3 Physical chemistry

Exam analysis

Most tested topics

(a) Energetics

Exothermic reactions transfer heat energy from the reacting chemicals to the surroundings; endothermic reactions take in heat energy from the surroundings. The direction of transfer defines the term.

Reaction type	Heat-energy transfer	Typical measured effect on the surroundings	Sign of $\Delta H$
exothermic	reaction $\rightarrow$ surroundings	temperature rises	negative
endothermic	surroundings $\rightarrow$ reaction	temperature falls	positive

Combustion and many neutralisation reactions are exothermic. Dissolving ammonium nitrate in water is endothermic: the solution cools because energy is taken from its surroundings.

A cold mixture has not 'released cold'. It has absorbed heat from the surroundings. Temperature change is evidence of the transfer, while exothermic and endothermic describe the reaction's energy direction.

Simple calorimetry estimates the heat transferred by measuring the temperature change of water or a solution. Insulation, stirring and prompt temperature readings make the measured change closer to the true change.

Change	Core method	Measurement
combustion	place a known mass of water in a metal can; record its initial temperature; burn a measured mass of fuel beneath it; stir and record the highest temperature	water mass, temperature rise and fuel mass burned
displacement, dissolving or neutralisation	put measured reactant in an insulated polystyrene cup with a lid; record initial temperature; add the other reactant, replace lid, stir and record the highest or lowest temperature	total solution mass and temperature change

Use the same reactant amounts and concentrations when comparing reactions. Keep the flame-to-can distance constant for combustion, start solutions at the same temperature, repeat trials and calculate a mean after investigating any anomaly.

No school calorimeter is perfectly insulated. Heat loss, heating the apparatus, incomplete combustion and fuel evaporation usually make an exothermic temperature rise—and the calculated energy magnitude—too small.

The heat energy transferred to or from a substance depends on its mass, specific heat capacity and temperature change.

Q=mc\Delta T

Symbol	Meaning	Common unit
$Q$	heat energy change	J
$m$	mass of water or solution	g
$c$	specific heat capacity	$\mathrm{J\,g^{-1}\,^{\circ}C^{-1}}$
$\Delta T$	final temperature minus initial temperature; use the magnitude when finding energy transferred	$^{\circ}\mathrm C$

For $25.0\,\mathrm g$ of solution with $c=4.18\,\mathrm{J\,g^{-1}\,^{\circ}C^{-1}}$ warming from $19.0$ to $31.5\,^{\circ}\mathrm C$ , $\Delta T=12.5\,^{\circ}\mathrm C$ and $Q=25.0\times4.18\times12.5=1306\,\mathrm J\approx1.31\,\mathrm{kJ}$ .

Match units before substituting: $Q$ is in joules when $c$ is in joules per gram per degree Celsius. For dilute aqueous solutions, volume in $\mathrm{cm^3}$ may be converted to mass using the supplied density; do not assume that conversion when no density is given.

Molar enthalpy change, $\Delta H$ , is the heat-energy change for one mole of the reaction quantity specified. Divide the measured energy by the relevant amount in moles, convert joules to kilojoules and attach the sign from the reaction direction.

\Delta H=\frac{Q}{n}

Stage	Action
1	calculate $Q$ and convert J to kJ
2	calculate the reacting amount $n$ in mol; use $n=m/M_r$ when needed
3	divide the energy magnitude by $n$
4	use $-$ for exothermic or $+$ for endothermic and report $\mathrm{kJ\,mol^{-1}}$

If an exothermic displacement reaction releases $1.30\,\mathrm{kJ}$ while $0.0125\,\mathrm{mol}$ reacts, $|\Delta H|=1.30/0.0125=104\,\mathrm{kJ\,mol^{-1}}$ , so $\Delta H=-104\,\mathrm{kJ\,mol^{-1}}$ .

Do not divide joules by moles and label the result $\mathrm{kJ\,mol^{-1}}$ . The mole quantity must match the reaction basis named in the question, and the sign is not supplied by the division alone.

An energy level diagram compares the total energy of reactants and products. Their vertical separation is the enthalpy change, $\Delta H$ ; lower products mean energy has been transferred out, while higher products mean energy has been taken in.

Reaction	Relative levels	$\Delta H$ arrow and sign	Explanation
exothermic	products below reactants	downward; negative	products store less energy, so the difference is released
endothermic	products above reactants	upward; positive	products store more energy, so the difference is absorbed

Draw a vertical energy axis, then horizontal labelled lines for the balanced reactants and products at the correct relative heights. Draw a vertical arrow from the reactant level to the product level and label it $\Delta H$ with its value and sign when supplied.

This objective needs energy levels, not a reaction-profile hump. Activation energy and profile curves belong to the later rates Topic. The $\Delta H$ arrow follows reactants to products; reversing it reverses the sign.

Breaking a covalent bond requires energy and is endothermic. Making a covalent bond releases energy and is exothermic because the bonded particles move to a lower-energy arrangement.

Overall reaction	Comparison of the two energy totals
exothermic	energy released making product bonds is greater than energy taken in breaking reactant bonds
endothermic	energy taken in breaking reactant bonds is greater than energy released making product bonds

Every reaction normally includes both processes: reactant bonds are broken and product bonds are made. The overall energy change depends on the difference between the totals, not on only one bond.

Bonds do not release energy when they break. Statements such as 'the reaction is exothermic because bonds are broken' reverse the energy direction; an exothermic result requires bond making to release the larger amount.

Bond energy is the energy needed to break one mole of a specified gaseous covalent bond. Count every bond in the balanced reaction, total the bonds broken and made, then subtract in the correct order.

\Delta H=\sum E(\text{bonds broken})-\sum E(\text{bonds made})

For $\ce{C2H4 + H2 -> C2H6}$ , unchanged C–H bonds cancel. Using $E(\ce{C=C})=612$ , $E(\ce{H-H})=436$ , $E(\ce{C-C})=348$ and $E(\ce{C-H})=412\,\mathrm{kJ\,mol^{-1}}$ : broken $=612+436=1048$ ; made $=348+2(412)=1172$ ; therefore $\Delta H=1048-1172=-124\,\mathrm{kJ\,mol^{-1}}$ .

Count bonds from displayed or structural formulae and include coefficients. A negative result is exothermic because making the product bonds releases more energy; do not reverse the subtraction or count bonds that remain unchanged on only one side.

A temperature-change investigation compares the initial temperature with the highest or lowest temperature after a controlled change. The method must capture the extreme temperature while limiting heat exchange with the surroundings.

Change investigated	Suitable reactants and method
salt dissolving	measure water in an insulated cup, record its temperature, add a measured salt mass, replace lid, stir and record the minimum or maximum
neutralisation	mix measured acid and alkali volumes in an insulated cup and record the maximum
displacement	add an excess measured metal to a measured metal-salt solution in an insulated cup and record the maximum
combustion	heat a known water mass with a measured fuel mass and record the maximum and mass burned

Change one independent variable at a time. Keep reactant amounts, concentrations, starting temperature, apparatus, stirring and—when burning fuel—the flame distance constant. Repeat each condition, identify anomalies before calculating a mean, and use eye protection and small quantities.

Record the highest or lowest temperature, not a reading at an arbitrary fixed time. A polystyrene cup reduces heat transfer for solution reactions, while a metal can conducts heat from a flame to water; these apparatus choices serve different purposes.

(b) Rates of reaction

To investigate a rate factor, change only that factor, measure how quickly a fixed amount of product forms or reactant disappears, and keep every other condition constant.

Factor changed	Suitable comparison	Rate measurement	Essential controls
surface area	equal masses of marble chips with different chip sizes	carbon dioxide volume or mass loss against time	acid volume and concentration, temperature, marble mass
concentration	different hydrochloric acid concentrations with equal marble samples	gas volume, mass loss or time to a fixed endpoint	total acid volume, marble size and mass, temperature
temperature	sodium thiosulfate and acid at several measured temperatures	time until a cross is obscured; rate may be compared using $1/t$	solution volumes and concentrations, flask, cross and observer position
catalyst	equal hydrogen peroxide samples with equal amounts of different solids	oxygen volume in a fixed time or time to a fixed volume	peroxide volume and concentration, catalyst mass and surface area, temperature

Repeat each condition and calculate a mean after checking anomalies. Begin timing at mixing, collect readings at regular intervals and use the initial gradient when comparing continuous rate graphs.

Do not change two variables together. A shorter completion time means a faster rate, while a steeper product–time or mass-loss–time gradient means a faster rate; the final amount may remain unchanged when only rate changes.

Reaction rate increases when particles can collide successfully more often. The required factors alter collision frequency or the energy barrier without necessarily changing the final quantity of product.

Change	Effect on rate
increase the surface area of a solid	increases
increase solution concentration	increases
increase gas pressure	increases
increase temperature	increases
add a suitable catalyst	increases

On a product–time graph, a faster reaction has a steeper initial gradient and reaches its plateau sooner. If the reacting amounts are unchanged and only rate changes, both experiments reach the same plateau.

Rate describes change per unit time, not final yield. Greater pressure is a factor for gases; it is not used as the explanation for particles in a liquid solution.

A reaction occurs only when reactant particles collide with enough energy to react. Rate therefore depends on the number of successful collisions per unit time.

Factor increased	Particle-level change	Why successful collisions become more frequent
solid surface area	more reactant particles are exposed at the surface	more collisions can occur at the solid surface each second
solution concentration	more reactant particles occupy the same volume	collision frequency increases
gas pressure	gas particles are closer together; there are more particles per unit volume	collision frequency increases
temperature	mean kinetic energy increases; particles move faster	collisions are more frequent and a larger fraction have energy at least equal to the activation energy

As acid reacts with excess marble, acid concentration falls. Fewer acid particles remain per unit volume, so successful collisions become less frequent and the graph becomes less steep before levelling off when the acid is used up.

Higher concentration or pressure does not make individual particles move faster or give them more energy. That energy explanation belongs specifically to increasing temperature.

A catalyst increases the rate of a reaction but is chemically unchanged at the end. It participates in reaction steps, yet is regenerated rather than consumed overall.

Test	Evidence expected if the solid is a catalyst
compare rate with and without the solid	the reaction is faster when the solid is present
recover the solid after completion	filter, wash if needed, dry and reweigh it
compare before and after	the same substance and approximately the same mass remain

For manganese(IV) oxide in hydrogen peroxide decomposition, oxygen is produced faster, but the manganese(IV) oxide can be recovered after the peroxide has reacted.

A catalyst is not a reactant and is not included as a consumed substance in the overall equation. 'Chemically unchanged' does not mean it performs no role during the reaction.

A catalyst provides an alternative reaction pathway with a lower activation energy, $E_a$ .

At the same temperature, particle energies are not raised by the catalyst. Lowering the energy barrier means a larger fraction of collisions already have enough energy to react, so successful collisions occur more often and rate increases.

Changed by a catalyst	Not changed by a catalyst
reaction pathway and activation energy	reactant and product energy levels
reaction rate and time to reach completion or equilibrium	$\Delta H$ and the final equilibrium position

A catalyst does not supply energy to particles, increase their speed or make $\Delta H$ more negative. It changes the route, not the starting and finishing energy levels.

A reaction profile plots energy vertically against progress of reaction horizontally. It shows reactant and product energy levels, the activation-energy barrier and the enthalpy change.

Feature	How to draw or label it
reactants and products	labelled horizontal levels at the start and end
pathway	a curve rising to one peak and falling to the product level
activation energy, $E_a$	upward arrow from reactant level to the top of the peak
enthalpy change, $\Delta H$	arrow from reactant level to product level; downward/negative for exothermic, upward/positive for endothermic
catalyst pathway	second curve with a lower peak but the same start and end levels

The peak represents the minimum energy barrier along that pathway. A catalyst lowers this peak; because the endpoints remain fixed, it does not alter $\Delta H$ .

Do not measure $E_a$ from zero or from the product level for the forward reaction. Do not move product energy when adding a catalyst, and keep the activation-energy arrow pointing upward to the peak.

Investigate marble and dilute hydrochloric acid by following carbon dioxide production while changing either marble surface area or acid concentration.

\ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)}

Stage	Method
1	place a measured acid volume and concentration in a conical flask
2	add a measured mass of marble, immediately fit a gas syringe or place the flask on a balance with a cotton-wool plug, and start timing
3	record gas volume or mass at regular times until it stops changing
4	repeat with equal-mass smaller chips while keeping acid constant; separately repeat with different acid concentrations while keeping marble mass and size constant
5	repeat trials and compare initial gradients or time to a fixed gas volume/mass loss

Control temperature, total acid volume, apparatus and mixing. A cotton-wool plug prevents acid spray leaving while allowing carbon dioxide to escape, so mass loss is attributable to gas.

Never change chip size and acid concentration in the same comparison. Smaller chips may make a steeper curve without changing the final gas amount when reactant quantities and limiting reactant are unchanged.

Compare different solid catalysts by measuring how quickly equal hydrogen peroxide samples produce oxygen under otherwise identical conditions.

\ce{2H2O2(aq) -> 2H2O(l) + O2(g)}

Stage	Method and control
1	place the same volume and concentration of hydrogen peroxide in a conical flask connected to a gas syringe
2	add the same mass and comparable particle size of the first solid, seal immediately and start the timer
3	record oxygen volume at regular intervals or the time to a fixed volume
4	clean the apparatus and repeat at the same temperature with each solid; repeat each catalyst trial
5	compare initial gradients or mean times; the steepest gradient or shortest time identifies the most effective catalyst

Wear eye protection, use dilute hydrogen peroxide and small quantities, and keep the apparatus unblocked so oxygen cannot build dangerous pressure.

Equal catalyst mass alone is insufficient if particle sizes differ greatly, because surface area would also change. Catalyst effectiveness is judged from rate, not from final oxygen volume when the peroxide amount is fixed.

(c) Reversible reactions and equilibria

The symbol ⇌ shows that a reaction can proceed in both directions: reactants can form products, and the products can react to reform the reactants.

\ce{NH4Cl(s) <=> NH3(g) + HCl(g)}

Read the left-to-right process as the forward reaction and the right-to-left process as the reverse reaction. Conditions such as heating or cooling can favour one visible direction in a reversible change.

The symbol ⇌ does not by itself mean that the mixture is at equilibrium. It states that both directions are possible; equilibrium requires additional conditions and equal forward and reverse rates.

A reversible reaction can be demonstrated by driving a chemical change in one direction and then changing the conditions so that the original substances reform.

Example	Forward observation and change	How the reverse is produced
hydrated copper(II) sulfate	heating blue hydrated crystals removes water and leaves white anhydrous copper(II) sulfate	add water; the blue hydrated solid reforms
ammonium chloride	heating the white solid forms ammonia and hydrogen chloride gases	cool the gases; they recombine to form white ammonium chloride

\ce{CuSO4.5H2O(s) <=> CuSO4(s) + 5H2O(g)}

\ce{NH4Cl(s) <=> NH3(g) + HCl(g)}

A colour change alone is not the explanation: identify which substances are formed and how changing the conditions restores the starting substance. These demonstrations show reversibility; they do not automatically show dynamic equilibrium.

A reversible reaction can reach dynamic equilibrium when it occurs in a sealed container, so reacting substances cannot enter or escape.

Stage	What happens in the sealed container
1	reactants form products, so the forward reaction is initially dominant
2	products accumulate and the reverse reaction becomes faster
3	forward and reverse rates become equal; dynamic equilibrium has been reached

Sealing matters because the full reacting system is retained. If a gas escapes, its concentration changes and the reverse reaction may be unable to balance the forward reaction.

A sealed container is necessary for this equilibrium description, but sealing an irreversible reaction does not make it reversible. Both a reversible reaction and a sealed system are required.

At dynamic equilibrium in a sealed container, the forward and reverse reactions continue at equal rates. Therefore reactant and product concentrations remain constant over time.

At equilibrium	Correct interpretation
forward rate = reverse rate	equal amounts are converted in opposite directions per unit time
concentrations are constant	there is no overall composition change, although both reactions continue
concentrations need not be equal	the equilibrium mixture may contain different amounts of reactants and products

Macroscopic properties such as colour and pressure can stay unchanged because there is no net change, while particles continue reacting in both directions at the microscopic level.

Equilibrium is dynamic, not static. Do not say that both reactions stop, and do not replace 'equal rates' with 'equal concentrations'.

A catalyst increases both the forward and reverse reaction rates by providing lower-activation-energy pathways. It does not change the position of equilibrium.

A catalyst changes	A catalyst does not change
how quickly equilibrium is reached	the equilibrium concentrations of reactants and products
both forward and reverse rates	the equilibrium yield or the direction in which equilibrium lies

Because both directions are accelerated, their rates become equal sooner. The same equilibrium composition is reached at the same temperature and pressure, whether or not a catalyst is present.

A catalyst can improve production rate without improving equilibrium yield. It does not shift equilibrium toward products and is not a way to obtain a greater final equilibrium amount.

Changing temperature or pressure can change the equilibrium composition. Predict the direction by identifying the endothermic direction for temperature and comparing gaseous mole ratios for pressure.

Change	Direction favoured at the new equilibrium
increase temperature	the endothermic direction
decrease temperature	the exothermic direction
increase pressure	the side with fewer moles of gas
decrease pressure	the side with more moles of gas
change pressure when both sides have equal gaseous moles	no change in equilibrium position

\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad \Delta H < 0

For ammonia formation, the forward direction is exothermic and changes four moles of gas into two. Lower temperature favours ammonia at equilibrium, while higher pressure favours ammonia because the product side has fewer gaseous moles.

Count only gaseous coefficients when predicting a pressure effect. A condition may increase equilibrium yield but slow the reaction, or decrease yield while speeding it up, so keep equilibrium position separate from reaction rate.

3 Physical chemistry

Exam analysis

Most tested topics

(a) Energetics

Distinguish exothermic from endothermic change

Measure temperature change by simple calorimetry

Calculate heat energy with Q = mcΔT

Convert heat energy into molar enthalpy change

Draw and explain energy level diagrams

Track energy through bond breaking and making

Calculate ΔH from bond energies

Investigate temperature changes fairly

(b) Rates of reaction

Investigate four factors that change reaction rate

Predict how five factors affect reaction rate

Explain rate changes with collision theory

Recognise what a catalyst does—and does not do

Link catalysts to a lower activation-energy pathway

Construct and explain a reaction profile

Compare marble reaction rates fairly

Compare catalysts for hydrogen peroxide decomposition

(c) Reversible reactions and equilibria

Read the reversible-reaction symbol precisely

Explain two observable reversible reactions

Know when dynamic equilibrium can be reached

Describe dynamic equilibrium without saying reactions stop

Separate catalyst speed from equilibrium position

Predict temperature and pressure effects on equilibrium