The first order reaction, SO_2Cl_2 rightarrow SO_2 + Cl_2, has a half-life of 8.

Question

The first order reaction, SO_2Cl_2 rightarrow SO_2 + Cl_2, has a half-life of 8.75 hours at 593 hours K. How long will it take for the concentration of SO_2Cl_2 to fall to 12.5% of its initial value? 26.2 hr 0.165 hr 6.06 hr 3.22 hr A plot of 1/[BrO^-] time is liner for the reaction: 3 BrO-(aq) rightarrow BrO_3^-(aq) + 2 Br^- (aq) What is the order of the reaction with respect to the hypobromite ion. BrO^-? 0 1 2 Hydrogen iodide decomposes at 800 K via a second-order process to produce hydrogen and iodine according to the following chemical equation. 2 HI(g) rightarrow H_2(g) + I_2(g) At 800 K it takes 142 seconds for the initial concentration of HI to decrease from 6.75 x 10^-2 M to 3.50 10^-2 M. What is the rate constant for the reaction at this temperature? 10.3 M^-1s^-1 1.95 x 10^3 M^-1s^-1 5.12 x 10^-4 9.69 x 10^-2 M^-1s^-l

Explanation / Answer

Ans. 8. Half- life of a second order reaction if given by-

            t1/2 = 0.693 / k                     - equation 1

                        where,

                                    t1/2 = half-life

                                    k = rate constant

Putting the values in equation 1-

            8.75 hr = 0.693 / k

            Or, k = 0.693 / 8.75 hr = 0.000792 hr-1

Thus, rate constant for the reaction = 0.000792 hr-1

Step 2: Now, using First order kinetics-

ln [A]t = - kt   + ln [A]0                        -equation 2

where, [A]t = Final concentration of reaction

            [A]0 = Initial concentration of reactant

            k = rate constant

            t = time of reaction

Let [A]0 = 100.

Then, [A]t = 12.5 % of 100 = 12.5

Putting the values in equation 2-

            ln (12.5) = - (0.000792 hr-1) t + ln 100

or, 2.5257 = - (0.000792 hr-1) t + 4.6051

or, 2.5257 – 4.6051 = - (0.000792 hr-1) t

or, t = -2.0794/ (0.000792 hr-1)

or, t = 2625.50 hr

Thus, required time = 2625.50 hr

#9. 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 with intercept of (1/[A]0).          

Thus, the reaction is second order.

Correct option. C. 2.

#10. Second order kinetics-

                                    (1/ [A]t) = - kt + (1/[A]0)               - equation 3

Putting the values in equation 3-

            [1/ (3.5 x 10-2 M)] = - k (142 s) + [ 1/ (6.75 x 10-2 M)]

            Or, 28.571 M-1 = - k (142 s) + 14.815 M-1

            Or, (28.571 – 14.815) M-1 = - k (142 s)

            Or, k = - (13.756 M-1 / 142s)

            Or, k = - 0.09687 M-1s-1

            Or, k = -9.69 x 10-2 M-1s-1

Thus, rate constant =-9.69 x 10-2 M-1s-1.

Note: the -ve sign indicates that concentration of reactant decrease with time.

Correct option. D.

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