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Reaction Rates & Equilibrium. Unit 9 *Start with visual from Phet. In order for a reaction to take place, the reacting molecules must collide into each other. Once molecules collide they may react together or they may not, depending on two factors -
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Reaction Rates & Equilibrium Unit 9 *Start with visual from Phet
In order for a reaction to take place, the reacting molecules must collide into each other. • Once molecules collide they may react together or they may not, depending on two factors - • Whether the collision has enough energy to "break the bonds holding reactant molecules together"; • Whether the reacting molecules collide in the proper orientation for new bonds to form. Collision Theory of Kinetics
Collisions in which these two conditions are met (and therefore the reaction occurs) are called effective collisions. • The higher the frequency of effective collisions the faster the reaction rate. • When two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex • It is a transition state between reactant and product • It has a very short lifetime (10-13 s) • Has to form for product to be formed Effective Collisions
The difference in potential energy between the reactant molecules and the activated complex is called the activation energy, Ea • This is the minimum amount of energy that particles must have in order to react. • The larger the activation energy, the slower the reaction • The energy to overcome the activation energy comes from the kinetic energy of the collision being converted into potential energy, or from energy available in the environment, i.e. heat. • Different reactions have different activated complexes and therefore different activation energies Activated Complex
Energy of products is lower than energy of reactants • energy lost, exothermic, -∆H • Energy of products is higher than energy of reactants • energy gained, endothermic, +∆H Energy Diagram
Nature of the Reactants • Cl2(g) + CH4(g) CH3Cl(g) + HCl(g) • Cl2 Cl + Cl (fast) • Cl + CH4 CH3Cl + H (slow) • H + Cl HCl (very fast) • individual steps = elementary steps • all steps together = reaction mechanism • the slowest step determines the rate of the reaction • called the rate determining step • eliminating the intermediates allows us write the balanced equation of the mechanism • Intermediates – product in one step, reactant in another Factors Affecting Reaction Rate
2) INCREASING CONCENTRATION Increasing concentration = more frequent collisions = increased rate of reaction Low concentration = fewer collisions Higher concentration = more collisions However, increasing the concentration of some reactants can have a greater effect than increasing others
3) INCREASING SURFACE AREA • Increasing surface area increases chances of a collision - more particles are exposed • Powdered solids react quicker than larger lumps • Catalysts (e.g. in catalytic converters) are in a finely divided form for this reason • + • In many organic reactions there are two liquid layers, one aqueous, the other non-aqueous. Shaking the mixture improves the reaction rate as an emulsion is often formed and the area of the boundary layers is increased giving more collisions. 1 1 CUT THE SHAPE INTO SMALLER PIECES 1 1 3 3 SURFACE AREA 9+9+3+3+3+3 = 30 sq units SURFACE AREA 9 x (1+1+1+1+1+1) = 54 sq units
4. Agitation • this puts more liquid/gas particles in contact with the solid = ↑ collisions = ↑ act. complex = ↑ product Factors Affecting Reaction Rate
5) INCREASING THE PRESSURE • increasing the pressure forces gas particles closer together • this increases the frequency of collisions so the reaction rate increases • many industrial processes occur at high pressure to increase the rate... but • it can adversely affect the position of equilibrium and yield • The more particles there are in a given volume, the greater the pressure • The greater the pressure, the more frequent the collisions • The more frequent the collisions, the greater the chance of a reaction
6. Temperature • most effective at speeding up a reaction • ↑ temp. = ↑ KE (particles moving faster) • particles move faster leading to more collisions • the collisions are also harder • these harder collisions contain the needed energy to overcome the Ea • therefore the reaction rate will increase Factors Affecting Reaction Rate
INCREASING TEMPERATURE According to KINETIC THEORY, all particles must have energy; the greater their temperature, the more energy they possess. The greater their KINETIC ENERGY the faster they travel. ZARTMANN heated tin in an oven and directed the gaseous atoms at a rotating disc with a slit in it. Any atoms which went through the slit hit the second disc and solidified on it. Zartmann found that the deposit was spread out and was not the same thickness throughout. This proved that there was a spread of velocities and the distribution was uneven. ZARTMANN’S EXPERIMENT
INCREASING TEMPERATURE MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULARENERGY MOLECULAR ENERGY Because of the many collisions taking place between molecules, there is a spread of molecular energies and velocities. This has been demonstrated by experiment. It indicated that ... no particles have zero energy/velocity some have very low and some have very high energies/velocities most have intermediate velocities.
INCREASING TEMPERATURE MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY T1 NUMBER OF MOLECUES WITH A PARTICULARENERGY T2 TEMPERATURE T2 > T1 MOLECULAR ENERGY • Increasing the temperature alters the distribution • get a shift to higher energies/velocities • curve gets broader and flatter due to the greater spread of values • area under the curve stays constant - it corresponds to the total number of particles
INCREASING TEMPERATURE T3 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY T1 NUMBER OF MOLECUES WITH A PARTICULARENERGY TEMPERATURE T1 > T3 MOLECULAR ENERGY • Decreasing the temperature alters the distribution • get a shift to lower energies/velocities • curve gets narrower and more pointed due to the smaller spread of values • area under the curve stays constant - it corresponds to the total number of particles
INCREASING TEMPERATURE T3 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY T1 NUMBER OF MOLECUES WITH A PARTICULARENERGY T2 TEMPERATURE T2 > T1 > T3 MOLECULAR ENERGY REVIEW no particles have zero energy/velocity some particles have very low and some have very high energies/velocities most have intermediate velocities as the temperature increases the curves flatten, broaden and shift to higher energies
INCREASING TEMPERATURE MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULARENERGY NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER Ea MOLECULAR ENERGY ACTIVATION ENERGY - Ea The Activation Energy is the minimum energy required for a reaction to take place The area under the curve beyond Ea corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react.
INCREASING TEMPERATURE TEMPERATURE T2 > T1 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY T1 T2 NUMBER OF MOLECUES WITH A PARTICULARENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER Ea MOLECULAR ENERGY Explanation increasing the temperature gives more particles an energy greater than Ea more reactants are able to overcome the energy barrier and form products a small rise in temperature can lead to a large increase in rate
7. Catalyst – Lowers Activation energy! • substance that speeds up a reaction, but isn’t used up in the reaction • provides a “different pathway” that requires lower Ea • lower Ea = more collisions having the proper amount of energy = ↑ act. complex = ↑ product Factors Affecting Reaction Rate
ADDING A CATALYST • Catalysts provide an alternative reaction pathway with a lower Activation Energy (Ea) • Decreasing the Activation Energy means that more particles will have sufficient • energy to overcome the energy barrier and react • Catalysts remain chemically unchanged at the end of the reaction. WITHOUT A CATALYST WITH A CATALYST
ADDING A CATALYST MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULARENERGY NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER Ea MOLECULAR ENERGY The area under the curve beyond Ea corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react. If a catalyst is added, the Activation Energy is lowered - Ea will move to the left.
ADDING A CATALYST MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULARENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER Ea MOLECULAR ENERGY The area under the curve beyond Ea corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react. Lowering the Activation Energy, Ea, results in a greater area under the curveafterEashowing that more molecules have energies in excess of the Activation Energy
CATALYSTS - A REVIEW • work by providing an alternative reaction pathway with a lower Activation Energy • using catalysts avoids the need to supply extra heat - safer and cheaper • catalysts remain chemically unchanged at the end of the reaction. • Types Homogeneous CatalystsHeterogeneous Catalysts • same phase as reactants different phase to reactants • e.g. CFC’s and ozone e.g. Fe in Haber process • Uses used in industry especially where an increase in temperature results in • a lower yield due to a shift in equilibrium (Haber and Contact Processes)
CATALYSTS - A REVIEW • work by providing an alternative reaction pathway with a lower Activation Energy • using catalysts avoids the need to supply extra heat - safer and cheaper • catalysts remain chemically unchanged at the end of the reaction. • Types Homogeneous Catalysts Heterogeneous Catalysts • same phase as reactants different phase to reactants • e.g. CFC’s and ozone e.g. Fe in Haber process • Uses used in industry especially where an increase in temperature results in • a lower yield due to a shift in equilibrium (Haber and Contact Processes) • CATALYSTS DO NOT AFFECT THE POSITION OF ANY EQUILIBRIUM • but they do affect the rate at which equilibrium is attained • a lot is spent on research into more effective catalysts - the savings can be dramatic • catalysts need to be changed regularly as they get ‘poisoned’ by other chemicals • catalysts are used in a finely divided state to increase the surface area
Reaction rate – how fast reactants disappear and how fast product appears Reaction Rate
NOTE: The rate of chemical reactions CAN ONLY be found by experimentation. There are no theoretical or mathematical ways to calculate the rate of a reaction. IT IS A PURELY EXPERIMENTAL SCIENCE Reaction Rate
Reaction Rate = ∆ [A] ∆ t • Example: CO(g) + NO2(g) CO2(g) + NO(g) - at t = 4.0 min, [CO2] = .12 mol◦dm-3 (mol/L) • at t = 8.0 min, [CO2] = .24 mol◦dm-3 • reaction rate = .24 mol/L - .12 mol/L 8.0 sec – 4.0 sec = 0.030 mol◦dm-3 ◦sec-1 (mol/L sec) • Unit for reaction rate = conc. with some time unit • Products have a (+) rate • Reactants have a (-) rate • The SI unit of rate of reaction is mol/dm3, • (moles per dm3 or moles per liter.) Reaction Rate
Reaction rate graphs will generally be graphed with time on the x-axis and some measure of how far the reaction has gone (ie concentration, volume, mass loss etc) on the y-axis. • This will generally produce a curve with, for example, the concentration of the reactants approaching zero. Interpretation of rate graphs.
gradient = y x MEASURING THE RATE RATE How much concentration changes with time. It is the equivalent of velocity. THE SLOPE OF THE GRADIENT OF THE CURVE GETS LESS AS THE REACTION SLOWS DOWN WITH TIME CONCENTRATION y x TIME • the rate of change of concentration is found from the slope (gradient) of the curve • the slope at the start of the reaction will give the INITIAL RATE • the slope gets less (showing the rate is slowing down) as the reaction proceeds
C CONCENTRATION A B TIME RATE CHANGE DURING A REACTION Reactions are fastest at the start and get slower as the reactants concentration drops. In a reaction such as A + 2B ——> C the concentrations might change as shown • Reactants (A and B) • Concentration decreases with time • Product (C) • Concentration increases with time • the steeper the curve the faster the • rate of the reaction • reactions start off quickly because • of the greater likelihood of collisions • reactions slow down with time as • there are fewer reactants to collide
MEASURING THE RATE • Experimental Investigation • the variation in concentration of a reactant or product is followed with time • the method depends on the reaction type and the properties of reactants/products • e.g. Extracting a sample from the reaction mixture and analysing it by titration. • - this is often used if an acid is one of the reactants or products • Using a colorimeter or UV / visible spectrophotometer. • Measuring the volume of gas evolved. • Measuring the change in conductivity. • More details of these and other methods can be found in suitable text-books.
THE RATE EQUATION Format links the rate of reaction to the concentration of reactants it can only be found by doing actual experiments it cannot be found by just looking at the equation the equation... A + B ——> C + D might have a rate equation like thisr = k [A] [B]2 r rate of reaction units of conc. / time usually mol dm-3 s-1 k rate constant units depend on the rate equation [ ] concentration units of mol dm-3 Interpretation The above rate equation tells you that the rate of reaction is... proportional to the concentration of reactant A doubling [A] doubles rate proportional to the square of the concentration of B doubling [B] quadruples (22) rate
ORDER OF REACTION Order tells you how much the concentration of a reactant affects the rate Individual order The power to which a concentration is raised in the rate equation Overallorder The sum of all the individual orders in the rate equation.
ORDER OF REACTION Order tells you how much the concentration of a reactant affects the rate Individual order The power to which a concentration is raised in the rate equation Overallorder The sum of all the individual orders in the rate equation. e.g. in the rate equation r = k [A] [B]2 the order with respect to A is 1 1st Order the order with respect to B is 2 2nd Order and the overall order is 3 3rd Order Value(s) need not be whole numbers can be zero if the rate is unaffected by how much of a substance is present
ORDER OF REACTION Order tells you how much the concentration of a reactant affects the rate Individual order The power to which a concentration is raised in the rate equation Overallorder The sum of all the individual orders in the rate equation. e.g. in the rate equation r = k [A] [B]2 the order with respect to A is 1 1st Order the order with respect to B is 2 2nd Order and the overall order is 3 3rd Order Value(s) need not be whole numbers can be zero if the rate is unaffected by how much of a substance is present NOTES The rate equation is derived from experimental evidence not by looking at an equation. Species appearing in the stoichiometric equation sometimes aren’t in the rate equation. Substances not in the stoichiometric equation can appear in the rate equation - CATALYSTS
THE RATE EQUATION Experimental determination of order Method 1 Plot a concentration / time graph qnd calculate the rate (gradient) at points on the curve Plot another graph of the rate (y axis) versus the concentration at that point (x axis) If it gives a straight line, the rate is directly proportional to concentration - 1st ORDER. If the plot is a curve then it must have another order. Try plotting rate v. (conc.)2. A straight line would mean 2nd ORDER. This method is based on trial and error.
THE RATE EQUATION Experimental determination of order Method 1 Plot a concentration / time graph qnd calculate the rate (gradient) at points on the curve Plot another graph of the rate (y axis) versus the concentration at that point (x axis) If it gives a straight line, the rate is directly proportional to concentration - 1st ORDER. If the plot is a curve then it must have another order. Try plotting rate v. (conc.)2. A straight line would mean 2nd ORDER. This method is based on trial and error. Method 2-The initial rates method. Do a series of experiments (at the same temperature) at different concentrations of a reactant but keeping all others constant. Plot a series of concentration / time graphs and calculate the initial rate (slope of curve at start) for each reaction. From the results calculate the relationship between concentration and rate and hence deduce the rate equation. To find order directly, logarithmic plots are required.
THE RATE CONSTANT (k) Units The units of k depend on the overall order of reaction. e.g. if the rate equation is... rate = k [A]2 the units of k will be dm3 mol-1 sec-1 Divide the rate by as many concentrations as appear in the rate equation. Overall Order 0 1 2 3 units of k mol dm-3 sec-1 sec-1 dm3 mol-1 sec-1 dm6 mol-2 sec-1 example in the rate equation r = k [A] k will have units of sec-1 in the rate equation r = k [A] [B]2 k will have units of dm6 mol-2 sec-1
RATE EQUATION - SAMPLE CALCULATION • In an experiment between A and B the initial rate of reaction was found for various starting concentrations of A and B. Calculate... • the individual orders for A and B • the overall order of reaction • the rate equation • the value of the rate constant (k) • the units of the rate constant [A] [B] Initial rate (r) 1 0.5 1 2 2 1.5 1 6 3 0.5 2 8 r initial rate of reaction mol dm-3 s-1 [ ] concentration mol dm-3
RATE EQUATION - SAMPLE CALCULATION CALCULATING ORDER wrt A Choose any two experiments where... [A] is changed and, importantly, [B] is KEPT THE SAME See how the change in [A] affects the rate As you can see, tripling [A] has exactly the same effect on the rate so... THE ORDER WITH RESPECT TO A = 1 (it is FIRST ORDER) [A] [B] Initial rate (r) 1 0.5 1 2 2 1.5 1 6 3 0.5 2 8 Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger rate [A] FIRST ORDER with respect to (wrt) A
RATE EQUATION - SAMPLE CALCULATION CALCULATING ORDER wrt B Choose any two experiments where... [B] is changed and, importantly, [A] is KEPT THE SAME See how a change in [B] affects the rate As you can see, doubling [B] quadruples the rate so... THE ORDER WITH RESPECT TO B = 2 It is SECOND ORDER [A] [B] Initial rate (r) 1 0.5 1 2 2 1.5 1 6 3 0.5 2 8 Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger rate [B]2 SECOND ORDER wrt B
RATE EQUATION - SAMPLE CALCULATION [A] [B] Initial rate (r) [A] [B] Initial rate (r) 1 0.5 1 2 1 0.5 1 2 2 1.5 1 6 2 1.5 1 6 3 0.5 2 8 3 0.5 2 8 Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger rate [A] FIRST ORDER with respect to (wrt) A Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger rate [B]2 SECOND ORDER wrt B OVERALL ORDER= THE SUM OF THE INDIVIDUAL ORDERS = 1 + 2 = 3
RATE EQUATION - SAMPLE CALCULATION [A] [B] Initial rate (r) [A] [B] Initial rate (r) 1 0.5 1 2 1 0.5 1 2 2 1.5 1 6 2 1.5 1 6 3 0.5 2 8 3 0.5 2 8 Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger rate [A] FIRST ORDER with respect to (wrt) A Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger rate [B]2 SECOND ORDER wrt B By combining the two proportionality relationships you can construct the overall rate equation rate = k [A] [B]2
RATE EQUATION - SAMPLE CALCULATION [A] [B] Initial rate (r) [A] [B] Initial rate (r) 1 0.5 1 2 1 0.5 1 2 2 1.5 1 6 2 1.5 1 6 3 0.5 2 8 3 0.5 2 8 Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger rate [A] FIRST ORDER with respect to (wrt) A Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger rate [B]2 SECOND ORDER wrt B rate = k [A] [B]2 re-arranging k = rate [A] [B]2 Chose one experiment (e.g. Expt. 3) and substitute its values into the rate equation k = 8 = 4 dm6 mol-2 sec-1 (0.5) (2)2
RATE EQUATION - SAMPLE CALCULATION SUMMARY [A] [B] Initial rate (r) [A] [B] Initial rate (r) 1 0.5 1 2 1 0.5 1 2 2 1.5 1 6 2 1.5 1 6 3 0.5 2 8 3 0.5 2 8 Compare Experiments 1 & 2 [B] same [A] 3 x bigger rate 3 x bigger rate [A] FIRST ORDER with respect to (wrt) A Compare Experiments 1 & 3 [A] same [B] 2 x bigger rate 4 x bigger rate [B]2 SECOND ORDER wrt B rate = k [A] [B]2 re-arranging k = rate [A] [B]2 Chose one experiment (e.g. Expt. 3) and substitute its values into the rate equation k = 8 = 4 dm6 mol-2 sec-1 (0.5) (2)2
RATE EQUATION QUESTIONS [A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.25 0.25 4 Expt 2 0.25 0.50 8 Expt 3 0.50 0.25 8 No 1 CALCULATE THE ORDER WITH RESPECT TO A THE ORDER WITH RESPECT TO B THE OVERALL ORDER OF REACTION THE FORMAT OF THE RATE EQUATION THE VALUE AND UNITS OF THE RATE CONSTANT ANSWER ON NEXT PAGE
ANSWER RATE EQUATION QUESTIONS [A] / mol dm-3 [B] / mol dm-3 Rate / mol dm-3 s-1 Expt 1 0.25 0.25 4 Expt 2 0.25 0.50 8 Expt 3 0.50 0.25 8 No 1 Expts 1&2 [A] is constant [B] is doubled Rate is doubled Therefore rate [B] 1st order wrt B Explanation: What was done to [B] had exactly the same effect on the rate