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A reaction is first order in A and second order in B
(i) Write differential rate equation.
(ii) How is the rate affected when the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?

(i) Differential rate equation of reaction is

dxdt=k[A]1[B]2 = k[A] [B]2

(ii)  When conc. of B is tripled, it means conc. of B becomes [3 x B]
∴  New rate of reaction,

dx'dt=k[A] [3B]2 = 9k[A] [B]2 = 9dxdt
i.e., rate of reaction will become 9 times.


(iii)  When conc. of A is doubled and that of B is also doubled, then conc. of A becomes [2A] and that of B becomes [2B] rate of reaction
dx''dt=k[2A] [2B]2 = 8k[A] [B]2
i.e., the rate of reaction will become 8 times the rate as in (1).
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A reaction is second order with respect to a reactant. How is the rate of rection affected if the concentration of the reactant is
(i) Doubled
(ii) Reduced to 1/2?


Let the reaction A → B is a 2nd order reaction w.r.t A and conc. of A is ‘a’ mol/L, then rate of reaction can be written as:

dxdt=k[A]2 =ka2

(i) When conc. [A] is doubled
i.e., [A’] = 2a mol/L
Then new rate of reaction

d'(x)dt=k[2a]2 = 4ka2 = 4dxdt

Thus rate of reaction will become four times where concentration is doubled.


(ii) Similarly, when conc. of A is reduced to 12 i.e., [A] is a/2 then new rate of reaction,

d'(x)dt = ka22 = 14ka2 = 14dxdt

The rate of reaction will become one-fourth of the initial rate of reaction.


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The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

exp.

 

[A]/
mol L–1

[B]/M

Initial rate of formation
of D/mol L–1 min–1

I

0.1

0.1


6.0 x 10–3

II

0.3

0.2

7.2 x 10–2

III

0.3

0.4

2.88 x 10–1

IV

0.4

0.1

2.40 x 10–2


Determine the rate law and the rate constant for the reaction.

Let rate law of reaction be,

rate = kAx By

where x and y are order of reaction w.r.t. A and B respectively.

From experiments I and IV, we can write
6.0 x 10–3 = k[0.1][0.1]y ...(i)
2.4 x 10–2 = k[0.4][0.1]y ...(ii)

Dividing (ii) by (i), we can write

               2.4×10-26.0×10-3 = 0.40.1x = 4x

or                 4=4x      or      x = 1.

   From experiments I and III, we can write
                   7.2×10-2 = k0.3x 0.2y              ...(iii)2.88×10-1 = k0.3x 0.4y            ...(iv)

Dividing (iv) by (iii), we can write

                2.88×10-17.2×10-1 = 0.40.2y = 2y
or                        4=2y
or                        y=2.
               Rate = kA B2

Order of reaction w.r.t. A = 1
Order of reaction w.r.t. B = 2
Overall order of reaction = 1 + 2 = 3.
Substituting the values of initial rate of formation of experiment (I) (equ. V) we can write 6.0 x 103 mol L–1 min–1.
= k[0.1 mol L–1] [0.1 mol L–1]2

or      
                 k=6.0×10-3M min-11×10-3M3   = 6 M-2 s-1.
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In a pseudo first order hydrolysis of ester in water the following results were obtained:

t / s

0

30

60

90

Ester / mol L–1

0.55

0.31

0.17


0.085

 

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

(i) Average rate of reaction between interval of time 30 to 60 second is given by
Average space rate space equals space fraction numerator increment straight x over denominator increment straight t end fraction equals fraction numerator straight C subscript 2 minus straight C subscript 1 over denominator increment straight t end fraction
space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 0.17 minus 0.3 over denominator 60 minus 30 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 0.14 over denominator 30 end fraction space equals space minus 0.00467
space space space space space space space space space space space space space space space space space space space space space space space equals space minus 4.67 space cross times space 10 to the power of negative 3 end exponent space Ms to the power of negative 1 end exponent
Minus sign shows that rate of reaction is decreasing with time as conc. of ester is decreasing with time.
(ii) Pseudo first order rate constant k is given by
straight k equals fraction numerator 2.303 over denominator straight t end fraction log fraction numerator straight a over denominator straight a minus straight x end fraction
where a is initial conc. and (a – x) conc. after time t. Here a = 0.55 M.

(i) Average rate of reaction between interval of time 30 to 60 second
The nearly equal values of k confirms that reaction is of first order.
The actual value of rate constant is the average of three values of k.
Therefore, rate constant of reaction
straight k equals fraction numerator straight k subscript 1 plus straight k subscript 2 plus straight k subscript 3 over denominator 3 end fraction
space space space equals fraction numerator left parenthesis 1.91 plus 1.96 plus 2.06 right parenthesis cross times 10 to the power of negative 2 end exponent over denominator 3 end fraction
space space space equals space 1.97 space cross times space 10 to the power of negative 2 end exponent space straight s to the power of negative 1 end exponent.
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In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:

A/mol L–1

0.20

0.20

0.40

B/mol L–1

0.30

0.10

0.05

r/mol L–1S–1

5.07 x 10–5

5.07 x 10–5

1.43 x 10–4

What is the order of the reaction with respect of A and B?


In a reaction A and B, Let order of reaction w.r.t. A is x and w.r.t. B is y. Then the rate of reaction can be written as

rate = k[A]x [B]y

From given table data, 1 and 2 we can write

5.07 x 10–5 = k[10.20]x [0.30]y ...(i)
5.07 x 10–5 = k[0.20]x [0.10]y ...(ii)

Dividing (ii) by (i), we get

5.07×10-55.07×10-5=  k0.20x 0.10yk[0.20]x 0.30y


or                1 = 0.100.30yor         y = 0

 From given table data, 2 and 3 we can write

      5.07 × 10-5 = k0.20x 0.20y                        = k0.20y × 1                  [  y= 1]    ...(iii)7.06 × 10-5 =  k[0.20]x [0.05]y                         = k0.40x × 1                                          ...(iv)                                  
 
Dividing (iv) by (iii), we get
       7.60 × 10-55.07×10-5 = k0.40xk0.20x = 0.40.2x = (2)2


or            (2)x = 3/2 = 1.5

or                    x=0.5

Thus the order of reaction w.r.t. A  is 12 and w.r.t. B is zero.
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