHhyd(Mg²⁺)+2(−350). - Rearranging gives magnesium-ion hydration−1900 kJ mol⁻¹. The total hydration term is −2600 and solution is −100. - An omitted coefficient can leave an algebraically neat but chemically incorrect result. Recheck the dissolved ion formula before accepting the answer. ## Interpret entropy through accessible states and energy dispersal - Entropy relates to the number of accessible microscopic arrangements and energy dispersal. Visual mess is only a limited analogy; entropy is not literally a count of arrangements. - Melting commonly increases a substance’s entropy, and forming gas from condensed reactants often gives a positive reaction-entropy contribution. Deposition from gas to solid commonly decreases it. - These are system trends under the specified conditions. The system’s entropy change alone does not decide spontaneity of system plus surroundings. ## Calculate a coefficient-weighted standard entropy change - Calculate standard reaction entropy as the product sum of stoichiometric coefficients times standard molar entropy, minus the corresponding reactant sum. - For the supplied CaCO₃ decomposition, ΔS°=40+214−93=+161 J K⁻¹ mol⁻¹ of reaction as written. - Absolute standard molar entropies of reference-state elements are not generally zero at ordinary temperatures. That zero convention belongs to formation enthalpies and formation free energies, not all thermodynamic properties. ## Apply the second law to system plus surroundings - The entropy of an isolated total system cannot decrease in an allowed spontaneous change. It increases for an irreversible spontaneous process and is unchanged in the reversible limit. - The reacting subsystem can lose entropy while its surroundings gain more. Add both contributions when evaluating the isolated overall boundary. - The supplied record has−50+80=+30 J K⁻¹ overall. It is consistent with an irreversible spontaneous change even though the subsystem entropy decreases. ## Use consistent units in the Gibbs equation - At a fixed temperature, G=H−TS leads to the reaction relation ΔG=ΔH−TΔS. Temperature must be in kelvin. - When enthalpy is in kJ mol⁻¹ and entropy in J K⁻¹ mol⁻¹, divide the entropy number by 1000 before combining the terms. Equivalently convert enthalpy to joules and keep everything in joules. - Entropy can legitimately be expressed in joules or kilojoules per kelvin; the requirement is dimensional consistency. Do not compare an entropy change directly with an enthalpy change without the temperature factor. ## Calculate a standard Gibbs energy change - Use the supplied decomposition model’s stated temperature, phases and per-reaction basis. The rounded values are treated as constant for this calculation. - At 1000 K, ΔG°=180−1000(160/1000)=+20 kJ mol⁻¹. - This is a standard-state result. The actual driving force in a particular gas/solid mixture can also depend on composition and partial pressure, so a positive ΔG° is not a claim that every possible mixture cannot react forward. ## Use standard formation free-energy data - Calculate a standard reaction Gibbs change by summing coefficient-weighted standard formation free energies of products and subtracting the reactant sum. - For the supplied CO oxidation, ΔrG°=−394−[−137+½(0)]=−257 kJ mol⁻¹. - The reference-element formation value is zero by convention. This does not mean that oxygen has zero absolute Gibbs energy or zero standard molar entropy. ## Compare the four enthalpy–entropy sign combinations - With fixed phases and constant ΔH and ΔS in the stated model, negative ΔH with positive ΔS gives negative ΔG for every positive temperature in the model range. - Positive ΔH with negative ΔS gives positive ΔG. When both signs match, a positive threshold may separate the temperature ranges with opposite Gibbs signs. - These statements are conditional on the model range. Real phase changes and temperature-dependent data can prevent extrapolation to every physical temperature. ## Find a high-temperature threshold - For the supplied positive-enthalpy, positive-entropy model, set ΔG=0 and use matching energy units: T=60.0/0.100=600 K. - Above 600 K within the valid range, the entropy term exceeds the enthalpy term and ΔG becomes negative. Below 600 K it is positive. - At the threshold there is no net thermodynamic driving force. It is a boundary, not evidence that the reaction proceeds rapidly or completely. ## Find a low-temperature threshold - For negative ΔH and negative ΔS, the−TΔS term is positive. In the supplied model ΔG=−40.0+0.100T. - The boundary is 400 K. Below 400 K the negative enthalpy dominates and ΔG is negative; above it ΔG is positive within the valid range. - Keep signs when dividing ΔH byΔS. A negative quotient would not represent a physically meaningful positive-temperature crossing in this simple model. ## Read entropy and enthalpy from a Gibbs graph - For constant ΔH and ΔS, the Gibbs equation has straight-line form ΔG=(−ΔS)T+ΔH. Gradient is−ΔS and the extrapolated vertical intercept isΔH. - The supplied graph falls by 20 kJ mol⁻¹ per 200 K, so its gradient is −0.100 kJ K⁻¹ mol⁻¹ and ΔS=+100 J K⁻¹ mol⁻¹. - The plotted range starts at 200 K. A vertical intercept inferred at 0 K is an algebraic extrapolation of the model, not a measured value proving the same phases exist there. ## Distinguish standard and actual reaction driving forces - As an extension linking equilibrium and thermodynamics, actual reaction Gibbs energy depends on composition: ΔrG=ΔrG°+RT lnQ with a dimensionless activity quotient. - The supplied record gives a positive standard contribution and a larger negative composition contribution, leaving actual ΔrG=−5.0 kJ mol⁻¹. - Composition or pressure can therefore change the actual forward driving force without a catalyst changing ΔrG°. This distinction resolves why a standard-state graph alone does not determine every industrial mixture’s behaviour. ## Separate thermodynamic feasibility from reaction rate - Negative actual ΔrG indicates a thermodynamically favoured forward direction at the stated composition and conditions. It does not specify a fast rate. - The supplied catalyst lowers the activation barrier while retaining the same endpoints, temperature and composition. The reaction Gibbs energy remains−10 kJ mol⁻¹. - A catalyst can speed approach to equilibrium but changes neither the equilibrium constant nor the reaction’s state-function difference. At equilibrium the actual reaction Gibbs energy is zero; the standard value need not be zero.