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THE RATE EQUATION A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

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INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard THE RATE EQUATION KNOCKHARDY PUBLISHING

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THE RATE EQUATION CONTENTS Collision theory Methods for increasing rate Rate changes during a reaction The rate equation Worked examples Graphical methods for determining rate Half-life Rate determining step

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Before you start it would be helpful to… know how the energy changes during a chemical reaction know the basic ideas of Kinetic Theory know the importance of catalysts in industrial chemistry THE RATE EQUATION

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COLLISION THEORY Collision theory states that... particles must COLLIDE before a reaction can take place not all collisions lead to a reaction reactants must possess at least a minimum amount of energy - ACTIVATION ENERGY plus particles must approach each other in a certain relative way - the STERIC EFFECT REVISION

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COLLISION THEORY Collision theory states that... particles must COLLIDE before a reaction can take place not all collisions lead to a reaction reactants must possess at least a minimum amount of energy - ACTIVATION ENERGY plus particles must approach each other in a certain relative way - the STERIC EFFECT According to collision theory, to increase the rate of reaction you therefore need... more frequent collisionsincrease particle speedor have more particles present more successful collisionsgive particles more energyor lower the activation energy REVISION

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INCREASING THE RATE INCREASE THE SURFACE AREA OF SOLIDS INCREASE TEMPERATURE SHINE LIGHT ADD A CATALYST INCREASE THE PRESSURE OF ANY GASES INCREASE THE CONCENTRATION OF REACTANTS INCREASE THE SURFACE AREA OF SOLIDS INCREASE TEMPERATURE SHINE LIGHT ADD A CATALYST INCREASE THE PRESSURE OF ANY GASES INCREASE THE CONCENTRATION OF REACTANTS The following methods may be used to increase the rate of a chemical reaction REVISION

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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 REVISION

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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

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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

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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

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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

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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 Order tells you how much the concentration of a reactant affects the rate

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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

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THE RATE EQUATION Experimental determination of order Method 1 Plot a concentration / time graph and 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.

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THE RATE EQUATION Experimental determination of order Method 1 Plot a concentration / time graph and 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.

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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 0 1 2 3 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

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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 0.512 1.516 0.528 1 2 3 [A][B] Initial rate (r) r initial rate of reaction mol dm -3 s -1 [ ] concentrationmol dm -3

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Compare Experiments 1 & 2 [B]same [A]3 x bigger rate3 x bigger rate [A] FIRST ORDER with respect to (wrt) A 0.512 1.516 0.528 1 2 3 [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

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0.512 1.516 0.528 1 2 3 [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

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Compare Experiments 1 & 2 [B]same [A]3 x bigger rate3 x bigger rate [A] FIRST ORDER with respect to (wrt) A 0.512 1.516 0.528 1 2 3 [A][B] Initial rate (r) 0.512 1.516 0.528 1 2 3 [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 = 1 + 2 = 3 RATE EQUATION - SAMPLE CALCULATION

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0.512 1.516 0.528 1 2 3 [A][B] Initial rate (r) 0.512 1.516 0.528 1 2 3 [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

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0.512 1.516 0.528 1 2 3 [A][B] Initial rate (r) 0.512 1.516 0.528 1 2 3 [A][B] Initial rate (r) 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 - 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

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0.512 1.516 0.528 1 2 3 [A][B] Initial rate (r) 0.512 1.516 0.528 1 2 3 [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

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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 1 0.250.25 4 Expt 2 0.250.50 8 Expt 3 0.500.25 8 No 1

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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 1 0.250.25 4 Expt 2 0.250.50 8 Expt 3 0.500.25 8 No 1 ANSWER

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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 1 0.250.25 4 Expt 2 0.250.50 8 Expt 3 0.500.25 8 No 1 ANSWER

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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 1 0.250.25 4 Expt 2 0.250.50 8 Expt 3 0.500.25 8 No 1 ANSWER

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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 1 0.250.25 4 Expt 2 0.250.50 8 Expt 3 0.500.25 8 No 1 ANSWER

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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 1 0.250.25 4 Expt 2 0.250.50 8 Expt 3 0.500.25 8 No 1 ANSWER

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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 1 0.400.400.16 Expt 2 0.200.400.04 Expt 3 0.401.201.44 No 2

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RATE EQUATION QUESTIONS [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt 1 0.400.400.16 Expt 2 0.200.400.04 Expt 3 0.401.201.44 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

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RATE EQUATION QUESTIONS [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt 1 0.400.400.16 Expt 2 0.200.400.04 Expt 3 0.401.201.44 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

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RATE EQUATION QUESTIONS [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt 1 0.400.400.16 Expt 2 0.200.400.04 Expt 3 0.401.201.44 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

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RATE EQUATION QUESTIONS [C] / mol dm -3 [D] / mol dm -3 Rate / mol dm -3 s -1 Expt 1 0.400.400.16 Expt 2 0.200.400.04 Expt 3 0.401.201.44 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 / 0.2 2 x 0.4 2 = 6.25 Units of krate / conc 2 x conc 2 = dm 9 mol -3 s -1 ANSWER

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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 1 0.400.400.16 Expt 2 0.800.800.32 Expt 3 0.801.200.32

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RATE EQUATION QUESTIONS [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt 1 0.400.400.16 Expt 2 0.800.800.32 Expt 3 0.801.200.32 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

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RATE EQUATION QUESTIONS [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt 1 0.400.400.16 Expt 2 0.800.800.32 Expt 3 0.801.200.32 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

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RATE EQUATION QUESTIONS [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt 1 0.400.400.16 Expt 2 0.800.800.32 Expt 3 0.801.200.32 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

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RATE EQUATION QUESTIONS [E] / mol dm -3 [F] / mol dm -3 Rate / mol dm -3 s -1 Expt 1 0.400.400.16 Expt 2 0.800.800.32 Expt 3 0.801.200.32 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

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GRAPHICAL DETERMINATION OF RATE RATE CALCULATION The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction Place a rule on the outside of the curve and draw a line as shown on the graph. y x gradient = y / x In the reaction… A(aq) + B(aq) ——> C(aq) + D(aq) the concentration of B was measured every 200 minutes. The reaction is obviously very slow! The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.

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GRAPHICAL DETERMINATION OF RATE RATE CALCULATION The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction Place a rule on the outside of the curve and draw a line as shown on the graph. y x gradient = y / x concentration = 1.20 mol dm -3 gradient = - 1.60 mol dm -3 1520 min rate = - 1.05 x 10 -3 mol dm - 3 The rate is negative because the reaction is slowing down The variation in rate can be investigated by measuring the change in concentration of one reactants or product, plotting a graph and then finding the gradients of tangents to the curve at different concentrations.

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GRAPHICAL DETERMINATION OF RATE RATE CALCULATION The rate of reaction at any moment can be found from the gradient of the tangent at that point. The steeper the gradient, the faster the rate of reaction Place a rule on the outside of the curve and draw a line as shown on the graph. y x gradient = y / x The gradients of tangents at several other concentrations are calculated. Notice how the gradient gets less as the reaction proceeds, showing that the reaction is slowing down. The tangent at the start of the reaction is used to calculate the initial rate of the reaction. The variation in rate can be investigated by measuring the change in concentration of one of the reactants or products, plotting a graph and then finding the gradients of the curve at different concentrations.

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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

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

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

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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

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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

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A useful relationship k t ½ = log e 2 = 0.693 where t ½ = the half life FIRST ORDER REACTIONS AND HALF LIFE Half life= 17 minutes k t ½ = 0.693 k = 0.693 t ½ k = 0.693 = 0.041 min -1 17

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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.

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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

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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.

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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.

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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.

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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.

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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.

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ORDER OF REACTION GRAPHICAL DETERMINATION Calculate the rate of reaction at 1.0, 0.75, 0.5 and 0.25 mol dm -3 Plot a graph of rate v [A] Calculate the time it takes for [A] to go from... 1.00 to 0.50 mol dm -3 0.50 to 0.25 mol dm -3 Deduce from the graph that the order wrt A is 1 Calculate the value and units of the rate constant, k

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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.

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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?

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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?

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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

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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

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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.

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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

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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 ]

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OTHER TOPICS Autocatalysis A small number of reactions appear to speed up, rather than slow down, for a time. This is because one of the products is acting as a catalyst and as more product is formed the reaction gets faster. One of the best known examples is the catalytic properties of Mn 2+ (aq) on the decomposition of MnO 4 ¯(aq). You will notice it in a titration of KMnO 4 with either hydrogen peroxide or ethanedioic (oxalic) acid. MolecularityThe number of individual particles of the reacting species taking part in the rate determining step of a reaction e.g. A + 2B C + D molecularity is 3 - one A and two B’s need to collide A 2B however has a molecularity of 1 - only one A is involved

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THE RATE EQUATION The End © 2009 JONATHAN HOPTON & KNOCKHARDY PUBLISHING

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