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738 2020202020 PHYSICAL CHEMISTRY
Differences Between Order and Molecularity
PSEUDO–ORDER REACTIONS
A reaction in which one of the reactants is present in a large excess shows an order different from the actual order. The experimental order which is not the actual one is referred to as the pseudo order. Since for elementary reactions molecularity and order are identical, pseudo-order reactions may also be called pseudo molecular reactions. Let us consider a reaction A + B ⎯⎯→^ products
in which the reactant B is present in a large excess. Since it is an elementary reaction, its rate law can be written as
rate = k [A] [B] As B is present in large excess, its concentration remains practically constant in the course of reaction. Thus the rate law can be written as
rate = k ′ [A]
where the new rate constant k ′ = k [B]. Thus the actual order of the reaction is second-order but in practice it will be first-order. Therefore, the reaction is said to have a pseudo-first order.
Examples of Pseudo-order Reactions
(1) Hydrolysis of an ester. For example, ethyl acetate upon hydrolysis in aqueous solution using a mineral acid as catalyst forms acetic acid and ethyl alcohol. CH 3 COOC 2 H 5 + H 2 O ⎯⎯→ CH 3 COOH + C 2 H 5 OH ethyl acetate (excess) acetic acid ethyl alcohol Here a large excess of water is used and the rate law can be written as rate = k [CH 3 COOH] [H 2 O] = k ′ [CH 3 COOH] The reaction is actually second-order but in practice it is found to be first-order. Thus it is a pseudo-first order reaction.
(2) Hydrolysis of sucrose. Sucrose upon hydrolysis in the presence of a dilute mineral acid gives glucose and fructose. C 12 H 22 O 11 + H 2 O ⎯⎯→ C 6 H 12 O 6 + C 6 H 12 O 6 sucrose (excess) glucose fructose If a large excess of water is present, [H 2 O] is practically constant and the rate law may be written as
Order of a Reaction Molecularity of a Reaction
- It is the sum of powers of the 1. It is number of reacting species concentration terms in the rate law undergoing simultaneous collision in expression. the elementary or simple reaction.
- It is an experimentally determined value. 2. It is a theoretical concept.
- It can have fractional value. 3. It is always a whole number.
- It can assume zero value. 4. It can not have zero value.
- Order of a reaction can change with the 5. Molecularity is invariant for a chemical conditions such as pressure, temperature, equation. concentration.
CHEMICAL KINETICS 739
rate = k [C 12 H 22 O 11 ] [H 2 O] = k [C 12 H 22 O 11 ] The reaction though of second-order is experimentally found to be first-order. Thus it is a pseudo- first-order reaction.
ZERO ORDER REACTIONS
In a zero order reaction, rate is independent of the concentration of the reactions. Let us consider a zero-order reaction of the type
A ⎯⎯→^ Products Initial conc. a 0 Final conc. a – x x
0
– [ ]
Rate of reaction [ ] d A k A dt
or (^0 )
dx d a x k a x k dt dt
On integrating we get
0
x k t
= or x = k 0 t
where k 0 is the rate constant of a zero-order reaction, the unit of which is concentration per unit time. In zero order reaction, the rate constant is equal to the rate of reaction at all concentrations.
FIRST ORDER REACTIONS
Let us consider a first order reaction A ⎯⎯→^ products Suppose that at the beginning of the reaction ( t = 0), the concentration of A is a moles litre–1. If after time t , x moles of A have changed, the concentration of A is a – x. We know that for a first order reaction, the rate of reaction, dx / dt , is directly proportional to the concentration of the reactant. Thus,
( – )
dx k a x dt
or
dx k dt a x
Integration of the expression (1) gives
dx k dt a x
or – 1n ( a – x ) = kt + I ...(2)
where I is the constant of integration. The constant k may be evaluated by putting t = 0 and x = 0.
Thus, I = – 1n a Substituting for I in equation (2)
1n
a kt a x
or
1n
a k t a x
CHEMICAL KINETICS 741
SOLVED PROBLEM. From the following data for the decomposition of N 2 O 5 in CCl 4 solution at 48°C, show that the reaction is of the first order t (mts) 10 15 20 ∞ Vol of O 2 evolved 6.30 8.95 11.40 34. SOLUTION For a first order reaction the integrated rate equation is 1 log
V
k t V V
∞ ∞
In this example, V ∞ = 34.
t V ∞– Vt
log
V
t V V
∞ ∞
= k
log 10 28.
log 15 25.
log 20 23.
Since the value of k is fairly constant, it is a first order reaction.
(2) Decomposition of H 2 O 2 in aqueous solution. The decomposition of H 2 O 2 in the presence of Pt as catalyst is a first order reaction. Pt H O 2 2 ⎯⎯→ H O 2 + O The progress of the reaction is followed by titrating equal volumes of the reaction mixture against standard KMnO 4 solution at different time intervals.
SOLVED PROBLEM. A solution of H 2 O 2 when titrated against KMnO 4 solution at different time intervals gave the following results : t (minutes) 0 10 20 Vol KMnO 4 used for 10 ml H 2 SO 4 23.8 ml 14.7 ml 9.1 ml Show that the decomposition of H 2 O 2 is a first order reaction. SOLUTION The integrated rate equation for first order reaction is
= log
a k t a x Since volume of KMnO 4 used in the titration is measure of concentration of H 2 O 2 in solution, a = 23.8 ml ( a – x ) = 14.7 when t = 10 mts ( a – x ) = 9.1 when t = 20 mts Substituting these values in the rate equation above, we have 2.303 23. = log 10 14.
k
= 0.2303 (log 23.8 – log 14.7)
742 2020202020 PHYSICAL CHEMISTRY
and
= log 20 9.
k
= 0.10165 (log 23.8 – log 9.1) = 0.10165 (1.3766 – 0.9595) = 0. Since the value of k is almost constant, the decomposition of H 2 O 2 is a first order reaction.
(3) Hydrolysis of an Ester. The hydrolysis of ethyl acetate or methyl acetate in the presence of a mineral acid as catalyst, is a first order reaction.
CH 3 COOC 2 H 5 + H 2 O H
⎯⎯→ CH 3 COOH + C 2 H 5 OH ethyl acetate acetic acid For studying the kinetics of the reaction, a known volume of ethyl acetate is mixed with a relatively large quantity of acid solution, say N/2 HCl. At various intervals of time, a known volume of the reaction mixture is titrated against a standard alkali solution. Hydrolysis of the ester produces acetic acid. Therefore as the reaction proceeds, the volume of alkali required for titration goes on increasing.
SOLVED PROBLEM. The following data was obtained on hydrolysis of methyl acetate at 25°C in 0.35N hydrochloric acid. Establish that it is a first order reaction. t (secs) 0 4500 7140 ∞ ml alkali used 24.36 29.32 31.72 47. SOLUTION For a first order reaction,
log
a k t a x
At any time, the volume of alkali used is needed for the acid present as catalyst and the acid produced by hydrolysis. The volume of alkali used for total change from t 0 to t ∝ gives the initial concentration of ester. Thus, a = 47.15 – 24.36 = 22.79 ml ( a – x ) after 4500 sec = 47.15 – 29.32 = 17.83 ml ( a – x ) after 7140 sec = 47.15 – 31.72 = 15.43 ml Substituting values in the rate equation above, we have 2.303 22. log 0. 4500 17.
k = =
2.303 22. log 0. 7140 15.
k = =
Since the values of k in the two experiments are fairly constant, the reaction is of the first order.
(4) Inversion of Cane sugar (sucrose). The inversion of cane sugar or sucrose catalyzed with dil HCl,
H+ C 12 H 22 O 11 + H O 2 ⎯⎯→ C H 6 12 O 6 + C 6 H 12 O 6 D-glucose D-fructose
follows the first order kinetics. The progress of the reaction is followed by noting the optical rotation of the reaction mixture with the help of a polarimeter at different time intervals. The optical rotation
