Reaction Rates -What is meant by rate of reaction? -How do we calculate reaction rate?

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 speed or have more particles present more successful collisionsgive particles more energy or lower the activation energy Collision theory

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 To increase the rate of a reaction we can: Increasing Rate

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 TIME CONCENTRATION B A C

Rate of Reaction The rate of reaction is the change in concentration of a reactant or product per unit time. rate= Units : mol dm -3 s -1 change in concentration of reactant or product time for the change to take place

Concentration Concentration in mol dm -3 is represented by square brackets. [HCl] means the ‘concentration of HCl in mol dm -3’.

Average rate The equation gives an average rate of reaction over the entire time period. Rate at a given time is measured using a graph. Rate at this point in time X Y = Y / X = Gradient of tangent Time / s Concn / mol dm -3

How to Draw a Tangent Line Choose a point on the curve. Adjust the angle of the ruler so that near the point it is equidistant from the curve on either side.

Initial Rate The initial rate of reaction is the change in concentration of reactant, or product, per unit time at the start of a reaction when t = 0. The rate of reaction changes as the reaction proceeds.

Measuring Rate from a Graph Sulphur dichloride, SO 2 Cl 2, decomposes according to the equation: SO 2 Cl 2 (g) SO 2(g) + Cl 2(g) The concentration of SO 2 Cl 2 was measured over time. Draw a concentration – time graph and calculate the rate of reaction at t = 0 and t = 3000 Time t/s [SO 2 Cl 2 ]/ mol dm

Measuring Rates Again 2H 2 O 2(aq) 2H 2 O (l) + O 2(g) Plot a concentration – time graph for this reaction. Calculate the initial rate, rate at t = 20 and t = 90 Time t/s [H 2 O 2 ]/mol dm

Rate Orders and Rate Constant -What is the order of reaction? -What is the rate constant?

Rate Orders In a chemical reaction the concentration of each reactant can affect the overall reaction rate. This effect is called the order of reaction. For a reactant A: – Rate [A] x – Where x = order with respect to A

Rate Orders Zero order – The rate is unaffected by concentration. Rate [A] o Any number to the power of zero = 1 First order – Order is 1 with respect to B. Rate [B] 1 – If [B] is increased by 1 then rate increases by 1 Second order – Order is 2 with respect to C. Rate [C] 2 – If [C] increases by 3 then rate increases by 3 2 =9

Rate Constant Combining the rates from the previous slide from the reaction A + B + C  products gives us: rate [A] 0 [B] 1 [C] 2 The rate constant, k, links the rate with the concentration, giving the equation: rate = k [B] 1 [C] 2 (remember [A] 0 = 1)

Overall Order The overall order of the reaction is the sum of the individual orders. For rate = k [B] 1 [C] 2 Overall order = = 3 The rate order comes from experimental results, not balanced equations!

Units of Rate Constants Units for k depend on the overall order.

Rate Determining Steps -What affects the order of reactions?

R.D.S Some reactions occur as a series of separate steps. Each step in the reaction has it’s own rate and rate constant. The overall rate is governed by the slowest step. This is the rate determining step (R.D.S.)

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 ]

R.D.S If the reactant appears in the rate equation, that reactant is involved in the rate determining step. rate = k [NO 2 ] 2 The R.D.S involves two molecules of NO 2 NO 2 + NO 2 Slow, R.D.S

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

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.

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.

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 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.

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) 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

[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 [C] 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