Explain reaction rates with collision theory and calculate energy changes with specific heat and bond energies

Learning goal: Use collision theory to explain why concentration, temperature, pressure of gases, surface area and catalysts change the rate of a reaction; draw and read potential energy diagrams with reactants, products, activated complex and activation energy; calculate thermal energy changes with specific heat in calorimetry; and predict whether a reaction is exothermic or endothermic from bond energies and from the flow of energy.

Rate and collisions. The rate of a chemical reaction is the change in the amount of a reactant or product in a given time, such as milliliters of gas per second or grams of solid dissolved per minute. Collision theory explains rates: particles react only when they collide with enough energy to break the necessary bonds and in an orientation that lets the new bonds form. Most collisions fail one test or the other. Anything that increases the number of collisions per second, or the fraction that are effective, increases the rate. Concentration: more particles in the same volume collide more often, so a more concentrated solution usually reacts faster (not always: when something else limits the rate, such as an enzyme or a catalyst surface that is already fully occupied, adding more reactant changes little). Pressure of a gaseous reactant: compressing a gas is the same as concentrating it. Temperature: hotter particles move faster, so they collide more often, and a larger fraction of collisions have energy above the minimum; the second effect is much the larger one, which is why a modest temperature rise can multiply a rate. Surface area: a solid reacts only at its surface, so a powder, with far more surface than a lump of the same mass, reacts faster. A catalyst is a substance that increases the rate without being used up, by providing an alternative pathway with a lower minimum energy so that a larger fraction of collisions succeed; enzymes in cells and the catalytic converter in a car exhaust are examples. Collision theory also explains why reactions between two solids are usually slow while the same substances react faster in solution or in the gas phase, where the particles are free to meet (free to meet does not guarantee fast: a reaction with a high activation energy is slow in any phase), and why a sugar cube dissolves faster when stirred or crushed. Dissolving rate is not solubility: stirring changes how fast sugar dissolves, not how much can dissolve. Mathematical rate laws are reserved for later study.

Potential energy diagrams. A reaction can be drawn as a graph of potential energy against the progress of the reaction. Reactants sit at the left at their energy, products at the right at theirs, and between them is a hump: the activated complex, the unstable arrangement partway between breaking the old bonds and forming the new ones. The height of the hump above the reactants is the activation energy, the minimum collision energy for reaction. If the products lie below the reactants the reaction releases energy and is exothermic; if above, it absorbs energy and is endothermic. A catalyst lowers the hump and leaves the start and end unchanged: it changes the rate, never the energy released. The diagram shows relative energies of reactants (50), activated complex (150 uncatalyzed, 110 catalyzed) and products (20) in arbitrary units, so the activation energy falls from 100 to 60 while the energy released stays 30.

Where the energy comes from. In a chemical system, potential energy is chemical energy stored in the arrangement of bonded atoms, and kinetic energy is thermal energy, the energy of the particles' motion. The total of either cannot be measured, but the change can. Energy must be supplied to break a bond, and energy is released when a bond forms. If the bonds formed in the products are, in total, stronger than the bonds broken in the reactants, the difference leaves the system and the reaction is exothermic; the surroundings warm up. If the bonds broken were stronger than those formed, energy must come in from the surroundings, the reaction is endothermic, and the surroundings cool. Ohio's standard describes molecules with weak bonds, such as ATP in cells, as less stable and tending to react to give more stable products, releasing energy. Read that with the rule above in mind: breaking the bond in ATP still requires energy, and energy is released overall only because the bonds and other interactions formed in the products of its reaction with water are stronger in total than those broken, so the products are more stable under the conditions in a cell. A student who says burning releases energy stored in the bonds of the fuel has the direction backwards: energy is released because stronger bonds form in carbon dioxide and water than were broken.

Bond energies as a prediction tool. A table of average bond energies, in kilojoules per mole of bonds, lets you estimate the energy change. Add the energies of all bonds broken (energy in), add the energies of all bonds formed (energy out), and subtract. For H₂ + Cl₂ → 2HCl, using table values H–H 436, Cl–Cl 243 and H–Cl 432 kJ/mol: broken 436 + 243 = 679 kJ; formed 2 × 432 = 864 kJ; change = 679 − 864 = −185 kJ per mole of reaction. The negative sign means net release, exothermic. For N₂ + 3H₂ → 2NH₃ with N≡N 946, H–H 436 and N–H 390: broken 946 + 3(436) = 2254; formed 6 × 390 = 2340; change −86 kJ, mildly exothermic. Because tabulated values are averages over many compounds, the result is an estimate, and different tables differ by a few kilojoules. Photosynthesis, 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, breaks strong C=O and O–H bonds and forms weaker C–C, C–H and O=O bonds overall, so it is endothermic and needs light; the combustion of sugar is the exact reverse and releases the same energy.

Calorimetry and specific heat. The energy change of a reaction or a process is measured by letting it warm or cool a known mass of water and applying conservation of energy: the energy lost by one part of the system equals the energy gained by the other. Thermal energy depends on the mass, the temperature and the substance. Specific heat, c, is the energy needed to raise the temperature of one gram of a substance by one degree Celsius. Water's is unusually high, 4.18 J/(g·°C), which is why water heats and cools slowly and helps regulate Earth's climate; copper's is 0.385 and iron's 0.449 J/(g·°C). The relation is q = m × c × ΔT, where q is the thermal energy transferred in joules, m the mass in grams and ΔT the change in temperature, final minus initial. Warming 50.0 g of water by 6.00 °C takes q = 50.0 × 4.18 × 6.00 = 1254 J, or 1.25 kJ to three significant figures. The equation can be rearranged to find any one of q, m, c, ΔT or a starting or final temperature.

A worked model: identify a metal by its specific heat. A 25.0 g piece of metal is heated to 100.0 °C in boiling water (a supervising adult handles the hot water) and dropped into 50.0 g of water at 22.0 °C in an insulated cup. The water warms to 25.4 °C. Energy gained by the water: q = 50.0 × 4.18 × (25.4 − 22.0) = 50.0 × 4.18 × 3.4 = 710.6 J. By conservation the metal lost 710.6 J while cooling from 100.0 to 25.4 °C, a change of 74.6 °C. Its specific heat is c = q / (m × ΔT) = 710.6 / (25.0 × 74.6) = 0.381 J/(g·°C), which is 0.38 to the two significant figures that the 3.4 °C temperature rise allows. That is close to copper's 0.385 and far from iron's 0.449, so the data support copper; they do not prove it, because other metals and alloys have similar specific heats. The method assumes no energy leaked to the cup or the air; a real cup loses some, which makes the calculated value a little low.

Investigation: rates and energy at the kitchen bench. Materials: effervescent antacid tablets, water at three temperatures (cold tap, room, warm tap, not above 50 °C), cups, a timer, a thermometer, baking soda, vinegar, calcium chloride ice-melt pellets from a hardware store, goggles. Part A, rate: drop one whole tablet into 100 mL of water at each temperature and time until fizzing stops; then compare a whole tablet with one crushed into powder in room-temperature water. Part B, energy: record the temperature of 50 mL of vinegar, add a spoon of baking soda and record the lowest temperature reached; record the temperature of 50 mL of water, add a spoon of calcium chloride, stir, and record the highest temperature. Record every time and temperature. A result that supports the claim: the tablet finishes fastest in the warmest water and the crushed tablet finishes faster than the whole one, matching collision theory; the vinegar mixture cools (endothermic) while the calcium chloride mixture warms (exothermic). What the model cannot show: timing the end of fizzing is a rough measure of rate, the temperature changes include energy exchanged with the cup and air, and calcium chloride dissolving is a process rather than a reaction between two compounds. Calcium chloride irritates eyes and skin; wear goggles, do not touch pellets with bare hands and wash up afterwards. No open flame is used.

Worked example 1 — explain a rate change with collisions. In a hypothetical trial, a strip of magnesium reacts with dilute hydrochloric acid in 90 seconds; the same mass of magnesium powder reacts in 15 seconds; the strip in warmer acid reacts in 40 seconds. Powder exposes far more surface, so more magnesium atoms are available for collisions with acid particles at any moment: faster. Warmer acid has particles moving faster, so collisions are more frequent and, more importantly, a larger fraction carry energy above the activation energy: faster. The amount of hydrogen produced is the same in all three because the same amount of magnesium reacted; only the rate changed.

Worked example 2 — read an energy diagram. A diagram shows reactants at 80 kJ, a peak at 200 kJ and products at 120 kJ (relative values). Activation energy: 200 − 80 = 120 kJ. Energy change: 120 − 80 = +40 kJ, positive, so the reaction is endothermic and the surroundings cool. With a catalyst the peak drops to 150 kJ: activation energy 70 kJ, faster reaction, but the energy change is still +40 kJ because the reactants and products are unchanged.

Worked example 3 — heat from a warm metal to cold water. How much energy is needed to warm 200.0 g of water from 18.0 °C to 65.0 °C? ΔT = 47.0 °C; q = 200.0 × 4.18 × 47.0 = 39 292 J = 39.3 kJ. Reversed: 39.3 kJ transferred into 1.00 kg of iron, specific heat 0.449, gives ΔT = q / (m c) = 39 292 / (1000 × 0.449) = 87.5 °C, so the same energy warms the iron almost twice as much as the water because iron's specific heat is lower.

Worked example 4 — exothermic or endothermic from bond energies. Is 2HCl → H₂ + Cl₂ exothermic or endothermic? Bonds broken: two H–Cl, 2 × 432 = 864 kJ. Bonds formed: H–H 436 and Cl–Cl 243, total 679 kJ. Change = 864 − 679 = +185 kJ: endothermic. It is exactly the reverse of the earlier reaction and the sign flips, as it must. The reactant bonds are stronger than the product bonds, so energy must come in.

Representation task: two diagrams and a table. Draw one potential energy diagram for an exothermic reaction and one for an endothermic reaction, each labeled with reactants, products, activated complex, activation energy and energy change, and add a dashed catalyzed pathway to the first. Then make a table with rows for concentration, temperature, pressure of a gas, surface area and catalyst, and columns for the effect on the number of collisions per second, the effect on the fraction of effective collisions, and one everyday example.

Independent performance: kitchen-bench rates and energy report. Carry out the investigation and write it up. Success criteria: a data table of times and temperatures with three trials for each condition; an explanation of each rate result using collision theory, naming frequency and energy of collisions; identification of each energy result as exothermic or endothermic with the direction of energy flow stated; a calorimetry calculation of the energy absorbed or released by the 50 mL of liquid in Part B using q = m c ΔT with the specific heat of water as an approximation, with units; a bond-energy calculation for one reaction using values your teacher supplies; and a statement of two sources of error in the temperature measurements.