Topic 5: Reactivity 2: How much, how fast and how far

This theme links three big questions about any chemical reaction: how much product forms, how fast it forms, and how far the reaction proceeds before settling. You will use the mole to count particles, collision theory to explain rate, and the equilibrium constant Kc to describe the balance reached at the end.

Amount of substance and stoichiometry

The mole is the chemist's counting unit: one mole contains the Avogadro constant, 6.02 x 10^23 particles. The amount in moles n equals the mass m divided by the molar mass M, so n = m / M. Balanced equations give mole ratios that let you convert between reactants and products. Worked example: how many moles are in 36.0 g of water? M(H2O) = 18.0 g/mol, so n = 36.0 / 18.0 = 2.00 mol, which is 2.00 x 6.02 x 10^23 = 1.20 x 10^24 molecules. For gases at STP (273 K, 100 kPa) the molar volume is 22.7 dm3/mol, so n = V / 22.7. For solutions, concentration c = n / V where V is in dm3, giving n = c x V.

Limiting reactant, theoretical and percentage yield

When reactants are not mixed in the exact stoichiometric ratio, one runs out first and limits how much product forms; this is the limiting reactant, and the others are in excess. Identify it by dividing the moles of each reactant by its coefficient and choosing the smallest result. The theoretical yield is the maximum product predicted from the limiting reactant. Real reactions give less, so percentage yield = (actual yield / theoretical yield) x 100. Worked example: 6.0 g of carbon (0.50 mol) reacts with 16.0 g of oxygen (0.50 mol) in C + O2 -> CO2. Both give 0.50 mol per coefficient, so neither is in large excess; 0.50 mol CO2 (22.0 g) is the theoretical yield. If only 19.8 g forms, percentage yield = 19.8 / 22.0 x 100 = 90.0%.

Atom economy

Percentage yield only counts the desired product made versus predicted; it ignores wasted by-products built into the equation. Atom economy measures how efficiently reactant atoms end up in the wanted product: atom economy = (molar mass of desired product / total molar mass of all products) x 100, using the balanced equation. A reaction can have high yield but poor atom economy if it produces large waste molecules. Green chemistry favours high atom economy because it reduces waste and cost. Worked example: in CaCO3 -> CaO + CO2, if CaO is the target, atom economy = 56.1 / (56.1 + 44.0) x 100 = 56.0%.

Titration calculations

A titration finds an unknown concentration by reacting it with a solution of known concentration until the equivalence point, signalled by an indicator colour change. Read the burette to find the volume delivered, then use moles. Worked example: 25.0 cm3 of NaOH is neutralised by 20.0 cm3 of 0.100 mol/dm3 HCl in NaOH + HCl -> NaCl + H2O. Moles of HCl = 0.100 x 0.0200 = 2.00 x 10^-3 mol. The 1:1 ratio means moles of NaOH = 2.00 x 10^-3 mol. Concentration of NaOH = n / V = 2.00 x 10^-3 / 0.0250 = 0.0800 mol/dm3. Always convert cm3 to dm3 by dividing by 1000.

Rate of reaction and how it is measured

Rate of reaction is the change in concentration of a reactant or product per unit time, with units such as mol/dm3/s. Because rate is fastest at the start and slows as reactants deplete, we often quote the average rate or take a tangent gradient from a concentration-time graph. Common ways to follow a reaction include measuring gas volume collected over time, the loss of mass as gas escapes, the time for a marked cross to disappear behind a precipitate (turbidity), changes in colour using a colorimeter, or changes in conductivity or pH. The method chosen depends on which property changes measurably during the reaction.

Collision theory and factors affecting rate

Collision theory states that particles must collide with sufficient energy (at least the activation energy, Ea) and the correct orientation for a reaction to occur. Anything that increases the frequency or energy of successful collisions speeds up the reaction. Increasing concentration or gas pressure packs particles closer, so collisions are more frequent. Increasing surface area of a solid exposes more particles to collide. Raising temperature gives particles more kinetic energy, so more collisions exceed Ea and they collide more often. A catalyst provides an alternative pathway with lower Ea, so a greater fraction of collisions succeed without the catalyst being consumed.

(HL) Rate laws, order and mechanisms

The rate law is found experimentally, not from the balanced equation: rate = k[A]^m[B]^n, where m and n are the orders with respect to each reactant and k is the rate constant. The overall order is m + n. Zero order means rate is independent of that reactant; first order means doubling it doubles the rate; second order means doubling it quadruples the rate. Worked example: if doubling [A] doubles the rate while [B] is constant, the reaction is first order in A. Reactions occur through a series of steps called the mechanism; the slowest step is the rate-determining step, and the rate law reflects only the species up to and including it. Units of k depend on the overall order; for a first-order reaction k has units of s^-1.

Dynamic equilibrium and Kc

A reversible reaction in a closed system reaches dynamic equilibrium when the forward and reverse reactions occur at equal rates, so concentrations stay constant although both reactions continue. The equilibrium constant Kc summarises the position: for aA + bB <=> cC + dD, Kc = ([C]^c[D]^d) / ([A]^a[B]^b), using equilibrium concentrations. Kc has a fixed value at a given temperature. A large Kc (>> 1) means products dominate at equilibrium; a small Kc (<< 1) means reactants dominate. Worked example: for H2 + I2 <=> 2HI, Kc = [HI]^2 / ([H2][I2]). Pure solids and liquids are omitted from the Kc expression.

Le Chatelier's principle

Le Chatelier's principle predicts how an equilibrium responds to a disturbance: the system shifts to partly oppose the change. Increasing the concentration of a reactant shifts the equilibrium toward products. Increasing pressure (by decreasing volume) shifts it toward the side with fewer moles of gas. Increasing temperature favours the endothermic direction, absorbing the added heat. A catalyst speeds up both directions equally, so it shortens the time to reach equilibrium but does not change the position or Kc. Worked example: in N2 + 3H2 <=> 2NH3 (exothermic), raising pressure favours the 2-mole product side and lowering temperature favours the forward exothermic reaction, both increasing ammonia yield.

(HL) Interpreting the magnitude of K

Only a change in temperature changes the value of K; concentration and pressure changes alter the position of equilibrium but leave K unchanged because the system re-establishes the same ratio. For an exothermic forward reaction, raising temperature decreases K (favouring reactants); for an endothermic forward reaction, raising temperature increases K. The reaction quotient Q has the same form as Kc but uses current, non-equilibrium concentrations: if Q < K the reaction proceeds forward, if Q > K it proceeds in reverse, and if Q = K the system is at equilibrium. Comparing Q with K therefore tells you the direction a mixture will shift to reach balance.