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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Rate of Reaction University of Lincoln presentation

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

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Why the difference? Is it the enthalpy change (Heat of combustion) ? Paraffin wax 42 MJ kg -1 Petrol 45 MJ kg -1 Is it the temperature? Yellow/white – 1300 o C Pale orange/yellow – 1100 o C What is it then?

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Approximately how long will a 2 litre pool of petrol burn for? Important values: Petrol density = 0.8 kg litre -1 Heat of combustion is 45 MJ kg -1 2 litres of petrol has a mass of 1.6 kg (from the density) Total energy available from 1.6 kg petrol = 1.6 kg x 45 MJ kg -1 = 72 MJ 2 litre petrol pool is a 1 MW fire (this is a measured value) 1 MW = 1 MJ s -1 so at this rate it would take 72 s

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License How do ignitable liquids burn?

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License 2 litre petrol bomb takes about 10s to burn. What is the rate of heat release? 72 MJ in 10 s = 7.2 MW 2 litre petrol fully evaporated takes about 1 s to burn. What is the rate of heat release? 72 MJ in 1s = 72 MW Conclusion: Same total energy available but released at a faster rate

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License How long to burn a 1.6 kg candle? 1.6 kg paraffin wax at 42 MJ kg -1 can release 67.2 MJ Candle flame has a heat release rate of 80 W (80 Js -1 )

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License A candle bomb? NASA are researching the paraffin rocket!! How can this work? Increase rate of combustion –Increase concentration of the oxidant; use 100% oxygen –Paraffin as small liquid droplets Study of the rates of reaction - kinetics

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Factors affecting the rate of a chemical reaction 1.Concentration ( hydrogen peroxide demo) 2.Pressure (gases) 3.Temperature (glowstick) 4.Surface area (dust explosion) 5.Catalysis

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Measuring Reaction Rate Use a characteristic of the products or the reactants that can be used as a measure of amount. –Volume of gas –Change in mass –Absorption of light rate of decrease of reactant or rate of increase of a product

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License H 2 O 2(l) H 2 O (l) + ½O 2(g)

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Calculating Rate of Reaction The gradient of tangent to the curve is the rate of reaction What happens to the reaction rate with time? Concentration = 0.3 mol dm -3 s -1 Rate = gradient = 0.0068 mol dm -3 s -1 Concentration = 0.1 mol dm -3 s -1 Rate = gradient = 0.0023 mol dm -3 s -1

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License A Mathematical Relationship Select two other points on the curve and calculate the rate of reaction at that concentration of H 2 O 2 Plot a graph of Rate of Reaction as a function of H 2 O 2 concentration

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License

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What does the graph show? Graph is a straight line through the origin The two variables have a linear mathematical relationship We can say: Rate of Reaction is directly proportional to Hydrogen Peroxide concentration − Easy to predict what happens to reaction when [H 2 O 2 ] is changed − [H 2 O 2 ] x2 Rate x 2 − First Order with respect to H 2 O 2 − k is the rate constant; first order reaction has units of s -1 when the rate of reaction is measured in mol dm -3 s -1. Show this by rearranging the rate equation and why are the units of rate important.

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Rate of reaction can be measured from the rate that oxygen gas is produced. Yeast suspension +hydrogen peroxide solution Inverted burette Water

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Vary the starting concentration and measure the initial rate [H 2 O 2 ]= 0.40 mol dm-3 [H 2 O 2 ]= 0.32 mol dm-3 [H 2 O 2 ]= 0.24 mol dm-3 [H 2 O 2 ]= 0.16 mol dm-3 [H 2 O 2 ]= 0.08 mol dm-3

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Initial Rate can be measured Initial gradient 0.51cm 3 s -1

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Plot Initial Rate as a function of starting concentration

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Summary –Decomposition of H 2 O 2 can be followed by measuring the decrease in H 2 O 2 concentration or the volume of O 2 evolved. –Rate of reaction can be calculated from the progress curve at different times or initial rate measurements. –Plots of rate as a function of reagent concentration can be used to determine the mathematical relationship –Order of reaction can be determined –Rate equation can be written

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License For you to do: Initial rate data [H 2 O 2 ]/mol dm -3 Rate/cm 3 O 2 s -1 0.080.1 0.160.215 0.240.32 0.41 0.40.51 Determine the order of reaction with respect to hydrogen peroxide and calculate the value of the rate constant.

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License General Rate equations Zero order Units of k ? First order Units of k ? Second order Units of k ?

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Further Analysis of data Logarithms can be very useful Plot of log rate as a function of log concentration (p439 Housecroft) Gradient is n; Intercept is log k Use this method on the initial rate data in slide 21 to determine order and the value of k

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Half-life Time taken for the concentration of reactant A at time t, [A] t to fall to half its value. A constant half-life for a first order reaction Progress curve and measure t½ at several different points.

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Constant half-life 27s 26s Going from 200 x 10 -5 mol to 100 x 10 -5 mol takes 27s Going from 100 x 10 -5 mol to 50 x 10 -5 mol takes 27s Going from 50 x 10 -5 mol to 25 x 10 -5 mol takes 26s

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Using Luminol to detect blood stains Exponential decay curve First order write rate equation Calculate half-life and why is it important Video clip or demo

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Reactions with more than one reactant A+B → products e.g. C 12 H 22 O 11 + H 2 O → C 6 H 12 O 6 + C 6 H 12 O 6 sucrose glucose fructose First order with respect to each reactant Second order reaction (sum of orders in rate equation)

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Determining order and rate equations Difficulties with more than one reactant? Initial rate method Isolation method Experimental Design Principle Vary one concentration and keep other(s) constant while measuring rate.

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License An Example Reaction peroxodisulfate (VI) and iodide ions Design the experiment 1. initial rate method (vary each concentration) 2.Plot a graph of log rate as a function of log initial concentration for each reactant. Gradient of each line is order of reaction for each reactant. 3.k is determined by rearranging the rate equation. Task: Determine the rate equation and a value for k

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Iodine clock data from experiment

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Collision theory Molecules have to collide if they are to react – increasing frequency of collisions? Increasing concentration increases the frequency of collisions Increasing pressure increases frequency of collisions Increasing temperature increases frequency of collision But not just about rate of collisions – how do we explain slow reactions?

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Activation Energy (Enthalpy) E a Energy of the collision must be above a certain value for reactants to react Why? Energy is needed to break bonds (remember bond enthalpies) This then creates reactive species to make new bonds The minimum energy required for a collision to result in chemical reaction is Ea

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Only those molecules with sufficient energy can react Number of molecules with kinetic energy E Kinetic energy (E) Activation enthalpy E a =50kJ mol -1

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Increasing Temperature increases Rate of Reaction Number of molecules with kinetic energy E Kinetic energy (E) Activation enthalpy E a =50kJ mol -1 Number of molecules with energy greater than 50kJ mol -1 at 300 K Number of molecules with energy greater than 50kJ mol -1 at 310 K 300 K 310 K

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Back to petrol Petrol vapour reacts with oxygen (air) But not spontaneous at room temperature Needs ignition. What does ignition do? –Provides energy to break bonds (endothermic) –Creates reactive species (free radicals) –Self-sustaining (can remove ignition source and it carries on). Why???? –Energy released from the reaction breaks more bonds and the reaction continues

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Combining Activation Energy and enthalpy Draw a diagram for an endothermic reaction Both can be shown on an enthalpy level diagram Add E a to the diagram A+B C+D ΔHΔH Positive enthalpy Negative enthalpy Reaction coordinate Exothermic reaction

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Rate equations and Temperature k is the rate constant; A is the pre-exponential factor; E a is the activation energy; R is the molar gas constant (8.314 J mol -1 K -1 ); T is the absolute temperature (Kelvin). increase temperatureincrease k increase rate decrease E a increase kincrease rate The Arrhenius equation How does it work? It might be easier to do this

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License The well known ‘rule of thumb’ Reaction rate doubles if temperature is increased by 10 o C Temp/KE a /kJ mol -1 A/L mol -1 s -1 k/L mol -1 s -1 313548.7 x 10 6 8.5 x 10 -3 323548.7 x 10 6 1.6 x 10 -2 Check the values of k by calculating them from the Arrhenius equation using the other values in the table Calculate k at 333 K. What is happening to the value of k? How will this affect the rate of this reaction?

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License An experiment to determine E a Determine order and rate equation for the reaction Measure the rate of reaction at different temperatures keeping the initial concentrations the same Calculate k at the different temperatures Plot lnk against 1/T: gradient = -E A /R; intercept = lnA

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Calculating Ea Temperature/Kk/dm 3 mol -1 s -1 2962.9 x 10 -3 3024.2 x 10 -3 3138.3 x 10 -3 3231.9 x 10 -2 Use the data below to calculate a value for the activation energy for this reaction

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License How do we explain catalysis?

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License What are catalysts? Definition and some examples; reactions and catalysts Hydrogen peroxide, metals and natural substances Enzymes Gases on metal surfaces What is a different reaction route?

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License A catalyst provides an alternative path for the reaction with a lower activation enthalpy Enthalpy Progress of reaction Reactants Products Activation enthalpy of catalysed reaction Activation enthalpy of uncatalysed reaction Uncatalysed reaction Catalysed reaction

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This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 2.0 UK: England & Wales License Acknowledgements JISC HEA Centre for Educational Research and Development School of natural and applied sciences School of Journalism SirenFM http://tango.freedesktop.org

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