Kinetics What are the factors affecting the rates of reactions in making new products? Why is it important to determine the rates of reactions? How does our knowledge of rates of reactions influence our ability to make wise decisions for safe products?

Rate of Reactions The speed at which reactants are used up or products are formed.

Rates of Reaction Really Quick – Firework Exploding What is meant by the rate of a chemical reaction. How to measure the rate of reaction. What factors affect the rate of reaction. Rates of Reaction Really Quick – Firework Exploding Really Slow - Rusting

Why are kinetics important? In order to control processes. speed up useful reactions that occur to slowly slow down reactions that are harmful Example: Catalysts are used in our cars to rapidly convert toxic substances into safer substances Refrigerators are used to slow the process of spoiling in food

How do we know a reaction has taken place? In most cases, there will be observable changes : the mass change per unit time (e.g. mass of product formed or reactant used up) the volume of gas given off per unit time during the reaction the change in temperature the formation of precipitate the change in pH the change in colour

Rates of reaction A measure of the “speed” of reactions. The rate of reaction is the speed at which reactants are used up or products are formed. How do we measure the rate of reactions?

Measuring Rate How could you measure the rate of these reactions? Use your definition to help you.

How do you measure the rate of reaction? Amount of product obtained per unit time. Speed of reaction = ________________ obtained time taken Amount of gas produced per unit time. = ________________ produced Mass of product Volume of gas

How do you measure the rate of reaction? Amount of reactant used up per unit time. Rate of reaction = ________________ used up time taken Mass of reactant

Example : Reaction that produces gases Reaction of magnesium strip with hydrochloric acid Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g)

Plot this graph of volume of gas produced against time Time taken(min) 1 2 3 4 5 6 7 8 9 Vol of Gas porduced(cm3) 17 25 30 34 35

Rate of Reaction at 4 minutes = (40-26)/(5.7-2.6) = 1.29 cm3/min

Region 1 (the 1st minute) What can say about the nature of the curve? What does this show? The rate of reaction is the ________. More reacting particles are present and the number of effective collisions per second is ___________. The slope of the gradient is the steepest fastest highest

Region 2 (1st – 4.5th minute) What does this show? The rate of reaction is ________ than in region 1. Fewer reacting particles are present and so the number of effective collisions per second will be ________. The gradient of the slope is less steep slower lesser

Region 3 (after 4.5th minute) What does this show? The reaction has ________ At least one of reactants has been _______ Hence no further ________ can occur between the two reactants. The graph is horizontal. There is no gradient. stopped used up collision

Amount of reactants used up against time CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l) Time taken(s) 10 20 30 40 50 60 70 80 90 Loss in Mass(g) 0.000 0.056 0.098 0.129 0.158 0.165

Mass loss(g) against time(min) What can you infer from the graph? The graph shows a decreasing curve As the time increases, the mass of calcium carbonate decreases as it is gradually used up in the reaction. Rate of loss of mass at time, t = g min-1 Mass of CaCO3(g) A 0.165 B C Time(min) CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)

Concentration of reactants against time Catalysed decomposition of hydrogen peroxide H2O2(l)  O2(g) + H2O(l) Time taken(min) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 Conc of H2O2(moldm-3) 0.159 0.129 0.101 0.080 0.063 0.050 0.040 0.035 0.025 0.020

Concentration(moldm-3) against time(min) What can you infer from the graph? Conc of H2O2(moldm-3) The graph shows a decreasing curve As the time increases, the concentration of hydrogen peroxide decreases as it decomposes. Rate of decompostion at time, t = mol dm-3 min-3 P 0.060 Q R Time(min) H2O2(l)  O2(g) + H2O(l)

Concentration(moldm-3) against time(min) Gradient of tangents, Rate Conc of H2O2(moldm-3) Conc of H2O2 (moldm-3) 0.16 0.12 0.08 0.04 Rate (moldm-3min -1) 0.076 0.049 0.033 0.019 [A]0 [A]1 [A]2 [A]3 [A]4 Time(min) [A]4 [A]3 [A]2 [A]1 [A]0 H2O2(l)  O2(g) + H2O(l) Conc of H2O2(moldm-3)

Rate of Decomposition against Concentration Draw the line of best fit The rate of reaction is proportional to the concentration of dinitrogen pentoxide

Not all collisions are successful in producing a reaction. Only a small fraction will result in a reaction = EFFECTIVE COLLISIONS All colliding particles need a minimum amount of energy for a reaction to occur = ACTIVATION ENERGY

Collision Theory How do reactions occur at the molecular level? Molecules collide with each other Form activated complex reactants molecules Must have sufficient energy Energy ≥ Activation energy (Ea - energy to break bonds) Right orientation (Animation)

Concentration If the concentration of the reactants is increased, there are more particles present in the same volume so that they are closer together. More likely to meet and react. Higher concentration Number of collisions increased In particular, doubling the conc of one of the reactants will double the rate of reaction. This explain why the greatest rate of reaction occurs as soon as the reactants are mixed, when they are at the highest conc. As the cnc of the chemicals decrease and the rate of reaction decreases as there are fewer collisions. Higher chance of successful collisions in a given amount of time  rate will increase

Pressure With __________ reactions, increasing pressure forces the particles to be closer together. This will increase the ______________ of the reacting particles. Hence, the rate will increase. gaseous concentration

Particle Size smaller Large lump of solid has _________surface area. Reaction can occur at the outside surface. Smaller lumps of solid have a ________surface area at which the reaction can happen. The rate is much faster. Increased surface area Number of collisions increased Animation larger

Temperature At a higher temperature, there is a higher rate of reaction as the particles are moving around with greater _____________ and ________ more frequently with energy greater than the activation energy. kinetic energy collide

Temperature Increased number of collisions per unit time More molecules have enough activation energy Maxwell-Boltzmann energy distribution curve Increased temperature, distribution flattens out More molecules have Ea Does activation energy change with temperature? Number of molecules At higher temp, the peak of the curve moves to the right so the most likely value of KE for the molecules increase. The curve flattens so the total area under it and therefore the total number of molecules remain constant. The area under the curve to the right of the activation energy Ea increases. This means at higher temp, a greater % of molecules have energies = or in excess of the Ea Kinetic energy

Light The rates of some reactions are increased by exposure to light. Example, photosensitive chemicals undergo partial decomposition in the presence of sunlight. 2AgX(s)  2Ag(s) + X2(g), X represents a halogen 2AgNO3(s)  2Ag(s) + 2NO2(g) + O2(g) 2H2O2(l)  O2(g) + 2H2O(l) 4HNO3(aq)  2H2O(l) + 4NO2(g) + O2(g)

Mixtures of hydrogen and bromine, or methane and chlorine react rapidly only in the presence of light. H2(g) + Br2(g)  2HBr(g) CH4 (g) + Cl2(g)  CH3Cl (g) + HCl(g)

Catalysts Substance that speeds up the rate of reaction but without being used up (remain chemically unchanged) . Hold molecules in the right _____________ for more successful collision per unit time. The activation energy is ________ by the catalyst. orientation lowered

Importance of catalysts Decomposition of hydrogen peroxide where manganese (IV) oxide is used to speed up the reaction. H2O2  H2O + O2 Manufacture of sulfur trioxide where vanadium(V)oxide is used to make sulfuric acid (Contact Process) 2SO2(g) + O2(g)  2SO3(g) Manufacture of ammonia (Haber Process) at a high pressure, 250 atmospheres at medium temperature, 4500C with the help of porous iron catalyst by reducing magnetite Fe3O4 N2(g) + 3H2 (g) 2NH3(g)

Catalysts hold molecules in right orientation Homogeneous catalyst (same physical state as the reactants - often both in solution) Catalysis by Co2+ Heterogeneous catalyst (different physical state from the reactants - often the reactants are gases and catalyst is a solid)

Biological catalyst - Enzymes Enzymes are present in all living organisms. E.g. Catalase found in blood, potato peeling can decompose hydrogen peroxide, H2O2. They are made of proteins Highly specific in their actions – act on 1 substance only. E.g. Amylase breaks down starch. Pepsin breaks down proteins. Can be made inactive by heating Works best between 35 – 400C.

The rate expression Rate α concentration of solution A Rate α [A] Rate = k [A] Rate expression is experimentally determined. xA + yB  C + D Rate = k[A]m[B]n Overall order = m + n

Example 1 Consider the reaction 2A  B Determine the order with respect to A the rate of reaction value of rate of reaction (with units) the rate of reaction when [A]=1.3moldm-3 Expt [A] /moldm-3 Rate / /moldm-3 1 1.0 0.60 2 2.0 1.2 3 5.0 3.0

Example 2 Consider the reaction 3A + B  C + D Determine the order with respect to A (b) order with respect to B the overall rate of reaction (d) the rate expression (e) The value of the rate constant (with units) the rate of reaction when [A]=1.60moldm-3 and [B]=0.30moldm-3 Expt [A] /moldm-3 [B] /moldm-3 Rate / /moldm-3 1 0.10 0.50 2 0.30 4.50 3 0.20

Example 3 Given that 2P + Q  R + S and the data determine the order with respect to P the order with respect to Q the overall rate of reaction the rate expression the value of the rate constant (with units) Expt [P]/moldm-3 [Q]/ moldm-3 Rate/moldm-3 s-1 1 1.20 2.00 5.00 x 10-3 2 2.40 1.00 x 10-2 3 6.00 8.00 0.100