Reactivity 2. How much, how fast and how far?

A balanced chemical equation conserves every element and charge. Its coefficients show the mole ratio of reactants and products.

Write the products and state symbols, balance atoms with coefficients rather than changing formula subscripts, then check every element and the charge.

For Al + O₂ → Al₂O₃, preserve the formulas and choose coefficients 4Al + 3O₂ → 2Al₂O₃. Changing O₂ to O or Al₂O₃ to another subscript would change the substances rather than balance them. Add state symbols from chemical evidence or stated conditions, not from atom counting.

Read the required-to-known ratio from the balanced equation. For gases at fixed temperature and pressure, the same coefficient ratio applies to volumes; always finish with the requested unit.

For N₂ + 3H₂ → 2NH₃, 0.50 mol N₂ corresponds to 1.00 mol NH₃ when H₂ is sufficient. Keep the conversion chain visible—given unit to moles, coefficient ratio, then requested unit—so molar mass, concentration or gas volume is used at the correct end.

Worked mass example: $\ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}$. For $1.00\,\mathrm{g}$ of $\ce{Mg(OH)2}$, $n=1.00/58.33=0.0171\,\mathrm{mol}$. The $1:2$ coefficient ratio gives $n(\ce{HCl})=2(0.0171)=0.0343\,\mathrm{mol}$, so $m(\ce{HCl})=0.0343\times36.46=1.25\,\mathrm{g}$. The result is the maximum neutralized mass when magnesium hydroxide reacts completely.

The limiting reactant is used up first and determines the maximum, or theoretical, yield. An excess reactant remains after the reaction is complete.

Convert each reactant to moles, divide by its balanced-equation coefficient, and identify the smallest normalized amount as limiting. Use that reactant's ratio to calculate product.

For 2H₂ + O₂ → 2H₂O with 3.0 mol H₂ and 2.0 mol O₂, compare n/coefficient: 1.5 for H₂ and 2.0 for O₂, so H₂ limits and forms 3.0 mol H₂O. A smaller starting mass is not necessarily limiting; the decision depends on moles relative to coefficients.

Find theoretical yield from the limiting reactant and stoichiometric ratio before comparing it with the measured experimental yield. Do not divide the product mass by a starting mass directly.

If stoichiometry predicts 10.0 g but 8.20 g is isolated, percentage yield is 82.0%. A value above 100% signals wet or impure product, measurement error or an incorrect theoretical yield; it is not evidence that the reaction created extra conserved matter.

Atom economy measures how much of the reactant mass is represented by the desired product. Low atom economy means more reactant mass becomes by-products and waste.

Apply stoichiometric coefficients to every formula mass before forming the ratio. Atom economy is fixed by the chosen equation and desired product, whereas percentage yield measures experimental recovery; a reaction can have high atom economy but poor yield, or the reverse.

Worked atom-economy example: $\ce{4CH3OH + 2CO + O2 -> 2(CH3O)2CO + 2H2O}$, with dimethyl carbonate as the desired product. $\text{Atom economy}=\frac{2(90.09)}{4(32.05)+2(28.01)+32.00}\times100=83.33\%$. Coefficients multiply every molar mass. The remaining $16.67\%$ of reactant mass becomes the water by-product for this equation; actual percentage yield is a separate experimental measure.

An instantaneous rate is the gradient of a tangent at the stated point on a concentration–time, volume–time or mass–time graph. Keep the units consistent.

Choose two well-separated points on the tangent, not on the curve, to calculate its gradient. A reactant concentration has a negative gradient, so report its disappearance rate as a positive magnitude unless a signed change is requested.

Worked tangent example: on a concentration–time graph for $\ce{HCl}$, two points on the tangent at $t=0$ are $(0\,\mathrm{s},0.250\,\mathrm{mol\,dm^{-3}})$ and $(14\,\mathrm{s},0.100\,\mathrm{mol\,dm^{-3}})$. The tangent gradient is $(0.100-0.250)/(14-0)=-0.0107\,\mathrm{mol\,dm^{-3}\,s^{-1}}$. For $\ce{Mg + 2HCl -> MgCl2 + H2}$, divide the positive disappearance-rate magnitude by the HCl coefficient: $v=0.0107/2=0.0054\,\mathrm{mol\,dm^{-3}\,s^{-1}}$.

A successful collision needs sufficient kinetic energy to overcome activation energy and a suitable orientation of the reacting particles.

Temperature raises average kinetic energy and changes collision frequency and the fraction of particles with energy at least Ea. Collision frequency alone does not guarantee reaction.

At one temperature, only the fraction of collisions above Ea and with a productive orientation can react. Raising temperature increases that fraction, not the energy of every particle by the same amount. Use collision frequency to explain concentration or pressure effects and the energy distribution to explain the stronger temperature effect.

Explain a predicted rate change through collision frequency or the fraction of effective collisions, not just by saying particles move faster.

Powdered CaCO₃ reacts faster than equal-mass chips because more surface sites are exposed, not because its particles have higher kinetic energy. For each changed condition, identify exactly what changes—collision frequency, energy distribution or pathway—and hold other variables constant in a fair comparison.

A Maxwell–Boltzmann curve shows the distribution of particle kinetic energies. At higher temperature the peak is lower and shifts right; the area beyond Ea is larger, so more particles can react.

Two Maxwell–Boltzmann curves for the same number of particles have equal total area. At higher T the curve is broader with a lower peak and a larger area to the right of a fixed Ea line; the peak does not move to Ea and no particle count is lost.

A catalyst provides an alternative reaction pathway with lower activation energy. It does not change the energy levels of the reactants or products.

Because the catalysed Ea is lower, a larger fraction of the same distribution lies beyond the threshold. The reaction therefore has more effective collisions at the same temperature.

At fixed temperature a catalyst does not change the Maxwell–Boltzmann distribution; it moves the threshold to a lower Ea, increasing the area beyond it. On an energy profile it changes the pathway and peak, not ΔH, reactant/product energies or the equilibrium constant.

A mechanism is a sequence of elementary steps. The slowest step is the rate-determining step; an intermediate is formed in one step and consumed in a later step, whereas a transition state is the high-energy configuration at a barrier.

A proposed mechanism must be compared with the experimental rate equation and stoichiometry. Matching a rate equation can support a mechanism but does not prove it, because different mechanisms may give the same expression.

Add all elementary steps and cancel intermediates to recover the overall equation. Then derive the rate dependence expected from the slow step, eliminating an intermediate when necessary, and compare with experiment. A catalyst consumed early and regenerated later cancels from the overall equation but is not an intermediate.

In a multistep profile, peaks are transition states and valleys between peaks are intermediates. Each step has its own Ea; the largest barrier from its preceding intermediate identifies the rate-determining step.

For each step, measure Ea upward from its own preceding reactant or intermediate valley to the next peak. The rate-determining barrier is the largest of those step barriers, which need not be the peak with the greatest absolute height above the page baseline.

Molecularity is the number of reacting particles in one elementary step: one is unimolecular, two is bimolecular and three is termolecular.

Count particles in the individual elementary step, not coefficients in the overall reaction equation.

A step A + 2B → products is termolecular because three reacting particles meet in that elementary event. Molecularity is always a positive whole-number description of one proposed step; reaction order is experimental and can be zero, fractional or unrelated to overall stoichiometric coefficients.

Use experimental trials that vary one concentration independently. Compare the rate factor with the concentration factor to infer each exponent; the balanced overall equation alone cannot determine the rate equation.

If doubling [A] while holding [B] constant quadruples rate, the order in A is 2; if doubling [B] leaves rate unchanged, the order in B is 0, giving rate = k[A]². Choose trial pairs with only one changed concentration before combining exponents.

Worked initial-rate deduction: doubling $[\ce{Fe^{3+}}]$ from $1.00\times10^{-2}$ to $2.00\times10^{-2}\,\mathrm{mol\,dm^{-3}}$ at constant $[\ce{I^-}]$ doubles rate, so the order in $\ce{Fe^{3+}}$ is 1. Doubling $[\ce{I^-}]$ at constant $[\ce{Fe^{3+}}]$ increases rate from $3.24\times10^{-5}$ to $1.30\times10^{-4}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$, approximately fourfold, so its order is 2. Therefore $v=k[\ce{Fe^{3+}}][\ce{I^-}]^2$, third order overall.

The exponent gives order with respect to that reactant. Use rate factors to identify zero, first or second order, then match the order to the concentration–time or rate–concentration graph.

Use multiple representations as cross-checks: zero-order rate is independent of concentration and [A] falls linearly; first-order rate is proportional to [A] and gives exponential decay; second-order rate curves upward on a rate-versus-concentration plot. Do not infer order from one balanced equation.

The units of k depend on the overall order and must cancel the concentration and time units in the rate equation. For a particular reaction, k changes with temperature.

Derive rather than memorize the units: first-order k has units s⁻¹, while an overall second-order law written with mol dm⁻³ and seconds gives dm³ mol⁻¹ s⁻¹. Concentration changes rate but does not change k at fixed temperature.

Worked $k$ example: for $v=k[\ce{Fe^{3+}}][\ce{I^-}]^2$, use $v=1.62\times10^{-5}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ and both concentrations $1.00\times10^{-2}\,\mathrm{mol\,dm^{-3}}$. $k=\frac{1.62\times10^{-5}}{(1.00\times10^{-2})(1.00\times10^{-2})^2}=16.2\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}$. Substitution returns the measured rate, and the units cancel those of a third-order concentration term.

On a plot of ln k against 1/T, gradient = −Ea/R and intercept = ln A. Use kelvin, preserve the negative gradient and convert J mol−1 to kJ mol−1 when required.

A steeper negative ln k versus 1/T gradient means a larger activation energy. When two temperatures are given, subtracting the two linear equations removes ln A and lets Ea be found without knowing the frequency factor.

Worked Arrhenius-plot example: using line points $(0.00302\,\mathrm{K^{-1}},-6.8)$ and $(0.00334\,\mathrm{K^{-1}},-10.8)$, the gradient is $[-6.8-(-10.8)]/(0.00302-0.00334)=-1.25\times10^4\,\mathrm{K}$. Since gradient $=-E_a/R$, $E_a=-(-1.25\times10^4)(8.31)=1.04\times10^5\,\mathrm{J\,mol^{-1}}=104\,\mathrm{kJ\,mol^{-1}}$. The negative plot gradient therefore produces a positive activation energy.

A is the frequency factor: it represents the frequency of collisions with proper orientation. In the linear Arrhenius form, the y-intercept is ln A.

Read the intercept, find A by exponentiating ln A, and keep A distinct from Ea: Ea describes the energy barrier, while A describes collision frequency and orientation.

From a fitted line, intercept b gives A = eᵇ, while gradient m gives Ea = −mR. Check the pair against k = Ae^(−Ea/RT) at one data temperature. A has the same units as k for the stated rate law; it is not an activation energy or a universal constant.

Worked intercept example: for the same first-order Arrhenius line, use $\ln k=-6.02$, $1/T=0.00296\,\mathrm{K^{-1}}$ and gradient $-E_a/R=-12500\,\mathrm{K}$. From $\ln k=(-E_a/R)(1/T)+\ln A$, $-6.02=(-12500)(0.00296)+\ln A$, so $\ln A=30.98$ and $A=e^{30.98}=2.85\times10^{13}\,\mathrm{s^{-1}}$. The unit is $\mathrm{s^{-1}}$ because this reaction is first order and $A$ has the same units as $k$.

Dynamic equilibrium occurs in a closed system when the forward and reverse processes continue at equal rates. Macroscopic amounts remain constant, but reactants and products need not be equal in amount.

The same rate-balance idea applies to physical equilibria such as vaporization and condensation as well as to reversible chemical reactions.

In a sealed liquid–vapour system at equilibrium, molecules continue evaporating and condensing at equal rates, so pressure and amounts are constant on average. Equal rates do not mean equal concentrations, and opening the system can prevent equilibrium by allowing matter to escape.

For a homogeneous reaction, place product concentrations over reactant concentrations and use each balanced-equation coefficient as the exponent.

Build the expression only after balancing the equation, and use equilibrium rather than initial concentrations. The numerical value of K changes with temperature; changing starting amounts can move the equilibrium composition without changing K.

K describes a ratio, not reaction speed: a very large K can still belong to a slow reaction. Reversing the equation gives 1/K, while multiplying every coefficient by a factor raises K to that factor. Interpret 'favoured' as equilibrium composition, not complete conversion.

Worked reading: if K = 0.0665 at 100 C, K < 1, so reactants are favoured and the forward reaction has a small extent. This describes equilibrium composition, not reaction speed. At the same temperature, reversing the equation gives K = 1/0.0665.

An equilibrium shifts to partially counteract an imposed change. Pressure favours the side with fewer gaseous molecules; temperature favours the endothermic direction; concentration changes alter composition.

At fixed temperature, concentration and pressure changes do not change K. Temperature changes K. A catalyst changes rates in both directions and does not change equilibrium position.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), compression favours the two-mole gas side, but K is unchanged if temperature is fixed. Heating favours the endothermic direction and changes K; a catalyst reaches the same equilibrium faster by accelerating both directions.

Q uses concentrations at any time, not necessarily equilibrium. Q < K means the forward direction is needed; Q > K means the reverse direction is needed; Q = K means equilibrium.

Calculate Q with the same expression as K but using the current concentrations. If Q = 0.20 and K = 5.0, too little product is present relative to equilibrium, so the forward direction lowers the mismatch. Recalculate after composition changes; Q is a snapshot, not a new constant.

Worked $Q$ example: for $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$, $Q=[\ce{NH3}]^2/([\ce{N2}][\ce{H2}]^3)$. If every current concentration is $0.50\,\mathrm{mol\,dm^{-3}}$, $Q=(0.50)^2/[(0.50)(0.50)^3]=4.0$. At 475 K, $K=0.59$, so $Q>K$: the mixture contains too much product relative to equilibrium and the reverse direction is favoured until $Q=K$.

Write the balanced reaction, record initial concentrations, express changes as coefficient multiples of x, and substitute the equilibrium row into Kc. Use a small-K approximation when justified; quadratic equations are not expected here.

Equilibrium reactions do not use the limiting-reactant idea: both directions remain possible, so calculate the equilibrium composition instead.

After using an approximation such as C₀ − x ≈ C₀, validate it by checking that x/C₀ is small, commonly below 5% for the stated course method. If the check fails, the approximation is not justified; revise the setup rather than treating equilibrium as a limiting-reactant completion.

Worked equilibrium example: for $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$, $K=3.0$, $[\ce{SO2}]_{eq}=0.12$ and $[\ce{SO3}]_{eq}=0.18\,\mathrm{mol\,dm^{-3}}$. From $3.0=(0.18)^2/[x(0.12)^2]$, $x=[\ce{O2}]_{eq}=0.75\,\mathrm{mol\,dm^{-3}}$. Forming $0.18\,\mathrm{mol\,dm^{-3}}$ of $\ce{SO3}$ consumes $0.18$ of $\ce{SO2}$ and $0.090$ of $\ce{O2}$, so their initial concentrations were $0.30$ and $0.84\,\mathrm{mol\,dm^{-3}}$, respectively.

K and ΔG both describe equilibrium position. K > 1 corresponds to product-favoured equilibrium and ΔG° < 0; K = 1 corresponds to ΔG° = 0 and approximately equal equilibrium tendencies.

Use a dimensionless K, T in kelvin and R = 8.31 J mol⁻¹ K⁻¹. The sign link applies to ΔG°: the actual ΔG away from standard conditions also depends on the reaction quotient.

Worked Gibbs example: for $\ce{2NO(g) <=> N2O2(g)}$ at 298 K, local spectroscopic data give $K=1.39\times10^{-5}$. $\Delta G^\circ=-(8.31\,\mathrm{J\,K^{-1}\,mol^{-1}})(298\,\mathrm{K})\ln(1.39\times10^{-5})=+2.77\times10^4\,\mathrm{J\,mol^{-1}}=+27.7\,\mathrm{kJ\,mol^{-1}}$. The positive value agrees with $K<1$: reactants are favoured under standard conditions at this temperature.