Question

1. For the reaction of 2A ---> B + C, the following data are obtained for [A] as a function of time
   
   time (min)             [A] (mol L^-1)
   0                            0.80
   8                            0.60
   24                          0.35
   40                          0.20
   
   a). Determine the order and the rate law of the reaction. (Hint: Consider the integrated rate laws shown on the first page).
   b). Calculate the value of the rate constant k
   

Answer 1

Answer:

Given reaction is 2A ----> B + C

(a) The integrated rate laws for zero order reaction is [A]= -kt + [A]₀ and the plot [A] vs time is linear and slope=-k.

For first order reaction, ln[A]= -kt + ln[A]₀ and the plot ln[A] vs time is linear and slope=-k.

For second order reaction, 1/[A]= kt + 1/[A]₀ and the plot 1/[A] vs time is linear and slope=k.

Where k=rate constant, [A]₀=initial concentration of A and [A]=concentration of A at time t.

The required calculations are tabulated below

Time(min)	[A] (mol/L)	ln[A]	1/[A]
0	8.00E-01	-0.223143551	1.25
8	6.00E-01	-0.510825624	1.666666667
24	3.50E-01	-1.049822124	2.857142857
40	2.00E-01	-1.609437912	5

The zero-order plot, [A] vs time is shown below

The first order plot, ln[A] vs time is shown below

The second-order plot, 1/[A] vs time is shown below

From the above plots, the plot ln[A] vs time shows a straight line with R²=0.9998.

Therefore the order of the reaction is the first-order reaction.

The rate law is

Rate=k[A]¹=k[A]

(b) From the plot, slope= -k = -0.0345 min^-1

Therefore rate constant, k=0.0345 min^-1 or 3.45 x 10^-2 min^-1.

Please let me know if you have any doubt. Thanks.