Question

A certain reaction has the following general form:

aA --> bB

At a particular temperature and [A]o = 0.100 M concentration versus time data were collected for this reaction and a plot of 1/[A] vs. time resulted in a straight line with a slope value of +4.15 x 10-3 L/mol.s.

a. Determine the rate law, the integrated law, and the value of the rate constant for this reaction.

b. Calculate the half life for this reaction.

c. How much time is required for this reaction to be 75% complete?

Answer 1

Ans.

#1A. Zero order kinetics-

                                                [A]_t = - kt + [A]₀      

- in form of      Y =   mx + C

A y= mX + C equation gives a linear graph.

Plotting for [A] vs time gives straight line.

#1B. First order kinetics-

ln [A]_t = - kt   + ln [A]₀

                in form of Y         =   mX + C

Plotting for ln[A] vs time gives straight line.

#1C. Second order kinetics-

                                    (1/ [A]_t) = - kt + (1/[A]₀)  

            - in form of        Y      = mx +    C

Plotting for (1/ [A]) vs time gives straight line.

#A. Comparison: Given that plotting for (1/ [A]) vs time gives straight line. So, reaction exhibits second order kinetics because a second order gives a straight line for plotting (1/ [A]) vs time.

Reaction is second order.

Rate law is given by - (1/ [A]_t) = kt + (1/[A]₀)  

Calculating rate constant: In the equation “y = mX + C ”, m = slope.

In (1/[A]) vs t graph, the slope m = k = rate constant.

Given, slope, m = 4.5 x 10^-3 L/(mol.s)

So, rate constant, k = slope, m = 4.5 x 10^-3 L/(mol.s)

                                    = 4.5 x 10^-3 M^-1 s^-1             ; [ M = mol/ L ; thus, L/mol = M^-1]

#B. Half- life of a second order reaction if given by-

            t_1/2 = 1 / (k [A]₀)

                        where,

                                    t_1/2 = half-life

                                    k = rate constant

                                    [A]₀ = Initial concertation

Putting the values in above equation-

            t_1/2 = 1 / [ (4.5 x 10^-3 M^-1 s^-1) x 0.100M] = 1/ (4.5 x10^-4 s^-1) = 2.22 x 10³ s

Thus, t_1/2 = 2.22 x 10³ s

#C. Let the time required for 75% completion = t

At 75% completion, 75% reactant (here, [A]) is converted into product. Only 25% (100% - 75% = 25%) [A] remains after time t (of 75% completion).

That is,

            [A]₀ at time t0 = 0.100 M                                        ; [A]₀ = [A] at time, t=0

            [A]_t at time t = 25% of 0.100M = 0.025 M           ; [A]_t = [A] at time, t=t

Putting the values in rate law equation-

            (1/ [A]_t) = kt + (1/[A]₀)                                        - for product formation

                or, (1/ 0.025M) = (4.5 x 10^-3 M^-1 s^-1) t + (1 / 0.100 M)

            or, (40 – 10) M^-1 = (4.5 x 10^-3 M^-1 s^-1) t

            or, t = 30 / (4.5 x 10^-3 s^-1) = 6.66 x 10³ s

Thus, required time, t = 6.66 x 10³ s