Reactivity 1. What drives chemical reactions?

A chemical reaction transfers energy between the system and surroundings. Total energy is conserved. Heat is energy transferred because of a temperature difference; temperature describes the thermal state.

State which part is the system, which part is the surroundings, and the direction of energy transfer before interpreting a temperature change.

For an exothermic hand-warmer reaction, define the reacting chemicals as the system: energy leaves that system and enters your hand and the air as surroundings. A temperature rise is evidence about the surroundings, while heat names the energy crossing the boundary; neither quantity is 'stored temperature'.

Classify from the direction of energy transfer, then check that the observed temperature change of the surroundings agrees.

Connect three representations: exothermic means energy flows out of the system, the surroundings warm, and ΔH for the system is negative; endothermic gives the opposite pattern and positive ΔH. State the observed part before inferring the reaction type.

Lower relative energy corresponds to greater relative stability. In an exothermic reaction the products are lower in energy than the reactants; in an endothermic reaction the products are higher.

Label the horizontal axis reaction coordinate and the vertical axis potential energy. Then read reactant and product levels, identify the sign and direction of ΔH, and distinguish ΔH from activation energy.

On a profile, ΔH is the vertical difference between product and reactant levels, while activation energy rises from reactants to the peak. A catalyst lowers the peak by changing the pathway but leaves the two energy levels, ΔH and the relative stability of reactants and products unchanged.

An energy profile identifies an activation barrier but does not by itself determine an observed rate: temperature, particle concentrations/collision frequency and the available pathway also matter. Use the diagram to compare energetic barriers only when the profiles and conditions make that comparison valid.

At constant pressure, calculate heat transferred from mass, specific heat capacity, and temperature change, then divide by reacting moles and apply the sign convention. Check standard conditions and units.

Worked calculation: if 100.0 g of solution warms by 6.0 K and $c=4.18\,\mathrm{J\,g^{-1}\,K^{-1}}$, $q(\mathrm{solution})=mc\Delta T=(100.0)(4.18)(6.0)=2.51\times10^3\,\mathrm{J}=+2.51\,\mathrm{kJ}$. Therefore $q(\mathrm{reaction})=-2.51\,\mathrm{kJ}$. If $0.0500\,\mathrm{mol}$ of limiting reactant reacted, $\Delta H=-2.51/0.0500=-50.2\,\mathrm{kJ\,mol^{-1}}$: the negative sign means the reaction released energy. Heat loss or ignored calorimeter heat capacity usually makes the measured magnitude too small.

A standard molar enthalpy change belongs to the balanced reaction as written, with substances in their stated standard states under standard conditions. A classroom calorimetry value is an experimental estimate: report its conditions and uncertainty separately from a data-book or theoretical standard value. Heat loss to the surroundings or ignored calorimeter heat capacity commonly makes the measured magnitude too small.

Breaking bonds absorbs energy; forming bonds releases energy. Count every bond with its stoichiometric multiplicity before applying the signed sum.

For H₂ + Cl₂ → 2HCl, break one H–H and one Cl–Cl bond, then form two H–Cl bonds. Average bond enthalpies give an estimate because the tabulated value averages that bond across different gaseous molecules.

Worked example — bond enthalpies: for $\ce{C2H4(g) + HBr(g) -> C2H5Br(g)}$, the local course book gives $\ce{C-H}=414$, $\ce{C=C}=614$, $\ce{H-Br}=366$, $\ce{C-C}=346$ and $\ce{C-Br}=285\,\mathrm{kJ\,mol^{-1}}$. $\Delta H=[4(414)+614+366]-[5(414)+346+285]=2636-2701=-65\,\mathrm{kJ\,mol^{-1}}$. The negative estimate means the bonds formed release more energy than the bonds broken absorb; it remains approximate because the values are gaseous averages.

Hess's law states that enthalpy change is independent of reaction pathway. Enthalpy values can therefore be combined through a balanced cycle.

Reverse an entire balanced equation by changing the sign of ΔH; scale every coefficient and ΔH by the same factor; then add equations and cancel identical species in identical physical states. Never change a chemical subscript, formula or state symbol merely to force cancellation. The surviving equation must exactly match the target before enthalpies are summed.

Treat chemical equations like algebra: reverse a step and reverse its ΔH sign; multiply all coefficients and ΔH by the same factor; then add and cancel species. The surviving overall equation must exactly match the target before the enthalpies are summed.

Worked example — Hess's law: target $\ce{C(s)+2H2(g)+1/2O2(g)->CH3OH(l)}$. Use $\ce{C+O2->CO2}$, $\Delta H=-394\,\mathrm{kJ\,mol^{-1}}$; double $\ce{H2+1/2O2->H2O(l)}$ to give $-572\,\mathrm{kJ\,mol^{-1}}$; reverse methanol combustion to give $+726\,\mathrm{kJ\,mol^{-1}}$. After cancelling $\ce{CO2}$ and $\ce{H2O}$, $\Delta H=-394-572+726=-240\,\mathrm{kJ\,mol^{-1}}$ for the target equation.

Use correct standard states and coefficients; do not mix formation and combustion definitions in one equation.

A formation equation must produce exactly one mole of the compound from elements in their standard states, so fractional coefficients may be necessary. A combustion equation must burn exactly one mole completely in O₂; keep the stated standard state of water because changing H₂O(l) to H₂O(g) changes ΔH°.

Worked scaling example: $\Delta H_c^\circ(\ce{C2H6})=-1560\,\mathrm{kJ\,mol^{-1}}$ refers to complete combustion of one mole of ethane. For the corresponding balanced combustion of two moles, multiply the complete equation and its enthalpy by two: $\Delta H^\circ=2(-1560)=-3120\,\mathrm{kJ}$. Do not report $\mathrm{kJ\,mol^{-1}}$ for the two-mole equation unless the result is normalized back to one mole.

Apply stoichiometric coefficients to every term, preserve signs, and use the formation or combustion formula that matches the supplied data.

For formation data, imagine every reactant and product connected to the same elements: product sums minus reactant sums gives the target. For combustion data the paths run toward common combustion products, reversing the subtraction. Write coefficients beside every tabulated value before calculating.

Worked example — formation data for pentane combustion: $\ce{C5H12(l)+8O2(g)->5CO2(g)+6H2O(l)}$. Using $\Delta H_f^\circ[\ce{C5H12(l)}]=-173$, $\Delta H_f^\circ[\ce{CO2(g)}]=-394$, $\Delta H_f^\circ[\ce{H2O(l)}]=-286$ and $\Delta H_f^\circ[\ce{O2(g)}]=0\,\mathrm{kJ\,mol^{-1}}$, $\Delta H^\circ=5(-394)+6(-286)-[-173+8(0)]=-3513\,\mathrm{kJ\,mol^{-1}}$. The negative sign identifies exothermic combustion.

A Born–Haber cycle tracks the energy changes that form an ionic solid from its elements. It includes atomization, ionization, electron affinity, and lattice enthalpy terms with the correct stoichiometry.

Create a state-and-particle ledger before summing: convert each element from its standard state to the required gaseous atoms (including sublimation/phase change and bond dissociation where needed), apply every ionization-energy and electron-affinity step with its coefficient, then form the lattice. For divalent ions include both electron steps, and fix whether lattice enthalpy means formation or dissociation before assigning its sign.

Build the cycle from physical steps and electron accounting: atomize each element, ionize the metal the required number of times, add electrons to the non-metal, then form the lattice. For a 2− ion include two electron-affinity terms, and confirm whether the supplied lattice enthalpy is defined for formation or dissociation before assigning its sign.

Worked example — KBr lattice enthalpy: use the DP lattice-dissociation convention $\ce{KBr(s)->K+(g)+Br-(g)}$. Following the alternative path in the local cycle, $\Delta H_\mathrm{lattice}^\circ=-(-392)+89+419+112-325=+687\,\mathrm{kJ\,mol^{-1}}$. The formation enthalpy is reversed, atomization and ionization are positive, and the first electron affinity is negative. The positive result is consistent with separating a solid lattice into gaseous ions.

In complete combustion with excess oxygen, carbon forms CO₂ and hydrogen forms H₂O. The fuel's other elements must also appear in the products.

Write the correct products first, then balance C, H and any other atoms before balancing O₂ last. Check every atom and state symbol.

For ethanol, write C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O by balancing C, then H, then oxygen. Complete combustion specifies products, not automatically conditions or state symbols; use the question's conditions to decide whether water is liquid or vapour.

When oxygen is limited, carbon may form CO and/or elemental carbon instead of only CO₂. The available oxygen and the fuel's elements determine the product set.

Include the products specified by the question, then balance all atoms. Do not replace a required CO or C product with CO₂ simply because the fuel contains carbon.

One valid carbon-monoxide equation is 2CH₄ + 3O₂ → 2CO + 4H₂O; with still less oxygen, elemental carbon can also form. Use the product set named by the question and rebalance from scratch—there is no single universal incomplete-combustion equation.

The product change has chemical consequences: carbon monoxide binds strongly to haemoglobin and reduces oxygen transport, while fine carbon particulates damage respiratory health and can alter atmospheric heating. Incomplete combustion also releases less usable energy per mole of fuel than complete conversion to CO₂ and H₂O; distinguish these separate health, environmental and energy claims.

Compare coal, crude oil and natural gas using carbon dioxide released per unit of useful energy, energy output, and pollutants such as CO, particulates and volatile organic compounds from incomplete combustion.

A lower carbon footprint is a specific comparison, not a blanket claim that a fuel causes no pollution. Include methane leakage when evaluating natural gas.

Compare fuels on a common service basis such as kilograms of CO₂ per megajoule of useful energy, not per mole of fuel. Natural gas can have a lower combustion CO₂ intensity than coal yet lose part of that advantage through upstream methane leakage; define the system boundary before ranking.

Greenhouse link: atmospheric CO₂ absorbs selected outgoing infrared wavelengths and re-emits energy in multiple directions, reducing net energy loss to space at those wavelengths. A sustained increase in CO₂ changes Earth's energy balance until a warmer state restores balance. This mechanism is separate from comparing absolute fuel emissions or emissions per unit useful energy.

Biofuels are renewable when their carbon is replenished through biological carbon fixation, such as photosynthesis, on a relevant timescale.

Evaluate both sides: possible benefits include renewable supply and lower net carbon or sulfur emissions; disadvantages include land competition with food production and other production impacts. State the evidence for each claim.

Test a carbon-neutral claim with a life-cycle boundary: count cultivation, fertilizer, processing, transport, land-use change and combustion, then credit carbon taken up during regrowth on the relevant timescale. Renewable describes replenishment, not automatically low impact or zero net emissions.

A fuel cell converts chemical energy from a spontaneous redox reaction directly into electrical energy. Oxidation occurs at the anode and reduction at the cathode.

Deduce each half-equation from the fuel-cell reactants and electrolyte context, balance atoms and charge, then add the half-equations to obtain the overall reaction. Hydrogen and methanol cells require different oxidation half-equations.

For a hydrogen fuel cell, oxidation of H₂ supplies electrons at the anode and O₂ gains electrons at the cathode; the half-equations must match the acidic or alkaline electrolyte before they are added to 2H₂ + O₂ → 2H₂O. Direct electrical conversion does not remove the need to evaluate fuel production and storage.

A fuel cell operates while fuel and oxidant are supplied continuously from outside; a conventional battery stores a finite set of reactants internally. Point-of-use water from a hydrogen cell is not a complete environmental assessment—fuel manufacture, transport, storage and electricity source remain inside a lifecycle comparison.

Worked equations — acidic hydrogen cell: anode $\ce{H2(g) -> 2H+(aq) + 2e-}$; cathode $\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}$. Double the anode equation before adding, giving $\ce{2H2 + O2 -> 2H2O}$. Direct-methanol cell: anode $\ce{CH3OH(aq) + H2O(l) -> CO2(g) + 6H+(aq) + 6e-}$; cathode $\ce{3/2O2(g) + 6H+(aq) + 6e- -> 3H2O(l)}$. Adding and cancelling gives $\ce{CH3OH + 3/2O2 -> CO2 + 2H2O}$. Match each half-equation to the stated electrolyte; proton-exchange-membrane construction details are not assessed.

Entropy describes the dispersal of matter and available energy. For the reaction system, calculate ΔS° = ΣS°(products) − ΣS°(reactants), including every coefficient, and report J K⁻¹ mol⁻¹ for the reaction as written. System ΔS° is not the same as total entropy change of system plus surroundings; state the boundary before using entropy to discuss spontaneity.

Use phase and particle count to predict a likely sign before calculating: producing more gas particles usually increases dispersal. Treat that prediction as a check, not a replacement for the standard-entropy sum.

Worked entropy example: for $\ce{H2(g) + Cl2(g) -> 2HCl(g)}$, use $S^\circ(\ce{HCl})=187$, $S^\circ(\ce{H2})=131$ and $S^\circ(\ce{Cl2})=223\,\mathrm{J\,K^{-1}\,mol^{-1}}$. $\Delta S^\circ=2(187)-[131+223]=+20\,\mathrm{J\,K^{-1}\,mol^{-1}}$. The small positive value is plausible because gas moles are unchanged; the tabulated values, not gas count alone, determine the sign.

Use an absolute temperature in kelvin and convert ΔS° to the same energy units as ΔH° before subtracting TΔS°. The result is the Gibbs energy change for the stated reaction.

Under the stated standard conditions, ΔG° < 0 is thermodynamically favourable, ΔG° = 0 marks equilibrium, and ΔG° > 0 favours the reverse direction. This criterion predicts feasibility, not how fast the change occurs.

Worked Gibbs example: for propane combustion to gaseous water, the local course book gives $\Delta H^\circ=-2045\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S^\circ=+103\,\mathrm{J\,K^{-1}\,mol^{-1}}$ at $5\,^{\circ}\mathrm{C}$. Convert $T=278.15\,\mathrm{K}$ and $\Delta S^\circ=0.103\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$; then $\Delta G^\circ=-2045-(278.15)(0.103)=-2074\,\mathrm{kJ\,mol^{-1}}$. Its negative sign means the stated reaction is spontaneous under those standard conditions, not necessarily fast.

For a temperature-dependent reaction, find the boundary by setting ΔG° = 0 in ΔG° = ΔH° − TΔS°, then use the signs of ΔH° and ΔS° to decide which temperature range is spontaneous.

Use the signs before calculating: ΔH < 0 and ΔS > 0 is favourable at all temperatures, while ΔH > 0 and ΔS < 0 is not; matching signs create a temperature threshold. At T = ΔH/ΔS, first put ΔH and ΔS in consistent units, then test one temperature on each side rather than guessing the inequality.

Worked threshold example: $\ce{CH4(g) + H2O(g) -> 3H2(g) + CO(g)}$ has $\Delta H^\circ=205\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S^\circ=216\,\mathrm{J\,K^{-1}\,mol^{-1}}=0.216\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$. At the boundary, $0=205-T(0.216)$, so $T=205/0.216=949\,\mathrm{K}$. Because both changes are positive, the forward reaction is spontaneous above 949 K and non-spontaneous below it, assuming $\Delta H^\circ$ and $\Delta S^\circ$ are approximately constant.

At equilibrium for the stated reaction and temperature, Q = K and the current ΔG = 0. Substitution into ΔG = ΔG° + RT ln Q shows that ΔG° is generally not zero; it equals −RT ln K. Away from equilibrium, Q < K favours the forward direction and Q > K favours the reverse. Compare Q and K only for the same balanced reaction orientation and fixed temperature.

Compare Q with K to predict the immediate direction: Q < K gives ΔG < 0 for the forward reaction, while Q > K gives ΔG > 0 and favours the reverse. Keep ΔG for the current composition distinct from ΔG°, which describes standard-state reactants and products and fixes K at that temperature.

Worked $K$ and $Q$ example: for ammonia synthesis at 298 K, the local course book gives $\Delta G^\circ=-31.8\,\mathrm{kJ\,mol^{-1}}=-31800\,\mathrm{J\,mol^{-1}}$. From $\ln K=-\Delta G^\circ/(RT)$, $K=3.77\times10^5$, so products are favoured at equilibrium. If $Q=1.0\times10^6$, $\Delta G=-31800+(8.31)(298)\ln(10^6)=+2410\,\mathrm{J\,mol^{-1}}=+2.41\,\mathrm{kJ\,mol^{-1}}$; the forward reaction is then non-spontaneous because the current mixture has $Q>K$.