Unit 4: Rates, Equilibria and Further Organic Chemistry A2

Rate commonly has units mol dm$^{-3}$ s$^{-1}$. Rearrange the measured rate equation to find $k$, then derive its units: zero order gives mol dm$^{-3}$ s$^{-1}$; first order gives s$^{-1}$; second order gives dm$^3$ mol$^{-1}$ s$^{-1}$.

A 100 mg dose with a 20 min half-life undergoes 12 half-lives in 4 h. The mass is $100(1/2)^{12}=0.0244$ mg, or 24.4 μg.

The measured signal must change monotonically with the chosen reactant or product and be fast to record compared with the reaction. State what is measured, how it relates to concentration, the sampling frequency and the main source of loss or delay.

A clock method uses the time to reach the same small, fixed amount of reaction. If the clock reagent is small compared with the main reactants, their concentrations change little before the endpoint, so $1/t$ is proportional to initial rate.

Keep total volume, temperature and all non-tested initial concentrations constant. Start timing at consistent mixing and use the same endpoint; otherwise $1/t$ compares different extents rather than rates.

Compare runs in which only one concentration changes. If concentration changes by factor $f$ and rate by factor $g$, solve $g=f^m$. When two concentrations change together, first account for the known order of one before isolating the other.

A concentration–time gradient gives rate, so compare gradients at chosen concentrations. A rate–concentration graph displays the dependence directly. State both the observed factor/shape and the resulting order.

The evidence supports a slow acid-catalysed formation of the enol from propanone, involving propanone and $\mathrm{H^+}$, followed by fast reaction of the enol with iodine. Iodine is absent from the rate-determining process, and $\mathrm{H^+}$ is regenerated.

Follow iodine loss by timed aliquots and titration, or by calibrated colorimetry. Use iodine in the smaller amount so propanone and acid remain nearly constant during a run.

If the observed rate is first order in H$_2$O$_2$, a slow elementary step containing one H$_2$O$_2$ molecule is consistent; a slow collision between two H$_2$O$_2$ molecules predicts second order and is inconsistent.

For an overall $\mathrm{A+B+C ightarrow P}$ reaction with $ext{rate}=k[\mathrm{A}][\mathrm{B}]$, a possible scheme is slow $\mathrm{A+B ightarrow X}$ followed by fast $\mathrm{X+C ightarrow P}$. Adding the steps cancels X and gives the overall equation.

Work from both ends: use the rate law to constrain the slow step, then add fast steps needed to account for the remaining overall reactants and products. Reject any scheme that leaves an intermediate uncancelled.

If doubling [RX] doubles rate and doubling [OH$^-$] also doubles rate, the reaction is first order in each and supports S$_N$2. If changing [OH$^-$] has no effect while rate follows [RX], the evidence supports S$_N$1.

Use $R=8.31$ J K$^{-1}$ mol$^{-1}$, so the calculated $E_a$ is initially in J mol$^{-1}$. The Arrhenius line has a negative gradient but activation energy is reported as a positive barrier.

A finely divided catalyst exposes more active surface area, so more reactant particles can be adsorbed at once. The catalyst is heterogeneous because solid and gas are different phases, and separation from the surface stops this catalytic route.

Thiosulfate removes iodine as it forms; it does not slow the main iodine-producing reaction. Once thiosulfate is exhausted, iodine remains and forms the blue-black starch complex. Keep temperature, total volume and non-tested concentrations constant.

Use consistent mixing/start time, repeat each condition and compare concordant endpoint times. In the aliquot method, sampling time and complete quenching are critical; in the clock, subjective colour judgement and delay in mixing affect precision.

At a fixed endpoint, the same small amount reacts in each run, so relative rate is proportional to $1/t$. Any constant proportionality factor changes the intercept of the Arrhenius plot but not its gradient.

Maintain temperature during each run, use the same observer/endpoint and minimise delay between mixing and timing. Include all valid points in a best-fit line and investigate rather than silently discard an anomalous result.

The complete criterion combines the reacting system with its surroundings. A change is thermodynamically feasible in the forward direction when $\Delta S_{total}$ is positive at the stated temperature.

With many molecules and quanta, dispersed arrangements vastly outnumber concentrated ones. Natural change therefore tends towards macrostates compatible with more microscopic arrangements.

On an entropy–temperature graph, entropy rises within each phase and jumps upward at melting and boiling. For water at standard pressure these transitions occur near 273 K and 373 K.

A perfect crystal at 0 K has one perfectly ordered arrangement and is assigned zero entropy. Imperfections or temperatures above 0 K introduce additional arrangements.

When a partition is removed, gas molecules spread into the available volume. The dispersed state corresponds to vastly more arrangements than all molecules confined to one side, so entropy increases and spontaneous remixing is overwhelmingly favoured.

For sodium chloride, ions fixed in a lattice become mobile and dispersed in water; rearrangement/disruption of water structure also contributes, so $\Delta S_{system}$ is positive. Other salts must be judged from evidence because strong hydration can order water.

Add the signed values after converting them to the same units. For example, $\Delta S_{total}=-78.7$ and $\Delta S_{surroundings}=+157.4$ J K$^{-1}$ mol$^{-1}$ give $\Delta S_{system}=-236.1$ J K$^{-1}$ mol$^{-1}$.

Multiply each standard molar entropy by its balanced-equation coefficient, total the products, total the reactants, then subtract reactants from products. Preserve state symbols because the same substance has different entropy in different states.

If product entropies total 738.2 and reactant entropies total 401.9 J K$^{-1}$ mol$^{-1}$, then $\Delta S_{system}^{\circ}=+336.3$ J K$^{-1}$ mol$^{-1}$. The positive sign means the products have greater entropy.

Use temperature in kelvin and make units consistent: convert $\Delta H$ from kJ mol$^{-1}$ to J mol$^{-1}$ when combining with entropy in J K$^{-1}$ mol$^{-1}$. Then add the signed surroundings and system values.

For $\Delta H=-67.2$ kJ mol$^{-1}$ at 298 K, $\Delta S_{surroundings}=+225.5$ J K$^{-1}$ mol$^{-1}$. The minus sign in the equation reverses the sign of the exothermic enthalpy.

At the threshold use kelvin and identical energy units. Decide which side of the threshold is feasible from the signs, not from the algebra alone. The equivalent $\Delta G=\Delta H-T\Delta S_{system}$ approach is permitted but not required.

If $\Delta S_{system}=-120$ and $\Delta S_{surroundings}=+180$ J K$^{-1}$ mol$^{-1}$, then $\Delta S_{total}=+60$ J K$^{-1}$ mol$^{-1}$ and the forward reaction is feasible despite the system becoming more ordered.

A reaction can be thermodynamically feasible ($\Delta S_{total}>0$) but extremely slow because a large activation barrier makes the reactant kinetically stable. Heating or a catalyst can increase rate without changing the thermodynamic balance.

For MgCl$_2$, include atomisation of one Mg and two Cl atoms, first and second ionisation energies of Mg, twice the first electron affinity of Cl, then lattice energy. Every line needs correct formulae, charges, electrons, coefficients and state symbols.

Apply Hess’s law around the closed cycle and rearrange only after all arrows have signs. A step used in reverse contributes the negative of its stated value.

The theoretical value treats ions as spherical charges interacting electrostatically. Its magnitude becomes larger for higher ionic charges and smaller ionic radii because attraction is stronger at shorter distance.

NaF shows close experimental/theoretical values, while MgCl$_2$ shows a larger discrepancy. The higher-charge Mg$^{2+}$ distorts the larger Cl$^-$ electron cloud more strongly, adding covalent character beyond the simple ionic model.

A fluoride ion is small and difficult to polarise, so calcium fluoride is close to the ionic model. Iodide is larger and readily polarised by Ca$^{2+}$, so calcium iodide has more covalent character and a larger experimental–theoretical lattice-energy difference.

Hydration is normally exothermic because ion–dipole attractions form between ions and water. Enthalpy of solution may be positive or negative because separating the lattice costs energy while hydrating the ions releases energy.

For MX$_2$, include one cation hydration term and twice the anion hydration term. If a missing anion value is required, subtract the known lattice/solution/cation terms, then divide by two.

Label every energy level with formulae, charges and state symbols. Arrow direction controls sign: the course lattice energy arrow points from gaseous ions down to solid.

Lattice energy depends on both ions’ charges and their separation; hydration enthalpy is quoted for one ion and follows its charge density. Down a group, anions become larger, so both their hydration enthalpy and the corresponding lattice energy generally become less exothermic.

For NH$_4$Cl versus NH$_4$Br, Br$^-$ is larger: its hydration is less exothermic, but the NH$_4$Br lattice energy is also less exothermic. Because both effects weaken, their difference—and hence $\Delta_{sol}H$—may remain similar.

KCl has an endothermic $\Delta_{sol}H$ of about +17 kJ mol$^{-1}$ yet is soluble because its positive system entropy change is large enough to make $\Delta S_{total}$ positive at 298 K.

Use supplied entropy and enthalpy data to explain the Unit 2 trends: Group 2 hydroxides become more soluble down the group, whereas Group 2 sulfates become less soluble. Compare the total entropy balance for each salt rather than relying on lattice energy alone.

Write equilibrium concentrations of products over reactants and use the balanced-equation coefficients as powers. Square brackets mean concentration in mol dm$^{-3}$; coefficients are exponents, not multiplying factors.

For $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$, $K_c=[\ce{CO2}]$. The two solids are present in the equilibrium but omitted from the expression. For a homogeneous mixture, include every participating species in that phase.

Use only gaseous species. Each $p(X)$ is the equilibrium partial pressure of gas X in atm, and each balanced coefficient becomes its power. Solids, liquids and aqueous species are omitted from $K_p$.

For $\ce{Mg(NO3)2(s) <=> MgO(s) + 2NO2(g) + 1/2O2(g)}$, $K_p=p(\ce{NO2})^2p(\ce{O2})^{1/2}$. The solid terms do not appear.

Keep initial, change and equilibrium amounts separate. If 1.60 mol $\ce{SO3}$ forms from $\ce{2SO2 + O2 <=> 2SO3}$, then 1.60 mol $\ce{SO2}$ and 0.80 mol $\ce{O2}$ are consumed before mole fractions are calculated.

$\Delta n$ is products minus reactants for the species actually present in the chosen constant expression. If $\Delta n=0$, the units cancel. State units only where appropriate and retain enough figures during intermediate steps.

Pressure has no effect on equilibrium composition when both sides contain the same total moles of gas. Count gaseous coefficients only; pure solids and liquids do not enter this comparison.

In a heterogeneous system, changing the amount or surface area of a pure solid does not change the equilibrium composition while some solid remains. It can change how quickly equilibrium is reached.

Industrial conditions balance equilibrium yield with rate and cost. A high temperature may reduce the equilibrium yield of an exothermic product yet be chosen because the low-temperature rate is too slow.

After a concentration or pressure disturbance, the reaction proceeds in the direction that restores the composition quotient to the unchanged equilibrium-constant value. The new equilibrium composition can differ even though K is identical.

A catalyst lowers the activation energy for both directions. It shortens the time taken to reach equilibrium but does not alter the equilibrium composition, yield or constant.

The direction also works backwards as evidence. If lowering temperature makes $K_c$ smaller, the forward reaction is endothermic. If raising temperature makes $K_p$ smaller, the forward reaction is exothermic.

A new K requires new equilibrium partial pressures or concentrations. For an exothermic forward reaction at higher temperature, product terms decrease relative to reactant terms, so the calculated constant becomes smaller.

Use a complete causal chain: identify whether the forward direction is endothermic or exothermic; state how the temperature change alters K; then state which side must become more abundant so the equilibrium expression attains that new K.

Hot pure water can have pH below 7 while remaining neutral because $[\ce{H+}]=[\ce{OH-}]$; the larger $K_w$ increases both concentrations.

For an exothermic forward reaction, $-\Delta H/T$ is positive. Raising T makes this positive surroundings contribution smaller, so $\Delta S_{total}$ and $\ln K$ decrease; K decreases.

For an endothermic forward reaction, $-\Delta H/T$ is negative. Raising T makes it less negative, so $\Delta S_{total}$ and $\ln K$ increase; K increases. This comparison assumes $\Delta S_{system}$ is approximately constant over the stated range.

Describe the equilibrium position as lying toward products or reactants. A large K says the equilibrium is product-favoured; it does not mean that the equilibrium is currently 'shifting' right.

When K is already very large, a pressure increase that favours products may deliver only a small extra equilibrium yield. The small gain may not justify higher compression cost or risk.

A value such as $K_p=1.55 imes10^6$ indicates an equilibrium far toward products. A value such as $K_c=4.85$ indicates more products than reactants for a comparable simple mixture, but appreciable amounts of both remain.

Here HA donates H$^+$ and is the acid. Water accepts H$^+$ and is the base. Writing only $\ce{HA <=> H+ + A-}$ hides the accepting species, so use water when the question asks how the acid behaves in aqueous solution.

A species can play different roles in different reactions. For example, $\ce{HCO3-}$ can accept a proton to form $\ce{H2CO3}$ or donate one to form $\ce{CO3^2-}$; classify the role from the actual proton transfer.

Match formulae first, then check charge: loss of H$^+$ makes the charge one unit more negative; gain makes it one unit more positive. The two members of a pair normally appear on opposite sides of the equation.

The concentration $[\ce{H+}]$ (or $[\ce{H3O+}]$) is in mol dm$^{-3}$. Because the scale is logarithmic, a decrease of one pH unit corresponds to a tenfold increase in hydrogen-ion concentration.

Use the equilibrium hydrogen-ion concentration in mol dm$^{-3}$, enter its base-10 logarithm, then change the sign. For $[\ce{H+}]=2.50\times10^{-3}$ mol dm$^{-3}$, $\mathrm{pH}=-\log_{10}(2.50\times10^{-3})=2.602$.

Keep unrounded concentration values during a multi-stage calculation. A pH commonly has as many decimal places as the concentration has significant figures when the data justify that precision.

For pH 1.125, $[\ce{H+}]=10^{-1.125}=7.50\times10^{-2}$ mol dm$^{-3}$. The operation is the inverse of the base-10 logarithm.

When pH data are used in dilution, convert both pH values to concentrations before applying conservation of moles or $c_1V_1=c_2V_2$. A pH increase of 1 means a tenfold decrease in $[\ce{H+}]$, not a decrease of 1 mol dm$^{-3}$.

The first dissociation of sulfuric acid is effectively complete, but its second dissociation is an equilibrium. Therefore a 0.100 mol dm$^{-3}$ solution need not contain exactly 0.200 mol dm$^{-3}$ H$^+$.

For 0.500 mol dm$^{-3}$ HCl, $[\ce{H+}]=0.500$ mol dm$^{-3}$ and pH $=0.301$. For 1.25 mol dm$^{-3}$ HCl, pH $=-0.097$: negative pH is mathematically possible.

If acid and base are mixed, calculate reacting moles first, identify the excess H$^+$, divide by the total volume, then calculate pH.

Place equilibrium concentrations of products over reactant and use square brackets. Liquid water is omitted because its effective concentration is constant. For ethanoic acid, replace HA and A$^-$ with $\ce{CH3COOH}$ and $\ce{CH3COO-}$, preserving formulae and charges.

At a fixed temperature, a larger $K_a$ means the dissociation equilibrium lies further toward ions and the weak acid is stronger.

For a weak monoprotic acid of initial concentration $c$, assume dissociation is small so $c-x\approx c$. Then $x\approx\sqrt{K_ac}$ and $\mathrm{pH}=-\log_{10}x$. Convert $pK_a$ first with $K_a=10^{-pK_a}$.

For $c=0.100$ mol dm$^{-3}$ and $pK_a=4.88$, $K_a=1.32\times10^{-5}$ mol dm$^{-3}$, $[\ce{H+}]=1.15\times10^{-3}$ mol dm$^{-3}$ and pH $=2.94$.

In neutral water, $[\ce{H+}]=[\ce{OH-}]=\sqrt{K_w}$. At 25 °C, $K_w=1.00\times10^{-14}$ mol$^2$ dm$^{-6}$, so neutral pH is 7.00.

Kw changes with temperature. If $K_w=5.5\times10^{-14}$ mol$^2$ dm$^{-6}$ at 50 °C, neutral water has pH about 6.6 because the two ion concentrations remain equal.

At 25 °C, 0.200 mol dm$^{-3}$ $\ce{Ba(OH)2}$ gives $[\ce{OH-}]=0.400$ mol dm$^{-3}$. Thus pOH $=0.398$ and pH $=14.000-0.398=13.602$.

For acid-base mixtures, use balanced reacting moles, calculate excess OH$^-$ concentration in the total volume, and only then convert to pH.

The inverse conversions are $K_a=10^{-pK_a}$ and $K_w=10^{-pK_w}$. Because of the negative logarithm, a smaller $pK_a$ corresponds to a larger $K_a$ and hence a stronger weak acid.

If $K_a=1.38\times10^{-4}$ mol dm$^{-3}$, $pK_a=3.86$. At 25 °C, $K_w=1.00\times10^{-14}$ mol$^2$ dm$^{-6}$ gives $pK_w=14.00$.

A tenfold dilution of a strong monoprotic acid makes $[\ce{H+}]$ ten times smaller, so pH rises by 1. A weak acid dissociates to a greater fraction after dilution, partly replacing the removed H$^+$; its pH therefore rises by less than 1 in the supplied comparison.

For the same concentration series, HCl pH may rise 1.00 → 2.00 → 3.00 on successive tenfold dilutions, while ethanoic acid might rise 2.88 → 3.38 → 3.88.

If a solution has $c=0.500$ mol dm$^{-3}$ and pH 1.20, then $x=0.0631$ mol dm$^{-3}$ and $K_a=(0.0631)^2/(0.500-0.0631)=9.11\times10^{-3}$ mol dm$^{-3}$.

State the relevant chemistry: one H$^+$ and one conjugate-base ion form per dissociated acid molecule, and any later dissociation is negligible when the question says so.

The equivalence volume is where stoichiometric acid and base amounts have reacted. For a diprotic acid titrated by a strong base, the second equivalence volume is twice the first when both protons react sequentially under the same conditions.

Use the supplied Data Booklet range or approximately $pK_{In}\pm1$, and state the endpoint colour if requested. A named indicator earns justification only when its range is compared with the actual graph.

A typical acidic buffer contains appreciable amounts of a weak acid and its conjugate base; an alkaline buffer contains a weak base and its conjugate acid. Both components are needed to consume the two kinds of added reagent.

Because HA and A$^-$ are both present as a large reservoir, removing a small added amount changes their concentration ratio only slightly. The corresponding $[\ce{H+}]$ and pH therefore change only slightly.

The same logic applies to $\ce{NH3/NH4+}$: ammonia accepts added H$^+$ to form $\ce{NH4+}$, while $\ce{NH4+}$ supplies acid capacity against added OH$^-$.