Energetics and kinetics AS

Definition: Enthalpy change is the amount of heat energy taken in or given out during any change in a system provided the pressure is constant.In an exothermic change energy is transferred from the system (chemicals) to the surroundings. The products have less energy than the reactants.If an enthalpy change occurs then energy is transferred between system and surroundings. The system is the chemicals and the surroundings is everything outside the chemicals. In an endothermic change, energy is transferred from the surroundings to the system (chemicals). They require an input of heat energy e.g. thermal decomposition of calcium carbonate The products have more energy than the reactantsIn an exothermic reaction the ∆H is negative. Common oxidation exothermic processes are the combustion of fuels and the oxidation of carbohydrates such as glucose in respiration. In an endothermic reaction the ∆H is positive. Standard Enthalpy Change of Formation When an enthalpy change is measured at standard conditions the symbol is used Eg ∆H The standard enthalpy change of formation of a compound is the enthalpy change when 1 mole of the compound is formed from its elements under standard conditions (298K and 100kpa), all reactants and products being in their standard states. Symbol ∆fH Mg (s) + Cl2 (g)
MgCl2 (s) 2Fe (s) + 1.5 O2 (g)
Fe2O3 (s) The enthalpy of formation of an element = 0 kJ mol-1 Standard Enthalpy Change of Combustion The standard enthalpy of combustion of a substance is defined as the enthalpy change that occurs when one mole of a substance is combusted completely in oxygen under standard conditions. (298K and 100kPa), all reactants and products being in their standard states. Symbol ∆cH CH4 (g) + 2O2 (g)
CO2 (g) + 2 H2O (l) Incomplete combustion will lead to soot (carbon), carbon monoxide and water. It will be less exothermic than complete combustion. Enthalpy changes are normally quoted at standard conditions. Standard conditions are: • 100 kPa pressure • 298 K (room temperature or 25oC) • Solutions at 1mol dm-3 • all substances should have their normal state at 298K

. Calculating the enthalpy change of reaction, ∆Hr from experimental data General method 1. Using q= m x cp x ∆T calculate energy change for quantities used 2. Work out the moles of the reactants used. For a reaction in solution we use the following equation energy change = mass of solution x heat capacity x temperature change Q (J) = m (g) x cp (J g-1K-1) x ∆T ( K) This equation will only give the energy for the actual quantities used. Normally this value is converted into the energy change per mole of one of the reactants. (The enthalpy change of reaction, ∆Hr ) 3. Divide q by the number of moles of the reactant not in excess to give ∆H 4. Add a sign and unit (divide by a thousand to convert Jmol-1 to kJmol-1 The heat capacity of water is 4.18 J g-1K-1. In any reaction where the reactants are dissolved in water we assume that the heat capacity is the same as pure water. Also assume that the solutions have the density of water, which is 1g cm-3. Eg 25cm3 will weigh 25 g Example 1. Calculate the enthalpy change of reaction for the reaction where 25cm3 of 0.2 M copper sulphate was reacted with 0.01mol (excess of zinc). The temperature increased 7oC . Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x cp x ∆T Q = 25 x 4.18 x 7 Q = 731.5 J Step 2 : calculate the number of moles of the reactant not in excess. moles of CuSO4 = conc x vol = 0.2 x 25/1000 = 0.005 mol If you are not told what is in excess, then you need to work out the moles of both reactants and work out using the balanced equation which one is in excess. Step 3 : calculate the enthalpy change per mole which is often called ∆H (the enthalpy change of reaction) ∆H = Q/ no of moles = 731.5/0.005 = 146300 J mol-1 = 146 kJ mol-1 to 3 sf Finally add in the sign to represent the energy change: if temp increases the reaction is exothermic and is given a minus sign e.g. –146 kJ mol-1 Remember in these questions: sign, unit, 3 sig figs. Example 2. 25cm3 of 2M HCl was neutralised by 25cm3 of 2M NaOH. The Temperature increased 13.5oC What was the energy change per mole of HCl? Step 1: Calculate the energy change for the amount of reactants in the test tube. Q = m x cp x ∆T Q = 50 x 4.18 x13.5 Q = 2821.5 J Step 2 : calculate the number of moles of the HCl. moles of HCl = conc x vol = 2 x 25/1000 = 0. 05 mol Step 3 : calculate ∆H the enthalpy change per mole which might be called the enthalpy change of neutralisation ∆H = Q/ no of moles = 2821.5/0.05 = 564300 J mol-1 = -56.4 kJ mol-1 to 3 sf Exothermic and so is given a minus sign Remember in these questions: sign, unit, 3 sig figs. Note the mass is the mass of the copper sulphate solution only. Do not include mass of zinc powder. Note the mass equals the mass of acid + the mass of alkali, as they are both solutions. Example 3. Calculate the enthalpy change of combustion for the reaction where 0.65g of propan-1-ol was completely combusted and used to heat up 150g of water from 20.1 to 45.5oC Step 1: Calculate the energy change used to heat up the water. Q = m x cp x ∆T Q = 150 x 4.18 x 25.4 Q = 15925.8 J Step 2 : calculate the number of moles of alcohol combusted. moles of propan-1-ol = mass/ Mr = 0.65 / 60 = 0.01083 mol Step 3 : calculate the enthalpy change per mole which is called ∆Hc (the enthalpy change of combustion) ∆H = Q/ no of moles = 15925.8/0.01083 = 1470073 J mol-1 = 1470 kJ mol-1 to 3 sf Finally add in the sign to represent the energy change: if temp increases the reaction is exothermic and is given a minus sign eg –1470 kJ mol-1 Remember in these questions: sign, unit, 3 sig figs. Note the mass is the mass of water in the calorimeter and not the alcohol Enthalpies of combustion can be calculated by using calorimetry. Generally the fuel is burnt and the flame is used to heat up water in a metal cup.  Measuring Enthalpies of Combustion using Calorimetry

Hess’s law states that total enthalpy change for a reaction is independent of the route by which the chemical change takes place Hess’s law is a version of the first law of thermodynamics, which is that energy is always conserved. 2H (g) + 2Cl(g) H2 + Cl2 2HCl (g) a b ΔH On an energy level diagram the directions of the arrows can show the different routes a reaction can proceed by. In this example one route is arrow ‘a’ The second route is shown by arrows ΔH plus arrow ‘b’ So a = ΔH + b And rearranged ΔH = a – b H+ (g) + Br – (g) H+ (aq) + Br – (aq) H (g) + Br (g) HBr (g) a c d ΔH Interconnecting reactions can also be shown diagrammatically. In this example one route is arrow ‘a’ plus ΔH The second route is shown by arrows ‘c’ plus arrow ‘d’ So a+ ΔH = c + d And rearranged ΔH = c + d – aThis Hess’s law is used to work out the enthalpy change to form a hydrated salt from an anhydrous salt. This cannot be done experimentally because it is impossible to add the exact amount of water and it is not easy to measure the temperature change of a solid. Often Hess’s law cycles are used to measure the enthalpy change for a reaction that cannot be measured directly by experiments. Instead alternative reactions are carried out that can be measured experimentally. Instead both salts are dissolved in excess water to form a solution of copper sulphate. The temperature changes can be measured for these reactions. ∆H reaction = Σ ∆fH products – Σ ∆fH reactants Mean Bond energies Definition: The Mean bond energy is the enthalpy needed to break the covalent bond into gaseous atoms, averaged over different molecules These values are positive because energy is required to break a bond. The definition only applies when the substances start and end in the gaseous state. We use values of mean bond energies because every single bond in a compound has a slightly different bond energy. E.g. In CH4 there are 4 CH bonds. Breaking each one will require a different amount of energy. However, we use an average value for the C-H bond for all hydrocarbons. In general (if all substances are gases) ∆H = Σ bond energies broken – Σ bond energies made ∆H values calculated using this method will be less accuate than using formation or combustion data because the mean bond energies are not exact Reaction profile for an EXOTHERMIC reaction Reaction profile for an ENDOTHERMIC reaction. Enthalpies of Combustion in a Homologous Series When comparing the enthalpies of combustion for successive members of a homologous series such as alkanes or alcohols there is a constant rise in the size of the enthalpies of combustion as the number of carbon atoms increases. As one goes up the homologous series there is a constant amount and type of extra bonds being broken and made e.g. 1C-C, 2C-H and 1.5 O=O extra bonds broken and 2 C=O and 2 O-H extra bonds made, so the enthalpy of combustion increases by a constant amount. Mr of alcohol experimental calculated mol combustion kJ Enthalpy of -1 If the results are worked out experimentally using a calorimeter the experimental results will be much lower than the calculated ones because there will be significant heat loss. There will also be incomplete combustion which will lead to less energy being released. Remember that calculated values of enthalpy of combustions will be more accurate if calculated from enthalpy of formation data than if calculated from average bond enthalpies. This is because average bond enthalpy values are averaged values of the bond enthalpies from various compounds.

Errors in this method • Heat losses from calorimeter • Incomplete combustion of fuel • Incomplete transfer of heat • Evaporation of fuel after weighing • Heat capacity of calorimeter not included • Measurements not carried out under standard conditions as H2O is gas, not liquid, in this experiment. Calorimetric method
General method • washes the equipment (cup and pipettes etc) with the solutions to be used • dry the cup after washing • put polystyrene cup in a beaker for insulation and support • clamp thermometer into place making sure the thermometer bulb is immersed in liquid • measure the initial temperatures of the solution or both solutions if 2 are used • transfers reagents to cup. If a solid reagent is used then add the solution to the cup first and then add the solid weighed out on a balance. • stirs mixture • Measures final temperature One type of experiment is one in which substances are mixed in an insulated container and the temperature rise measured. This could be a solid dissolving or reacting in a solution or it could be two solutions reacting together • heat transfer from surroundings (usually loss) • approximation in specific heat capacity of solution. The method assumes all solutions have the heat capacity of water. • neglecting the specific heat capacity of the calorimeter- we ignore any heat absorbed by the apparatus. • reaction or dissolving may be incomplete or slow. • Density of solution is taken to be the same as water If the reaction is slow then the exact temperature rise can be difficult to obtain as cooling occurs simultaneously with the reaction To counteract this we take readings at regular time intervals and extrapolate the temperature curve/line back to the time the reactants were added together. We also take the temperature of the reactants for a few minutes before they are added together to get a better average temperature. If the two reactants are solutions then the temperature of both solutions need to be measured before addition and an average temperature is used.

The Activation Energy is defined as the minimum energy which particles need to collide to start a reaction Maxwell Boltzmann Distribution The Maxwell-Boltzmann energy distribution shows the spread of energies that molecules of a gas or liquid have at a particular temperature Learn this curve carefully The energy distribution should go through the origin because there are no molecules with no energy The energy distribution should never meet the x axis, as there is no maximum energy for molecules The mean energy of the particles is not at the peak of the curve The area under the curve represents the total number of A few have low particles present energies because collisions cause some particles to slow down Only a few particles have energy greater than the EA Most molecules have energies between the two extremes but the distribution is not symmetrical (normal) Q. How can a reaction go to completion if few particles have energy greater than Ea? A. Particles can gain energy through collisions Emp this is the most probable energy (not the same as mean energy) Increasing Temperature As the temperature increases the distribution shifts towards having more molecules with higher energies The total area under the curve should remain constant because the total number of particles is constant At higher temperatures the molecules have a wider range of energies than at lower temperatures. At higher temps both the Emp and mean energy shift to high energy values although the number of molecules with those energies decrease 1.5 Kinetics Ea Collision energy
Fraction of molecules with energy higher temperature Lower temperature Ea Collision energy
Fraction of molecules with energy mean Collision energy
T= 25OC T= 1067OC Fraction of molecules with energy reactants products Activation Energy: EA ∆H Progress of Reaction Energy Reactions can only occur when collisions take place between particles having sufficient energy. The energy is usually needed to break the relevant bonds in one or either of the reactant molecules. This minimum energy is called the Activation Energy 1 Emp N Goalby chemrevise.org Measuring Reaction Rates The rate of reaction is defined as the change in concentration of a substance in unit time Its usual unit is mol dm-3s -1 When a graph of concentration of reactant is plotted vs time, the gradient of the curve is the rate of reaction. The initial rate is the rate at the start of the reaction where it is fastest Reaction rates can be calculated from graphs of concentration of reactants or products Effect of Increasing Concentration and Increasing Pressure At higher concentrations(and pressures) there are more particles per unit volume and so the particles collide with a greater frequency and there will be a higher frequency of effective collisions. If concentration increases, the shape of the energy distribution curves do not change (i.e. the peak is at the same energy) so the Emp and mean energy do not change They curves will be higher, and the area under the curves will be greater because there are more particles More molecules have energy > EA (although not a greater proportion) Comparing rate curves Need to calculate/ compare initial moles of reactants to distinguish between different finishing volumes. e.g. the amount of product is proportional to the moles of reactant Different volumes of the same initial concentrations will have the same initial rate (if other conditions are the same) but will end at different amounts The higher the concentration/ temperature/ surface area the faster the rate (steeper the gradient) In the experiment between sodium thiosulphate and hydrochloric acid we usually measure reaction rate as 1/time where the time is the time taken for a cross placed underneath the reaction mixture to disappear due to the cloudiness of the Sulphur . Na2S2O3 + 2HCl
2NaCl + SO2 + S + H2O This is an approximation for rate of reaction as it does not include concentration. We can use this because we can assume the amount of Sulphur produced is fixed and constant. Higher concentration Ea Collision energy
Number of molecules with energy lower concentration Initial rate = gradient of tangent time concentration Amount of product e.g. Volume of gas Time (secs) A B C D Note: If a question mentions a doubling of concentration/rate then make sure you mention double the number of particles per unit volume and double the frequency of effective collisions. N Goalby chemrevise.org 2 Effect of Increasing Surface area Increasing surface area will cause successful collisions to occur more frequently between the reactant particles and this increases the rate of the reaction. Effect of Catalysts Definition: Catalysts increase reaction rates without getting used up. Explanation: They do this by providing an alternative route or mechanism with a lower activation energy Comparison of the activation energies for an uncatalysed reaction and for the same reaction with a catalyst present. If the activation energy is lower, more particles will have energy > EA, so there will be a higher frequency of effective collisions. The reaction will be faster Effect of Increasing Temperature At higher temperatures the energy of the particles increases. They collide more frequently and more often with energy greater than the activation energy. More collisions result in a reaction As the temperature increases, the graph shows that a significantly bigger proportion of particles have energy greater than the activation energy, so the frequency of successful collisions increases higher temperature Lower temperature Ea Collision energy
Fraction of molecules with energy Ea catalysed Fraction of molecules with energy Collision energy
Ea un catalysed With a lower activation energy more particles have energy greater than the activation energy reactants Activation Energy: uncatalysed ∆H Progress of Reaction EA catalysed 3 products

A-level Chemistry exemplar for required practical No. 3 Investigation of how the rate of a reaction changes with temperature: To investigate how the rate of the reaction of sodium thiosulfate with hydrochloric acid changes as the temperature of the reaction is changed. Introduction Sodium thiosulfate reacts with hydrochloric acid according to the equation Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + SO2(g) + S(s) The reaction produces a precipitate of sulfur. The rate of this reaction can be monitored by measuring the time taken for a fixed amount of sulfur to be produced. An easy method to do this is by timing how long it takes for a cross, marked on the bottom of the reaction vessel, to disappear as it is obscured by the sulfur precipitate. Dilute hydrochloric acid will be added to sodium thiosulfate solution at different temperatures in a series of experiments. This table shows the approximate temperatures for five experiments. Experiment 1 2 3 4 5 Approximate temperature / °C room* ~25 ~35 ~45 ~55** [* The temperature of the room is likely to be 15 to 18 °C] [** The temperature must not exceed 55 °C] It is not necessary for these exact temperatures to be used although the temperature used must not exceed 55 °C. However, the actual temperature at which each experiment is carried out must be known as accurately as possible. One way that this can be achieved is to measure both the initial temperature and the final temperature and then use a mean temperature when plotting your graph. Requirements In addition to general laboratory apparatus, each student needs:  thermometer (–10 °C to 110 °C)  400 cm3 beaker (for use as a water bath)  plastic container with lid  2 glass tubes to hold 12–14 cm3 of liquid  0.05 mol dm–3 (or 40 g dm –3 ) sodium thiosulfate solution  1.0 mol dm–3 hydrochloric acid (or 0.5 mol dm–3 sulfuric acid)  10 cm3 measuring cylinder  plastic graduated pipette  stopwatch  graph paper. A lid is advised in this experiment. Two holes should be made in the lid using a hot wide cork borer. These holes should securely hold the glass tubes and vertically in the plastic container. A cross should be marked on the inside base of the plastic container below one of the larger holes using a permanent black marker pen. Alternatives include using clear A4 plastic wallets to use as a lid over the plastic container. Also, rather than marking the bottom of the plastic container directly, a laminated sheet marked with a black cross could be used Caution: the CLEAPSS Hazcard states ‘sulfur dioxide is produced in this reaction’ and ‘known sufferers of asthma should be closely monitored’. Centres are advised to ensure that the investigation is carried out in a well-ventilated room and that appropriate measures are taken to dispose of waste solutions. Stop baths – containers of sodium carbonate solution and phenolphthalein should be available to students so that the acid and sulfur dioxide can be neutralised (immediately, if required, during the practical and) after the experiment has finished. Once the colour of the solution in the stop bath changes, the sodium carbonate has been used up and the stop bath will need to be replenished. The stop bath should be placed in a fume cupboard, if available. Analysing the data In these experiments at different temperatures, the concentrations of all the reactants are the same. You are investigating the time taken to produce the same amount of sulfur at different temperatures. If you were to plot a graph of the amount of sulfur produced against time, it would initially be a straight line because the reaction has only just started. Therefore, the initial rate of reaction = (amount of sulfur)/time so the initial rate of reaction is proportional to 1/time ( 1 ). AS analysis  calculate the mean temperature of each reaction mixture  for each of the five temperatures, calculate 1 to 3 significant figures, where t is the time taken for the cross to be obscured  plot a graph of 1 on the y-axis against average temperature  the plotting of the points may be more straightforward if you multiply all of the values for 1 t by a common factor (eg 104 ).