Unit 2: Energetics, Group Chemistry AS, Halogenoalkanes and Alcohols

Standard conditions use a pressure of 100 kPa and a specified temperature, usually 298 K. Every substance must be in its standard state at those conditions unless another state is stated.

Physical state matters because changing state also transfers energy. For example, combustion data for liquid pentane include a different starting enthalpy from data for gaseous pentane.

The sign refers to the reacting system. If an insulated solution warms during a reaction, the solution gains heat but the reaction system releases it, so the reaction $\Delta H$ is negative.

A value of $\Delta H=-200$ kJ mol$^{-1}$ means 200 kJ is released per mole of reaction as written. A value of +50 kJ mol$^{-1}$ means 50 kJ is absorbed.

Label the vertical axis enthalpy, write the correct species and states on horizontal reactant and product levels, and draw the $\Delta H$ arrow directly between those levels with its value and units.

The vertical separation represents the enthalpy change. For SO$_3$(g)+H$_2$O(l)$\rightarrow$H$_2$SO$_4$(aq), $\Delta H=-200$ kJ mol$^{-1}$, reactants must be 200 kJ mol$^{-1}$ above products.

Definitions determine equation coefficients. For formation of one mole of water(l): H$_2$(g)+$\tfrac12$O$_2$(g)$\rightarrow$H$_2$O(l). For atomisation of chlorine: $\tfrac12$Cl$_2$(g)$\rightarrow$Cl(g).

Before choosing data, check: correct amount (usually 1 mol of the named product or substance), complete combustion where required, elemental standard states for formation/atomisation, and every physical state.

For the material whose temperature is measured, $q=mc\Delta T$, where $q$ is in J, $m$ in g, $c$ in J g$^{-1}$ °C$^{-1}$ and $\Delta T$ in °C.

If 51.0 g of solution warms by 6.9 °C and $c=4.18$ J g$^{-1}$ °C$^{-1}$, $q=51.0\times4.18\times6.9=1.47\times10^3$ J. If 0.0273 mol reacted, $\Delta H=-1.47/(0.0273)=-53.9$ kJ mol$^{-1}$.

Write the target equation, arrange known equations so unwanted species cancel, reverse any equation that runs the wrong way and change the sign of its $\Delta H$, multiply equations and enthalpies by the same factor, then add.

Apply stoichiometric coefficients to every enthalpy value and preserve physical states. Confirm that adding the manipulated chemical equations gives exactly the target before adding their numbers.

For CaCO$_3$(s)$\rightarrow$CaO(s)+CO$_2$(g), measure $\Delta H_1$ for CaCO$_3$+2HCl and $\Delta H_2$ for CaO+2HCl to their common CaCl$_2$(aq)+H$_2$O(l) destination. Then $\Delta H_{target}=\Delta H_1-\Delta H_2$.

Use the same acid concentration and comparable total solution mass, ensure the chosen reagent is fully reacted, stir consistently, and repeat measurements. Record enough temperature-time data for a cooling correction.

Record temperature at fixed times before mixing, add reactants at a known time, continue readings after the maximum, plot temperature against time, fit the post-reaction cooling line and extrapolate it back to the mixing time. Use the extrapolated temperature to obtain corrected $\Delta T$.

State the mechanism of each error, its direction where defensible, and whether it is random or systematic. Compare repeats and calculate percentage uncertainty from apparatus uncertainties rather than calling every difference 'human error'.

Use $\Delta H\approx\sum E(bonds\ broken)-\sum E(bonds\ formed)$. Breaking bonds requires energy and contributes positively; forming bonds releases energy and is subtracted.

Draw complete structures, count each bond in the stoichiometric equation, multiply by its mean value, total reactant bonds broken and product bonds formed, then subtract with units kJ mol$^{-1}$.

Mean values average different molecular environments and refer to gaseous species. They therefore give an estimate; state changes and the actual bond environment can make the value differ from an experimental standard enthalpy.

Start with a balanced reaction and $\Delta H=\sum E_{broken}-\sum E_{formed}$. Count the unknown bond as $nX$, where $n$ is the total number of those bonds formed or broken in the stoichiometric reaction.

If the products contain 48 S–F bonds in total, keep their contribution as $48X$ until all known bond totals and the reaction $\Delta H$ have been inserted, then divide by 48 to obtain the mean S–F bond enthalpy.

The result should be a positive energy per mole of bonds. Recount bonds, coefficients and whether the unknown is on the broken or formed side if the sign or size is implausible.

A smaller bond enthalpy means less energy is required for homolytic bond breaking, so that bond may break more readily and may help a reaction proceed faster at room temperature. A larger value indicates a stronger bond that is harder to break.

Rate depends on the complete mechanism, activation energy, collision geometry, temperature and catalysts. Bond enthalpies are averaged gas-phase data and indicate only one energetic contribution.

London forces strengthen as electron clouds become larger and more polarisable, and as molecular shapes allow more surface contact. They are present even when stronger named interactions also occur.

For comparable small molecules, hydrogen bonding is usually the strongest of these interactions, then permanent-dipole attraction, then London forces; real comparisons must also consider how many contacts and electrons are present.

The highly electronegative N, O or F atom polarises the X–H bond. The exposed H$^{\delta+}$ can approach a lone pair on another molecule closely, producing a particularly strong intermolecular attraction.

Water and ammonia have similar electron counts, so their London forces are comparable. Water can form a more extensive hydrogen-bond network, helping explain its much higher boiling temperature.

For the same mass, the larger volume of the open ice structure gives a lower density. Ice therefore floats on liquid water.

Draw the full local structure, mark $\delta+$ on the qualifying H, mark the acceptor lone pair, then place the dotted interaction from that lone pair to H. Use the actual bonding, not merely the molecular formula.

First list every force present, then compare the factor that changes—electron number/polarisability, contact surface or hydrogen-bond capability. More energy needed to separate molecules means a higher boiling temperature and lower volatility.

Dissolving requires separation of some solute particles and solvent molecules, followed by formation of solute–solvent attractions. 'Like dissolves like' is a summary of that energetic competition, not a replacement for naming the forces.

Not every ionic compound is soluble: the balance between lattice attraction and hydration differs. For dissolved ions, describe the orientation of water dipoles around each charge.

Assign the fixed values first, multiply each oxidation number by its atom count, set the total equal to the species charge, and solve for the unknown.

Write one unknown for the requested element, include every subscript and coefficient, and make the weighted total equal to 0 for a neutral compound or to the ion charge for a polyatomic ion.

Apply peroxide and metal-hydride exceptions before using the usual O = −2 and H = +1 rules. Reinsert the result and verify the total charge.

Write the numeral immediately after the relevant element name in parentheses. Use I, II, III, IV, V, VI or VII—not an Arabic numeral or a signed ionic charge.

Treat the given oxidation numbers as signed contributions. Choose the smallest whole-number ratio that makes their total zero, then write the electropositive element first and simplify the subscripts.

Keep a polyatomic ion intact and use brackets when more than one is required. Confirm both overall neutrality and the lowest ratio.

For Mg + 2H$^+$$\rightarrow$Mg$^{2+}$+ H$_2$, magnesium changes 0 to +2 and loses two electrons, so it is oxidised. Hydrogen changes +1 to 0 and gains electrons, so it is reduced.

In s- and p-block reactions, assign oxidation numbers before and after, identify each change, then use the electron count to confirm that total electron loss equals total electron gain.

In Mg + 2H$^+$$\rightarrow$Mg$^{2+}$+ H$_2$, H$^+$ gains electrons and is the oxidising agent; Mg loses electrons and is the reducing agent.

Identify the species containing the atom whose oxidation number decreases: that whole reactant species is the oxidising agent. The reactant whose oxidation number increases is the reducing agent.

In Cl$_2$ + H$_2$O $\rightleftharpoons$ HCl + HClO, chlorine starts at 0. It becomes −1 in HCl and +1 in HClO, so the same Cl$_2$ is both reduced and oxidised.

Confirm one starting oxidation number, at least two products containing that element, one increase and one decrease. Then check the balanced equation.

Cr$_2$O$_7^{2-}$$\rightleftharpoons$2CrO$_4^{2-}$changes colour with conditions, but chromium remains +6 in both ions. It is not redox. Chlorine reacting with cold alkali gives Cl$^-$and ClO$^-$, so Cl changes 0 to −1 and +1: disproportionation.

Assign only the oxidation numbers needed to test change, but compare the same element in reactants and products and cite the numerical changes.

Metals generally form positive ions by losing valence electrons. Their oxidation number increases from 0 in the element to a positive value: Na$\rightarrow$Na$^+$+e$^-$ and Mg$\rightarrow$Mg$^{2+}$+2e$^-$.

Non-metals generally form negative ions by gaining electrons. Their oxidation number decreases from 0: Cl$_2$+2e$^-$$\rightarrow$2Cl$^-$and O$_2$+4e$^-$\rightarrow$2O$^{2-}$.

These opposite electron changes allow metals to act as reducing agents and non-metal molecules such as halogens to act as oxidising agents in many s- and p-block reactions.

Write the changing species, balance its atoms, then add electrons to the more positive side until total charge is equal. For a full ionic equation, multiply half-equations so electron numbers match, add them and cancel electrons and any identical species.

Adding the last pair gives Fe(s)+2Fe$^{3+}$(aq)$\rightarrow$3Fe$^{2+}$(aq). Atoms and total charge are both balanced and no electron remains in the overall equation.

The outer electron feels weaker effective attraction and requires less energy to remove. Group 2 values remain generally higher than neighbouring Group 1 values because a Group 2 atom has a greater nuclear charge with a similar shell pattern.

Down the group, an extra shell increases distance and shielding, effective nuclear attraction for the outer electron falls, ionisation energy decreases, and forming M$^+$ or M$^{2+}$ becomes easier. Reactions with water or oxygen therefore become faster and more vigorous.

Group 2 atoms must lose two electrons rather than one, but the down-group trend is governed by the decreasing energies required to remove their outer electrons.

Magnesium reacts much more readily with steam: Mg(s)+H$_2$O(g)$\rightarrow$MgO(s)+H$_2$(g). Calcium, strontium and barium form increasingly alkaline hydroxide solutions or suspensions with cold water.

Typical evidence includes metal burning in oxygen/chlorine, hydrogen effervescence with water, and an alkaline indicator colour when hydroxide forms.

Soluble products give alkaline solutions. With acid, the basic oxide or hydroxide is neutralised to a salt and water; a solid may dissolve and the mixture may warm.

Reaction with water depends on accessibility and solubility: MgO reacts slowly, while heavier Group 2 oxides react more readily. Acid neutralisation is the common chemical pattern.

Increasing hydroxide solubility makes saturated solutions more alkaline down the group. Decreasing sulfate solubility means adding sulfate ions increasingly gives a white precipitate, with BaSO$_4$ especially useful for sulfate testing.

When comparing observations, separate solubility from reaction rate and use the correct anion. A cloudy mixture or precipitate indicates limited solubility, not absence of ions.

A small and/or highly charged cation has high charge density and strongly polarises the electron cloud of NO$_3^-$ or CO$_3^{2-}$. This weakens bonds within the anion and makes thermal decomposition easier. Down a group, cation radius increases, polarising power falls and thermal stability rises.

For similarly sized cations, a 2+ ion polarises more strongly than a 1+ ion. Lithium resembles magnesium because Li$^+$ is unusually small, so both have appreciable polarising power.

Heat excites electrons in atoms or ions to higher energy levels. When they fall to lower levels, they emit photons whose energies match the level differences. Characteristic wavelengths combine to give a diagnostic flame colour.

The observed colour identifies the metal ion because its allowed energy-level separations are characteristic. Sodium contamination can mask other colours because its yellow emission is intense.

Clean a nichrome/platinum wire loop with concentrated HCl and heat until no colour remains. Moisten it with HCl, pick up the sample, place it in a non-luminous blue flame, record the colour, and clean between samples.

Use similar sample amounts, particle sizes, heating positions and flame conditions. Record both the temperature/heating needed and verified gases rather than inferring decomposition from appearance alone.

CO$_2$ is confirmed by Ca(OH)$_2$+CO$_2$$\rightarrow$CaCO$_3$+H$_2$O. Acidifying the sulfate test removes carbonate interference before Ba$^{2+}$ is added.

Methyl orange changes red in acid through orange at the endpoint to yellow in alkali. Phenolphthalein is colourless in acid and pink in alkali; detect the first permanent very pale endpoint colour appropriate to titration direction.

With carbonate in the flask and HCl in the burette, methyl orange changes from yellow towards the first permanent orange endpoint. Read the burette at eye level to the nearest calibrated precision.

Use a volumetric pipette and filler, do not rinse the conical flask with analyte, and wash flask walls down with distilled water without changing moles present.

For quantities multiplied or divided, add percentage uncertainties. For a burette with ±0.05 cm$^3$ per reading, a titre has ±0.10 cm$^3$; percentage uncertainty is $0.10/\text{titre}\times100$%. A larger sensible titre reduces this percentage.

Obtain concordant titres and average them to improve precision. Report a result to precision supported by the apparatus and include dominant systematic limitations separately.

Pipette a known aliquot of standard acid into a conical flask, add the suitable indicator (commonly phenolphthalein for ethanedioic acid/NaOH), titrate with NaOH to concordant endpoints, and use the balanced equation to find NaOH concentration.

For 100.0 cm$^3$ of 0.0500 mol dm$^{-3}$ H$_2$C$_2$O$_4$$\cdot$2H$_2$O, moles required = 0.00500 mol; multiply by the hydrate's molar mass when calculating the solid mass.

The elements also darken down the group: fluorine pale yellow, chlorine pale green, bromine red-brown and iodine grey-black in standard states.

A more reactive halogen oxidises a less reactive halide: X$_2$+2Y$^-$$\rightarrow$2X$^-$+Y$_2$. X$_2$gains electrons and is reduced; Y$^-$ loses electrons and is oxidised.

HClO/chlorate(I) produced in water is an oxidising disinfectant that kills microorganisms, so chlorine is used in water treatment. Dose must be controlled because chlorine chemistry can also be hazardous.

Bromine and iodine can undergo analogous reactions, with the same need to balance atoms and track the halogen from 0 into lower and higher oxidation states.

Reducing ability increases HCl < HBr < HI because the H–X bond weakens and X$^-$ is increasingly easy to oxidise down the group.

Ag$^+$+X$^-$$\rightarrow$AgX(s). Hydrogen halides form white ammonium halide smoke with ammonia, HX(g)+NH$_3$(g)$\rightarrow$NH$_4$X(s), and form acidic solutions in water, HX+H$_2$O$\rightarrow$H$_3$O$^+$+X$^-$.

F$_2$ should displace every lower halide, while At$_2$ should displace none of Cl$^-$, Br$^-$ or I$^-$. Compounds and displacement behaviour are predicted by continuing the same electron-gain, shielding and polarisability trends.

State the observed trend, identify its cause, place F or At beyond the known sequence, and give a directional prediction. Keep state, colour, electronegativity and redox strength as separate claims.

When comparing one factor, keep the others constant: use the same amounts and temperature for a surface-area experiment, or the same gas temperature when changing pressure.

Reactant bonds or electron arrangements must be disturbed before product bonds can form. Collisions below $E_a$ separate without reaction; collisions at or above it can access the reaction pathway.

On a reaction-profile diagram, forward $E_a$ is the vertical energy difference from the reactant level to the top of the energy barrier. Reverse $E_a$ is measured from the product level to that same barrier.

For initial rate, draw a tangent at $t=0$. For rate at a stated time, draw a tangent touching the curve at that point. Choose two well-separated points on the tangent—not necessarily data points—form a large triangle and calculate rise/run.

A product graph has a positive gradient; a reactant-amount graph has a negative gradient. Reaction rate is normally quoted as the positive magnitude unless a signed species rate is specifically requested.

Write units from the graph axes and retain scale factors. A steeper tangent has a larger rate; a plateau has gradient zero.

A Maxwell–Boltzmann graph plots number of molecules against molecular energy. The curve starts at the origin, rises to a peak and approaches the energy axis asymptotically; total area represents the fixed number of molecules.

The increase in the high-energy tail is much more important than the modest increase in collision frequency. More particles have $E\geq E_a$, so successful collisions per unit time rise and the reaction is faster.

At the same temperature, the molecular energy distribution is unchanged. Lowering $E_a$ means a larger existing fraction of collisions has enough energy, so successful collisions occur more frequently.

Plot enthalpy on the vertical axis and reaction progress on the horizontal axis. Put reactants and products at the same levels for both paths; their vertical difference is the unchanged $\Delta H$.

Label both activation energies with upward arrows from the relevant starting level to a peak, label the catalyst intermediate at the valley, and label $\Delta H$ directly between reactant and product levels.

Benefits must be compared with catalyst manufacture, toxicity, scarcity, cost, lifetime, recovery and recycling. A valuable catalyst may still be economical because it is regenerated and used for many cycles.

For one fixed balanced reaction, a catalyst does not change its stoichiometric atom economy. Higher atom economy arises only when catalysis makes a different, more selective reaction route possible.

At one fixed temperature, draw one Maxwell–Boltzmann curve with axes number of molecules and energy. Mark the uncatalysed $E_a$ line and a catalysed $E_a$ line farther left.

The area to the right of a threshold represents molecules with $E\geq E_a$. Because the catalysed line is lower, its right-hand area is larger, so a greater proportion of collisions is energetic enough to react.

The alternative pathway therefore increases successful-collision frequency and rate. The curve itself, its peak and its total area remain the same because temperature and molecule count have not changed.

Before equilibrium, concentrations change as the two rates approach equality. At equilibrium, concentration–time curves become horizontal, but their values need not be equal.

Write the equilibrium equation, mark the forward reaction as exothermic or endothermic, count gaseous coefficients only for pressure, identify the imposed change, then state both shift direction and the specific reason.

If gaseous mole totals are equal, pressure has no effect on position. A catalyst speeds both directions and does not shift equilibrium. Pure solid amounts do not determine a homogeneous gas/solution equilibrium position in this qualitative model.

Use the supplied enthalpy sign, gas mole ratio, rate/yield data and cost information. Select a moderate temperature and pressure where extra yield or speed still justifies marginal energy, equipment and safety costs.

An industrial optimum maximises viable output per time and cost, not equilibrium percentage alone. Catalyst choice, feedstock conversion, separation and recycle can change the best compromise.

Compare reactant and product connectivity, identify bonds broken and formed, and classify the specified step. Conditions help distinguish competition: aqueous KOH favours substitution/hydrolysis, while ethanolic KOH with heat favours elimination.

Every curly arrow starts at an electron pair and ends where a new bond or lone pair forms. Atom count and total charge must be conserved through every step.

For C$^{\delta+}$–Br$^{\delta-}$, both C–Br electrons move to bromine. Br$^-$ is electron-rich and can act as a nucleophile; the electron-deficient carbon species can be attacked by a nucleophile and is electrophilic.

Bond polarity makes one heterolytic direction more plausible: the more electronegative atom takes the pair, stabilising negative charge, while the other centre becomes electron-deficient.

Draw the curly arrow from the actual lone pair (or the atom bearing it) to the $\delta+$ carbon. The nucleophile is attracted by charge but defined by pair donation.

Mark the relevant dipole, identify the electron-pair donor and acceptor, then start each curly arrow at the donor pair. Bond strength and reaction conditions still affect whether the predicted route is fast or favoured.

Choose the longest carbon chain, number it to give substituents the lowest set of locants, name F/Cl/Br/I as fluoro-, chloro-, bromo- and iodo-, alphabetise different prefixes, and use di-, tri- or tetra- for repeats.

A structural formula groups connected atoms; a displayed formula shows every atom and bond; a skeletal formula uses vertices/line ends for carbon and omits attached C–H bonds while showing halogen symbols.

Locate the carbon directly bonded to the halogen, ignore the halogen and any hydrogens, then count how many other carbon atoms are directly attached to that carbon.

For an unsymmetrical secondary halogenoalkane, elimination can remove H from either adjacent carbon and may give positional and E/Z alkene products. Aqueous conditions instead favour alcohol formation.

For R–X + OH$^-$, mark C$^{\delta+}$–X$^{\delta-}$. Draw a curly arrow from the O lone pair to the C–X carbon and another from the C–X bond to X. The one-step substitution gives ROH and X$^-$.