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

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.

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.

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.

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.

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.

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: $v=0.0107/2=0.0054\,\mathrm{mol\,dm^{-3}\,s^{-1}}$.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $k$.

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.

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.

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.

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 $Q=K$.

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.

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.