Elementos del sistema de gestión

Determination o f rate equation from initial rates Consider the reaction:

A + B+ C ^� products

The initial rates of reaction with different concentrations of A, B and C are found, and by taking the experiments two at a time, the order with respect to each substance can be found.

Look for two experiments where the concentration of only one substance varies. If doubling that concentration causes the rate to double, the order with respect to that substance is 1 .

Note that for the first-order graph, the half-life remains constant throughout the reaction, whereas for the second-order the half-life increases as the concentration falls.

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From experiments 1 and 2: [A] doubles and rate doubles. Therefore order with respect to A = 1

From experiments 1 and 3: [B] doubles and rate x 4. Therefore order with respect to B = 2

From experiments 3 and 4: [C] doubles and rate unaltered. Therefore order with respect to C = 0

Rate of reaction = k [Af [Bf [C]0, and the reaction is 3rd order.

Experimental techniques for following a reaction 1 Titration method. Mix the chemical and start the clock. t inten·al

withdraw a sample, add it to i ed water to low the reaction. Then titrate one of the ub tan e in the reaction. Thi method can u ed if an acid, an alkali or iodine i a reactant or a product.

2 Colorimetric method. This can be used when either a r actant or a product is coloured (iodine or potassium manganate VII) are example ).

The colorimeter must first be calibrated using solutions of the coloured substance of known concentrations. Then the reactants are mixed and the clock started. The intensity of the colour is measured as a function of time. The concentration of the coloured substance is proportional to the amount of light absorbed.

3 Gas volume method. If a gas is produced in the reaction, its volume can be measured at intervals of time. The gas can be collected in a

horizontal gas syringe or by bubbling it into an inverted measuring cylinder filled with water.

Graphs of results

If the concentration of a reactant is plotted against time, three shapes of graph are likely depending on the order of the reaction.

Mechanisms

A suggested mechanism must be consistent with the order of reaction. The partial order of any species which occurs in the mechanism after the rate determining step will be zero. The rate-determining step is the slowest step, e.g.

As the CO enters the mechanism after the rate determining step, the 1nechanism is consistent with the fact that the reaction is zero order with respect to CO.

T R A N S I T I O N M E TA L S , Q U A N T I TAT I V E K I N E T I C S A N D A P P L I E D O R G A N I C C H E M I S T R Y

Energy profile diagrams

The first diagram is for a reaction that takes place in a single step, and the second diagram is for this reaction with a catalyst (see Figure 5.6).

- - - Transition state

Intermediate

Uncatalysed Fig 5. 6 Energy profile diagrams

Catalysed

An example of a reaction with a transition state is the S 2 reaction of bromoethane with hydroxide ions.

An example of a catalysed reaction forming an intermediate is Fe2+ ions as a catalyst in:

S20t(aq) + Zr(aq) - 2SO/-(aq) + 12(aq) In this mechanism the catalyst first reduces the S2082- ions:

2Fe2+(aq) + S2082-(aq) - 2Fe3+(aq) + 2SO/-(aq) then the Fe3+ ions are reduced back to Fe2+ ions:

2Fe3+(aq) + Zr(aq) - 2Fe2+(aq) + 12(aq)

Checklis t

Before attempting the questions on this topic, check that you:

Can define rate constant, order of reaction and half-life.

Can deduce rate equations from initial rate data.

Understand the concept of activation energy and its relation to the rate constant.

Understand that infonnation about mechanisms can be deduced from the partial orders of the reactants.

Can recall the energy profiles of reactions with and without catalysts.

Can suggest suitable methods for following reactions.

Can deduce the order of a reaction from concentration/time graphs.

a From experiments 1 and 2, order with respect to N02 ⁼ 2

b From experiments 1 and 3, order with respect to CO ⁼ 0

c Order of reaction is 2 + 0 ⁼ 2 d Rate ⁼ k [N02]2

a Reaction 8 b Reaction 8

Step 2 as it has the largest fact val ue

1 00 to 6.25 ⁼ 4 half-lives, therefore time ⁼ 4 x 25 ⁼ 1 00 m i n utes

The answers to the num bered q uestions are on page 1 38.

� Tes ting your knowledge and understanding

For the following set of questions, cover the margin before you answer, then check to see if you are correct.

The following results were obtained from a study of the reaction:

NOz(g) + CO(g) - NO(g) + C02(g)

Experiment [N02]/mol dm-3 [CO]/mol dm-3

1 0.02 0.02

2 0.04 0.02

3 0.02 0.04

a What is the order with respect to N02?

b What is the order with respect to CO?

c What is the order of reaction?

d State the rate expre sion.

Relative rate 1 4 1

Reaction ha a high ,·alue of E , and rea ti n B ha a lower Ea ' alue.

a Which reaction has the larg r rate con tant?

b Which is the faster reaction?

A reaction takes place in 3 steps. The EalkJ 1nor1 for each tep are:

step 1: 32, step 2: 5 1 , step 3: 20 Which step determines the rate?

The decomposition of N205 is first order. At 200 °C the reaction has a half-life of 25 minutes. Calculate how long it wil l take for the concentration of N205 to fall to 6.25% of its original value.

1 The rate of the second order reaction :

2HI(g) - H2(g) + I2(g)

is 2.0 x 10-4 mol dm-3 s-1 when [HI] = 0.050 mol dm-3 at 785 K. Calculate the value of the rate constant, giving its units.

2 Describe how you would follow, at 60 °C, the rate of the reaction:

CH3COOH(aq) + CH30H(aq) - CH3COOCHll) + H20(1).

3 The decomposition of 3-oxobutanoic acid, CH3COCH2COOH, was studied:

CH3COCH2COOH - CH3COCH3 + C02 The results, at 40 °C, are tabulated below.

Time/min [3-oxobutanoic acid]/mol dm-3

0 1 .6

26 0.8

52 0.4

78 0.2

Deduce the order of the reaction.

T R A N S I T I O N M E TA L S , Q U A N T I TAT I V E K I N E T I C S A N D A P P L I E D O R G A N I C C H E M I S T R Y