Reaction Rates & Equilibrium Unit 9 *Start with visual from Phet

Collision Theory of Kinetics 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.

Effective Collisions 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 ( s) ◦Has to form for product to be formed

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Activated Complex The difference in potential energy between the reactant molecules and the activated complex is called the activation energy, E a 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

Energy Diagram 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

Factors Affecting Reaction Rate 1. Nature of the Reactants ◦Cl 2(g) + CH 4(g)  CH 3 Cl (g) + HCl (g)  Cl 2  Cl + Cl(fast)  Cl + CH 4  CH 3 Cl + 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

Increasing concentration = more frequent collisions = increased rate of reaction 2) INCREASING CONCENTRATION However, increasing the concentration of some reactants can have a greater effect than increasing others Low concentration = fewer collisionsHigher concentration = more collisions

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. CUT THE SHAPE INTO SMALLER PIECES SURFACE AREA = 30 sq units SURFACE AREA 9 x ( ) = 54 sq units 1 1 1

Factors Affecting Reaction Rate 4. Agitation ◦this puts more liquid/gas particles in contact with the solid = ↑ collisions = ↑ act. complex = ↑ product

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 5) INCREASING THE PRESSURE

Factors Affecting Reaction Rate 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 E a ◦therefore the reaction rate will increase

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

Stern-Zartman Experiment (1)silver-coated platinum wire, (2)slit forming a beam of silver atoms, (3)plate on which the silver atoms were deposited, The shift of the plate increases as the angular velocity of the apparatus increases and decreases as the speed v of the silver atoms increases. Since the atoms move with different speeds, the band spreads and becomes broader as the apparatus rotates. The density of the silver deposit at a given point of the band is proportional to the number of atoms moving with a given speed.

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 MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY MAXWELL-BOLTZMANN DISTRIBUTION OF 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 T1T1 T2T2 TEMPERATURE T 2 > T 1 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY INCREASING TEMPERATURE MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR 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 T1T1 T3T3 TEMPERATURE T 1 > T 3 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY INCREASING TEMPERATURE MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR 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 T1T1 T2T2 T3T3 TEMPERATURE T 2 > T 1 > T 3 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY INCREASING TEMPERATURE MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY

EaEa ACTIVATION ENERGY - E a The Activation Energy is the minimum energy required for a reaction to take place The area under the curve beyond E a corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react. MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER INCREASING TEMPERATURE MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY

Explanation increasing the temperature gives more particles an energy greater than E a 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 T1T1 T2T2 TEMPERATURE T 2 > T 1 EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY INCREASING TEMPERATURE MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER

Factors Affecting Reaction 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 E a ◦lower E a = more collisions having the proper amount of energy = ↑ act. complex = ↑ product

Catalysts provide an alternative reaction pathway with a lower Activation Energy (E a ) 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. ADDING A CATALYST WITHOUT A CATALYSTWITH A CATALYST

The area under the curve beyond E a 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 - E a will move to the left. MOLECULAR ENERGY EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER ADDING A CATALYST NUMBER OF MOLECUES WITH A PARTICULAR ENERGY

The area under the curve beyond E a corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react. Lowering the Activation Energy, E a, results in a greater area under the curve after E a showing that more molecules have energies in excess of the Activation Energy EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY ADDING A CATALYST EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY

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 reactantsdifferent phase to reactants e.g. CFC’s and ozone e.g. Fe in Haber process Usesused 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 CATALYSTS - A REVIEW

Reaction Rate 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) + NO 2(g)  CO 2(g) + NO (g) - at t = 4.0 min, [CO 2 ] =.12 mol ◦ dm -3 (mol/L) -at t = 8.0 min, [CO 2 ] =.24 mol ◦ dm -3 -reaction rate =.24 mol/L -.12 mol/L 8.0 sec – 4.0 sec = 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/dm 3, (moles per dm 3 or moles per liter.)

Interpretation of rate graphs. 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.

RATEHow much concentration changes with time. It is the equivalent of velocity. MEASURING THE RATE y CONCENTRATION gradient = y x 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 THE SLOPE OF THE GRADIENT OF THE CURVE GETS LESS AS THE REACTION SLOWS DOWN WITH TIME

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 RATE CHANGE DURING A REACTION 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 TIME CONCENTRATION B A C

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. MEASURING THE RATE

THE RATE EQUATION Formatlinks 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 this r = k [A] [B] 2 rrate of reactionunits of conc. / time usually mol dm -3 s -1 krate 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 (2 2 ) rate

ORDER OF REACTION Individual order The power to which a concentration is raised in the rate equation Overall orderThe sum of all the individual orders in the rate equation. Order tells you how much the concentration of a reactant affects the rate

ORDER OF REACTION Individual order The power to which a concentration is raised in the rate equation Overall orderThe 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 andthe 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 Order tells you how much the concentration of a reactant affects the rate

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) UnitsThe 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 dm 3 mol -1 sec -1 Divide the rate by as many concentrations as appear in the rate equation. Overall Order units of k mol dm -3 sec -1 sec -1 dm 3 mol -1 sec -1 dm 6 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 dm 6 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) r initial rate of reaction mol dm -3 s -1 [ ] concentrationmol dm -3

Compare Experiments 1 & 2 [B]same [A]3 x bigger rate3 x bigger  rate  [A] FIRST ORDER with respect to (wrt) A [A][B] Initial rate (r) 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) RATE EQUATION - SAMPLE CALCULATION

[A][B] Initial rate (r) 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 RATE EQUATION - SAMPLE CALCULATION Compare Experiments 1 & 3 [A]same [B]2 x bigger rate4 x bigger  rate  [B] 2 SECOND ORDER wrt B

Compare Experiments 1 & 2 [B]same [A]3 x bigger rate3 x bigger  rate  [A] FIRST ORDER with respect to (wrt) A [A][B] Initial rate (r) [A][B] Initial rate (r) Compare Experiments 1 & 3 [A]same [B]2 x bigger rate4 x bigger  rate  [B] 2 SECOND ORDER wrt B OVERALL ORDER = THE SUM OF THE INDIVIDUAL ORDERS = = 3 RATE EQUATION - SAMPLE CALCULATION

[A][B] Initial rate (r) [A][B] Initial rate (r)  rate = k [A] [B] 2 RATE EQUATION - SAMPLE CALCULATION By combining the two proportionality relationships you can construct the overall rate equation Compare Experiments 1 & 3 [A]same [B]2 x bigger rate4 x bigger  rate  [B] 2 SECOND ORDER wrt B Compare Experiments 1 & 2 [B]same [A]3 x bigger rate3 x bigger  rate  [A] FIRST ORDER with respect to (wrt) A

[A][B] Initial rate (r) [A][B] Initial rate (r) SUMMARY RATE EQUATION - SAMPLE CALCULATION Compare Experiments 1 & 3 [A]same [B]2 x bigger rate4 x bigger  rate  [B] 2 SECOND ORDER wrt B Compare Experiments 1 & 2 [B]same [A]3 x bigger rate3 x bigger  rate  [A] FIRST ORDER with respect to (wrt) A Chose one experiment (e.g. Expt. 3) and substitute its values into the rate equation k = 8= 4 dm 6 mol -2 sec -1 (0.5) (2) 2  rate = k [A] [B] 2 re-arranging k = rate [A] [B] 2

RATE EQUATION QUESTIONS CALCULATETHE 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 [A] / mol dm -3 [B] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 1

RATE EQUATION QUESTIONS Expts 1&2[A] is constant[B] is doubledRate is doubled Thereforerate  [B]1st order wrt B Explanation:What was done to [B] had exactly the same effect on the rate [A] / mol dm -3 [B] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 1 ANSWER

RATE EQUATION QUESTIONS Expts 1&2[A] is constant[B] is doubledRate is doubled Thereforerate  [B]1st order wrt B Explanation:What was done to [B] had exactly the same effect on the rate Expts 1&3[B] is constant[A] is doubledRate is doubled Therefore rate  [A] 1st order wrt A Explanation:What was done to [A] had exactly the same effect on the rate [A] / mol dm -3 [B] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 1 ANSWER

RATE EQUATION QUESTIONS Expts 1&2[A] is constant[B] is doubledRate is doubled Thereforerate  [B]1st order wrt B Explanation:What was done to [B] had exactly the same effect on the rate Expts 1&3[B] is constant[A] is doubledRate is doubled Therefore rate  [A] 1st order wrt A Explanation:What was done to [A] had exactly the same effect on the rate r = k[A][B] Rate equation isr = k[A][B] [A] / mol dm -3 [B] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 1 ANSWER

RATE EQUATION QUESTIONS Expts 1&2[A] is constant[B] is doubledRate is doubled Thereforerate  [B]1st order wrt B Explanation:What was done to [B] had exactly the same effect on the rate Expts 1&3[B] is constant[A] is doubledRate is doubled Therefore rate  [A] 1st order wrt A Explanation:What was done to [A] had exactly the same effect on the rate r = k[A][B] Rate equation isr = k[A][B] Value of kSubstitute numbers from Exp 1 to get value of k k = rate / [A][B] = 4 / 0.25 x 0.25= 64 [A] / mol dm -3 [B] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 1 ANSWER

RATE EQUATION QUESTIONS Expts 1&2[A] is constant[B] is doubledRate is doubled Thereforerate  [B]1st order wrt B Explanation:What was done to [B] had exactly the same effect on the rate Expts 1&3[B] is constant[A] is doubledRate is doubled Therefore rate  [A] 1st order wrt A Explanation:What was done to [A] had exactly the same effect on the rate r = k[A][B] Rate equation isr = k[A][B] Value of kSubstitute numbers from Exp 1 to get value of k k = rate / [A][B] = 4 / 0.25 x 0.25= 64 Units of krate / conc x conc= dm 3 mol -1 s -1 [A] / mol dm -3 [B] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 1 ANSWER

RATE EQUATION QUESTIONS CALCULATETHE ORDER WITH RESPECT TO C THE ORDER WITH RESPECT TO D THE OVERALL ORDER OF REACTION THE FORMAT OF THE RATE EQUATION THE VALUE AND UNITS OF THE RATE CONSTANT ANSWER ON NEXT PAGE [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 2

RATE EQUATION QUESTIONS [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 2 Expts 1&3[C] is constant [D] is tripledRate is 9 x bigger Therefore rate  [D] 2 2nd order wrt D Explanation:Squaring what was done to D affected the rate (3 2 = 9) ANSWER

RATE EQUATION QUESTIONS [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 2 Expts 1&3[C] is constant [D] is tripledRate is 9 x bigger Therefore rate  [D] 2 2nd order wrt D Explanation:Squaring what was done to D affected the rate (3 2 = 9) Expts 1&2[D] is constant [A] is halvedRate is quartered Therefore rate  [C] 2 2nd order wrt C Explanation:One half squared = one quarter ANSWER

RATE EQUATION QUESTIONS [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 2 Expts 1&3[C] is constant [D] is tripledRate is 9 x bigger Therefore rate  [D] 2 2nd order wrt D Explanation:Squaring what was done to D affected the rate (3 2 = 9) Expts 1&2[D] is constant [A] is halvedRate is quartered Therefore rate  [C] 2 2nd order wrt C Explanation:One half squared = one quarter r = k[C] 2 [D] 2 Rate equation isr = k[C] 2 [D] 2 ANSWER

RATE EQUATION QUESTIONS [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 2 Expts 1&3[C] is constant [D] is tripledRate is 9 x bigger Therefore rate  [D] 2 2nd order wrt D Explanation:Squaring what was done to D affected the rate (3 2 = 9) Expts 1&2[D] is constant [A] is halvedRate is quartered Therefore rate  [C] 2 2nd order wrt C Explanation:One half squared = one quarter r = k[C] 2 [D] 2 Rate equation isr = k[C] 2 [D] 2 Value of kSubstitute numbers from Exp 2 to get value of k k = rate / [C] 2 [D] 2 = 0.04 / x = 6.25 Units of krate / conc 2 x conc 2 = dm 9 mol -3 s -1 ANSWER

RATE EQUATION QUESTIONS CALCULATETHE ORDER WITH RESPECT TO E THE ORDER WITH RESPECT TO F THE OVERALL ORDER OF REACTION THE FORMAT OF THE RATE EQUATION THE VALUE AND UNITS OF THE RATE CONSTANT No 3 ANSWER ON NEXT PAGE [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt

RATE EQUATION QUESTIONS [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 3 Expts 2&3[E] is constant [F] is x 1.5Rate unchanged Rate is UNAFFECTED ZERO order wrt F Explanation:Concentration of [F] has no effect on the rate ANSWER

RATE EQUATION QUESTIONS [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 3 Expts 2&3[E] is constant [F] is x 1.5Rate unchanged Rate is UNAFFECTED ZERO order wrt F Explanation:Concentration of [F] has no effect on the rate Expts 1&2[E] is doubled [F] is doubledRate is doubled Therefore rate  [E] 2 2nd order wrt E Explanation:Although both concentrations have been doubled, we know [F] has no effect. The change must be all due to [E] ANSWER

RATE EQUATION QUESTIONS [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 3 Expts 2&3[E] is constant [F] is x 1.5Rate unchanged Rate is UNAFFECTED ZERO order wrt F Explanation:Concentration of [F] has no effect on the rate Expts 1&2[E] is doubled [F] is doubledRate is doubled Therefore rate  [E] 2 2nd order wrt E Explanation:Although both concentrations have been doubled, we know [F] has no effect. The change must be all due to [E] r = k[E] Rate equation isr = k[E] ANSWER

RATE EQUATION QUESTIONS [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt Expt Expt No 3 Expts 2&3[E] is constant [F] is x 1.5Rate unchanged Rate is UNAFFECTED ZERO order wrt F Explanation:Concentration of [F] has no effect on the rate Expts 1&2[E] is doubled [F] is doubledRate is doubled Therefore rate  [E] 2 2nd order wrt E Explanation:Although both concentrations have been doubled, we know [F] has no effect. The change must be all due to [E] r = k[E] Rate equation isr = k[E] Value of kSubstitute numbers from Exp 1 to get value of k k = rate / [E] = 0.16 / 0.4= 0.40 Units of krate / conc= s -1 ANSWER

FIRST ORDER REACTIONS AND HALF LIFE One characteristic of a FIRST ORDER REACTION is that it is similar to radioactive decay. It has a half-life that is independent of the concentration. It should take the same time to drop to one half of the original concentration as it does to drop from one half to one quarter of the original. The concentration of a reactant falls as the reaction proceeds

The concentration drops from 4 to 2 in 17 minutes 2 to 1 in a further 17 minutes 1 to 0.5 in a further 17 minutes FIRST ORDER REACTIONS AND HALF LIFE The concentration of reactant A falls as the reaction proceeds

A useful relationship k t ½ = log e 2 = where t ½ = the half life FIRST ORDER REACTIONS AND HALF LIFE Half life= 17 minutes k t ½ = k = t ½ k = = min -1 17

ORDER OF REACTION – GRAPHICAL DETERMINATION The order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order.

ORDER OF REACTION – GRAPHICAL DETERMINATION The order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. PLOTTING RATE AGAINST CONCENTRATION RATE OF REACTION / mol dm -3 s -1 CONCENTRATION / mol dm -3 ZERO ORDER – the rate does not depend on the concentration. The line is parallel to the x axis. FIRST ORDER – the rate is proportional to the concentration so you get a straight line of fixed gradient. The gradient of the line equals the rate constant for the reaction. SECOND ORDER – the rate is proportional to the square of the concentration. You get an upwardly sloping curve.

ORDER OF REACTION – GRAPHICAL DETERMINATION The order of reaction can be found by measuring the rate at different times during the reaction and plotting the rate against either concentration or time. The shape of the curve provides an indication of the order. PLOTTING RATE AGAINST TIME RATE OF REACTION / mol dm -3 s -1 TIME / s ZERO ORDER A straight line showing a constant decline in concentration. FIRST ORDER A slightly sloping curve which drops with a constant half-life. SECOND ORDER The curve declines steeply at first then levels out.

RATE DETERMINING STEP Many reactions consist of a series of separate stages. Each step has its own rate and rate constant. The overall rate of a multi-step process is governed by the slowest step (like a production line where overall output can be held up by a slow worker). This step is known as the RATE DETERMINING STEP. If there is more than one step, the rate equation may not contain all the reactants in its format.

RATE DETERMINING STEP THE REACTION BETWEEN PROPANONE AND IODINE Iodine and propanoneCH 3 COCH 3 + I 2 CH 3 COCH 2 I + HI react in the presence of acid The rate equation is...r = k [CH 3 COCH 3 ] [H + ] Why do H + ions appear in the rate equation? Why does I 2 not appear in the rate equation?

RATE DETERMINING STEP THE REACTION BETWEEN PROPANONE AND IODINE Iodine and propanoneCH 3 COCH 3 + I 2 CH 3 COCH 2 I + HI react in the presence of acid The rate equation is...r = k [CH 3 COCH 3 ] [H + ] Why do H + ions appear inThe reaction is catalysed by acid the rate equation?[H + ] affects the rate but is unchanged overall Why does I 2 not appear The rate determining step doesn’t involve I 2 in the rate equation?

RATE DETERMINING STEP THE REACTION BETWEEN PROPANONE AND IODINE Iodine and propanoneCH 3 COCH 3 + I 2 CH 3 COCH 2 I + HI react in the presence of acid The rate equation is...r = k [CH 3 COCH 3 ] [H + ] Why do H + ions appear inThe reaction is catalysed by acid the rate equation?[H + ] affects the rate but is unchanged overall Why does I 2 not appear The rate determining step doesn’t involve I 2 in the rate equation? The slowest step of any multi-step reaction is known as the rate determining step and it is the species involved in this step that are found in the overall rate equation. Catalysts appear in the rate equation because they affect the rate but they do not appear in the stoichiometric equation because they remain chemically unchanged

RATE DETERMINING STEP HYDROLYSIS OF HALOALKANES Haloalkanes (general formula RX) are RX + OH - ROH + X - hydrolysed by hydroxide ion to give alcohols. With many haloalkanes the rate equation is... r = k [RX][OH - ]SECOND ORDER This is because both the RX and OH - must collide for a reaction to take place in ONE STEP

RATE DETERMINING STEP HYDROLYSIS OF HALOALKANES Haloalkanes (general formula RX) are RX + OH - ROH + X - hydrolysed by hydroxide ion to give alcohols. With many haloalkanes the rate equation is... r = k [RX][OH - ]SECOND ORDER This is because both the RX and OH - must collide for a reaction to take place in ONE STEP but with others it only depends on [RX]... r = k [RX]FIRST ORDER The reaction has taken place in TWO STEPS... - the first involves breaking an R-X bondi) RX R + + X - Slow - the second step involves the two ions joiningii) R + + OH - ROH Fast The first step is slower as it involves bond breaking and energy has to be put in. The first order mechanism is favoured by tertiary haloalkanes because the hydroxide ion is hindered in its approach by alkyl groups if the mechanism involves the hydroxide ion and haloalkane colliding.

RATE DETERMINING STEP The reaction H 2 O 2 + 2H 3 O + + 2I¯ I 2 + 4H 2 O takes place in 3 steps Step 1H 2 O 2 + I¯ IO¯ + H 2 O SLOW Step 2IO¯ + H 3 O + HIO + H 2 O FAST Step 3HIO + H 3 O + + I¯ I 2 + 2H 2 O FAST The rate determining step is STEP 1 as it is the slowest

RATE DETERMINING STEP The reaction H 2 O 2 + 2H 3 O + + 2I¯ I 2 + 4H 2 O takes place in 3 steps Step 1H 2 O 2 + I¯ IO¯ + H 2 O SLOW Step 2IO¯ + H 3 O + HIO + H 2 O FAST Step 3HIO + H 3 O + + I¯ I 2 + 2H 2 O FAST The rate determining step is STEP 1 as it is the slowest The reaction 2N 2 O 5 4NO 2 + O 2 takes place in 3 steps Step 1N 2 O 5 NO 2 + NO 3 SLOW Step 2NO 2 + NO 3 NO + NO 2 + O 2 FAST Step 3NO + NO 3 2NO 2 from another Step 1 FAST The rate determining step is STEP 1 rate = k [N 2 O 5 ]

Equilibrium

Reaction Dynamics If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up. However, if the products are allowed to accumulate; they will start reacting together to form the original reactants - called the reverse reaction. We show this reverse reaction by using a double arrow (H 2(g) + I 2(g) ↔ 2HI (g) )

Reaction Dynamics The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing At the same time the reverse reaction speeds up as the concentration of the products increases. Eventually the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium. ◦rate forward = rate reverse

Chemical Equilibrium Equilibrium only occurs in a closed system!! When a system reaches equilibrium, the amounts of reactants and products in the system stays constant ◦the forward and reverse reactions still continue, but because they go at the same rate the amounts of materials don't change. There is a mathematical relationship between the amounts of reactants and products at equilibrium ◦regardless of the amounts of reactant and product that you start with

Equilibrium Expression Capital letters (A,B,C,D) – reactants or products Lowercase letter (a,b,c,d) – coefficients from the equation NOTICE – products on top, reactants on bottom In this expression, K eq is a number called the equilibrium constant. ◦ratio of product concentration to reactant concentration at equilibrium Do not include solids or liquids, only solutions and gases The value of K eq depends on temp. of the reaction – if temp. changes then the value of K eq changes. aA + bB  cC + dD = [C] c [D] d [A] a [B] b K eq

Example – Determine the value of the Equilibrium Constant for the Reaction 2 SO 2(g) + O 2(g)  2 SO 3(g) ¬ Determine the Equilibrium Expression ­ Plug the equilibrium concentrations into to Equilibrium Expression ® Solve the Equation SO O2O2 2.00SO 2 [Equilibrium][Initial]Chemical So what does this K eq value tell us???

Position of Equilibrium The size of the equilibrium constant shows whether products or reactants are favored at equilibrium. K eq > 1, products are favored at equilibrium K eq < 1, reactants are favored at equilibrium

Example – If the value of the Equilibrium Constant for the Reaction 2 SO 2 + O 2  2 SO 3 is 4.36, Determine the Equilibrium Concentration of SO 3 ¬ Determine the Equilibrium Expression ­ Plug the equilibrium concentrations and Equilibrium Constant into the Equilibrium Expression ® Solve the Equation ?3.00SO O2O2 2.00SO 2 [Equilibrium][Initial]Chemical

More Equilibrium Practice 1.Write the equilibrium constant expression for the following reaction. 3H 2(g) + N 2(g) ↔ 2NH 3(g) 2.An analysis of an equilibrium mixture for the previous reaction in a 1.0 L flask at 300 o C gave the following results: 0.15 mol H 2, 0.25 mol N 2 and 0.10 mol NH 3. Calculate the K eq for this reaction. 3.2BrCl (g) ↔ Cl 2(g) + Br 2(g) The equilibrium constant for this reaction is The equilibrium mixture contains 4.00 mol Cl 2 and 4.00 moles of Br 2. How many moles of BrCl are present?

Le Ch âtelier’s Principle Le Châtelier's Principle guides us in predicting the effect various changes have on the position of equilibrium it says that if stress is applied to a system in dynamic equilibrium, the system will change to relieve the stress. ◦The position of equilibrium moves to counteract the change. Three common stressors: ◦Concentration ◦Temperature ◦Pressure

Concentration Changes and Le Châtelier’s Principle A+ B ↔ C + D ◦Adding a reactant – equilibrium shifts right ◦Removing a reactant – equilibrium shifts left ◦Adding a product – equilibrium shifts left ◦Removing a product – equilibrium shifts right

Changing Temperature and Le Châtelier’s Principle Increasing the temperature causes the reaction to shift away from the heat. For exothermic reactions - ◦Think of heat as a product of the reaction ◦Therefore shift the position of equilibrium toward the reactant side For endothermic reactions - ◦Think of heat as a reactant ◦The position of equilibrium will shift toward the products Cooling an exothermic or endothermic reaction will have the opposite effects.

Changing Pressure and Le Châtelier’s Principle Only affects a reaction involving gases with an unequal number of mole of reactants & products. Increasing pressure causes equilibrium to shift toward the side with fewer gas molecules Decreasing pressure causes a shift toward the side with more gas molecules Example 3H 2(g) + N 2(g)  2NH 3(g) ↑ Pressure – shifts to the right ↓ Pressure – shift to the left

Examples – Le Chatelier’s Principle What effect do the following changes have on the equilibrium position for the following reaction? 1.PCl 5(g) + heat ↔ PCl 3(g) + Cl 2(g) a. addition of Cl 2 b. increase in pressure c. removal of heat d. removal of PCl 3 as formed 2. C (s) + H 2 O (g) + heat ↔ CO (g) + H 2(g) a. Lowering the temperature b. Increasing the pressure c. Removal of H 2 as formed

Examples – Le Chatelier’s Principle 3. At 425 K – Fe 3 O 4(s) + 4H 2(g) ↔ 3Fe (s) + 4H 2 O (g) How would the equilibrium concentration of H 2 O be affected by the following: a. Adding more H 2 b. Adding more Fe (s) c. Decreasing the pressure d. Adding a catalyst

Spontaneous vs. Non- Spontaneous Spontaneous rxn – will produce products when reactants come together Non-spontaneous rxn – no products formed when reactants come together Spontaneity does not refer to how fast a reaction proceeds. Rxns tend toward spontaneity when: 1.rxn is exothermic 2.rxn increases entropy

Entropy Entropy is the disorder of a system ◦dependent upon temperature +ΔS = more disorder, less order -ΔS = less disorder, more order Which has a higher entropy? ◦New pack of cards / cards in use ◦Cube of sugar in coffee / cube of sugar ◦Cup of water at room temp. / cup of water at 50 o C ◦I 2(g) / I 2(s)

Gibb’s Free Energy We can relate enthalpy & entropy quantitatively in a relationship known as….. ΔG = ΔH – T ΔS ◦ΔG = Gibb’s Free Energy (energy available to do work) [kJ/mol] ◦ΔH = Enthalpy (endo is + & exo is -) [kJ/mol] ◦ΔS = Entropy [J/Kmol] ◦T = Temperature in Kelvin -ΔG = spontaneous rxn +ΔG = nonspontaneous rxn