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Advanced level chemistry kinetics: Data  deducing orders of reaction & rate expression
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Doc Brown's Advanced A Level Chemistry Advanced A Level Chemistry Kinetics Part 6 5.3 Rate data, orders of reaction and rate expressions (rate equation) for more advanced kinetics analysis Experimental techniques i.e. examples of how to obtain rate data, explaining and deducing orders of reactants/reactions, calculating rate constants and deducing their units, with some exemplar rates questions involving deducing and using orders of reactants and rate expressions including graphical analysis. Advanced A Level Kinetics Index Doc Brown's Chemistry Advanced Level PreUniversity Chemistry Revision Study Notes for UK IB KS5 A/AS GCE advanced level physical theoretical chemistry students US K12 grade 11 grade 12 physical theoretical chemistry courses topics including kinetics rates of reaction speeds
5.3 Obtaining rate data, interpreting rate data, orders of reaction and rate expressions 5.3a. Examples of obtaining rate data
5.3b. Rate expressions and orders of reaction
5.3c. Deducing orders of reaction
Connect the graphs with the following:
From the point of view of coursework projects the detailed analysis described above is required, but quite often in examination questions a very limited amount of data is given and some clear logical thinking is required. So simplified rate data questions and their solution is given below. 5.3d. Simple exemplar rates questions to derive rate expressions A FEW EXAMPLES OF PROBLEM SOLVING IN CHEMICAL KINETICS From the point of view of coursework projects the detailed analysis described above is required, but quite often in examination questions a very limited amount of data is given and some clear logical thinking is required. So simplified rate data questions and their solutions avoiding graphical analysis are given below. These examples do NOT involve graphs directly, but a 'graphical' section of examples has been added in section 5.3e. I've made the numbers quite simple to follow the logic of the argument. I've also shown how to calculate the rate constant. These example calculations below are based on the initial rate of reaction analysis  so we are assuming the variation of concentration with time for each experimental run has been processed in some way e.g. by graphical analysis, to obtain an initial rate of reaction. The graph on the left illustrates the initial rate method for the formation of product. The gradients A and B would be for two different concentrations of a reactant, the concentration for A would be greater than the concentration of B. The initial rate is taken as the positive tangent  gradient for the curve at the point 0,0. The same argument applies if you imagine the graph inverted and you were following the depletion of a reactant. Then you would get two negative gradients one steeper than the other for the greater concentration. Reminder [x] means concentration of x, usually mol dm^{3} Example 1. The table below gives some initial data for the reaction: A + B ===> products
From runs (i) and (ii), keeping [B] constant, by doubling [A], the rate is doubled, so 1st order with respect to reactant A. From runs (ii) and (iii), keeping [A] constant, by doubling [B], the rate is unchanged, so zero order with respect to reactant B. Therefore the reaction is 1st order overall, and the rate expression is ... rate = k [A] ([B] can be omitted, anything to the power zero is 1) To calculate the rate constant, rearrange the rate expression and substitute appropriate values into it. Therefore, using run (iii) ... k = rate constant = rate / [A] = 6.0 x 10^{4}/0.2 = 3.0 x 10^{3} s^{1} In reality the results would be not this perfect and you would calculate k for each set of results and quote the average! Example 2. The table below gives some initial data for the reaction: A + B ===> products
From runs (i) and (ii), keeping [B] constant, by doubling [A], the rate is unchanged, so zero order with respect to reactant A. From runs (ii) and (iii), keeping [A] constant, by doubling [B], the rate is quadrupled, so 2nd order with respect to reactant B. Therefore the reaction is 2nd order overall, and the rate expression is ... rate = k [B]^{2} ([A] can be omitted, anything to the power zero is 1) To calculate the rate constant, rearrange the rate expression and substitute appropriate values into it. Therefore, using run (iii) ... k = rate constant = rate / [B]^{2} = 1.2 x 10^{3}/0.2^{2} = 3.0 x 10^{2} mol^{1} dm^{3} s^{1} In reality the results would be not this perfect and you would calculate k for each set of results and quote the average! Example 3. The table below gives some initial data for the reaction: A + B ===> products
From runs (i) and (ii), keeping [B] constant, by doubling [A], the rate is doubled, so 1st order with respect to reactant A. From runs (ii) and (iii). keeping [A] constant, by doubling [B], the rate is doubled, so 1st order with respect to reactant B. Therefore the reaction is 2nd order overall, and the rate expression is ... rate = k [A] [B] To calculate the rate constant, rearrange the rate expression and substitute appropriate values into it. Therefore, using run (iii) ... k = rate constant = rate / ([A][B] = 8 x 10^{3}/(2.0 x 2.0) = 2 x 10^{3} mol^{1} dm^{3} s^{1} In reality the results would be not this perfect and you would calculate k for each set of results and quote the average! Example 4. The following rate data was obtained at 25^{o}C for the reaction: A + 2B ==> C
(a) Deduce the order of reaction with respect to reactant A. Compare expts. 2 and 3. [B] is kept constant, but on halving A the rate is reduced by a factor of ^{1}/_{4} (^{1}/_{2} x ^{1}/_{2}) so the rate is proportional to [A]^{2}. Therefore the order with respect to A is 2 or 2nd order. You can think the other way round i.e. from expt. 3 to 2, [A] is doubled and the rate quadruples. (b) Deduce the order of reaction with respect to reactant B. Compare expts. 1 and 2. [A] is kept constant but doubling [B] doubles the rate, so the reaction is directly proportional to [B]. Therefore the order with respect to B is 1 or 1st order. Similarly, comparing expts. 1 and 3 (2x [B] gives 2x rate) or comparing expts. 1 and 4 (4x [B] gives 4x rate). (c) What is the overall order of the reaction between A and B? Total order = 2 + 1 = 3, 3rd order overall for the reaction as a whole. (d) Write out the full rate expression. rate = k [A]^{2} [B] (e) Calculate the value and units of the rate constant. rearranging rate = k [A]^{2} [B] gives k = rate / [A]^{2} [B] and the units will be mol dm^{–3}s^{–1 }/ (mol dm^{–3})^{2}(mol dm^{–3}) = s^{–1} / (mol dm^{–3})^{2} = mol^{–2} dm^{6} s^{–1} so, using the data from expt. 1 (or any set, assuming data perfect) gives k = 0.02 x 10^{2} / (0.1^{2} x 0.05) = 4 x 10^{3} mol^{–2} dm^{6} s^{–1} or expt. 3 gives 0.01 x 10^{2} / (0.05^{2} x 0.1) = 4 x 10^{3} etc. check the same for expts. 2 and 4. (hope they work out ok) Its not a bad idea to repeat the calculation with another set of data as a double check! (f) What will be the rate of reaction if the concentration of A is 0.20 mol dm^{–3} and the concentration of B is 0.30 mol dm^{–3}? You just substitute the values into the full rate expression: rate = k [A]^{2} [B] = 4 x 10^{3} x 0.2^{2} x 0.3 = 0.48 x 10^{2} mol dm^{–3 }s^{–1} Note: The reacting mole ratio is 2 : 1 BUT that does not mean that the orders are a similar ratio (since here, it happens to be the other way round for the individual orders). Orders of reaction can only be obtained by direct experiment and their 'complication' are due to complications of the actual mechanism, which can be far from simple. 5.3e More datagraph work  deducing orders of a reaction by graphical analysis Example 1. The inversion (hydrolysis) of sucrose sucrose + water ===> glucose + fructose
In simple terms: C_{12}H_{22}O_{12} + H_{2}O ====> C_{6}H_{12}O_{6} + C_{6}H_{12}O_{6} Some rate data for the inversion of sucrose is given below.
A linear graph, the gradient of the graph of concentration versus time does not change, therefore a zero order reaction. Therefore the rate expression is ... rate = k (mol dm^{–3} s^{–1}) A nonlinear graph of concentration versus time would suggest first or second order kinetics. This zero order reaction occurs when the enzyme (invertase) concentration is low and the substrate (sucrose) concentration is high. The maximum number of enzyme sites are occupied, which is itself a constant at constant enzyme concentration. For more details see section 7.3 Enzyme kinetics Example 2. Analysing a single set of data to deduce the order of reaction The data below are for the hydrolysis of 2–chloro–2–methylpropane in an ethanol–water mixture. The water concentration is effectively constant. The reaction is: 2–chloro–2–methylpropane + water ===> 2–methyl–propan–2–ol + 'hydrochloric acid'
A graph is drawn of (CH_{3})_{3}CCl concentration versus time.
From the graph the gradient (relative rate) was measured at 6 points. Since the gradient (rate) changes with concentration, it cannot be a zero order reaction.
A graph is then drawn of rate versus concentration of (CH_{3})_{3}CCl (RX)
The graph is 'reasonably linear' suggesting it is a 1st order reaction. Therefore the rate expression is ... rate = k[(CH_{3})_{3}CCl]
To put this graph in perspective, a 2nd order plot is done below of rate versus [RX]^{2}. (see data table above)
Wherever you draw a straight line, the data does not express itself as a linear plot and cannot be a 2nd order reaction.
Nucleophilic substitution by water/hydroxide ion [S_{N}1 or S_{N}2, hydrolysis to give alcohols] for a wider ranging mechanistic and kinetic analysis of the hydrolysis of haloalkanes. Example 3. Kinetics of the thermal decomposition of hydrogen iodide A single set of reaction rate data at a temperature of 781K Reaction: 2HI_{(g)} ==> H_{2(g)} + I_{2(g)}
The concentration of hydrogen iodide was measured every 500 seconds for 4000 seconds.
A plot of HI concentration versus time (above) was curved showing it could not be a zero order reaction with respect to the concentration of HI. From the [HI] versus time graph (above) the tangent angle (gradient = rate) was accurately measured from 500 s to 3500 s at 500 second intervals (results tabulated above).
The rate of reaction was is then plotted against HI concentration to test for 1st order kinetics. This proved to be a curve  compare the blue rate data curve with the black 'best straight line' (courtesy of Excel!), so not 1st order either.
The third graph is a plot of HI decomposition rate versus [HI] squared, and, proved to linear  the blue data line was pretty coincident with a black 'best straight line'. This proved that the decomposition of hydrogen iodide reaction is a 2nd order reaction. Therefore the rate expression is ... rate = k [HI]^{2} For more on this reaction see KINETICS section 7.4
Advanced A Level Kinetics Index
Advanced Level Theoretical Physical Chemistry of chromium – A level Revision notes to help revise for GCE Advanced Subsidiary Level AS Advanced Level A2 IB Revise AQA GCE Advanced Level Chemistry OCR GCE Advanced Level Chemistry Edexcel GCE Advanced Level Chemistry Salters AS A2 Chemistry CIE Chemistry, WJEC GCE AS A2 Chemistry, CCEA/CEA GCE AS A2 Chemistry revising courses for pre–university students (equal to US grade 11 and grade 12 and AP Honours/honors level courses) Experimental techniques to get rate data and how to derive rate expressions how to deduce orders of reactions with some worked out questions involving deducing and using orders of reactants and rate expressions. 
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