Exercise 15.54 with feedback Correct The half-life of a second order reaction is

Question

Exercise 15.54 with feedback Correct The half-life of a second order reaction is calculated by the formula The following reaction was monitored as a function of time: AB A + B A plot of 1/[AB] versus time yields a straight line with slope 5.4x10-2 (M s) From the equation, note that for a second order reaction, the half-life depends on the initial concentration. with slope 5.4× You may want to reference ( pages 568-575) section 15.5 while completing this problem Part D If the initial concentration of AB is 0.210 M, and the reaction mixture initially contains no products, what are the concentrations of A and B after 80 s? Express your answers numerically using two significant figures, separated by a comma Aj, [B] Submit My Answers Give Up Incorrect; One attempt remaining; Try Again Provide FeedbacK Continue

Explanation / Answer

Initial concentration [AB]o = 0.210 M

Time, t = 80 sec

Slope of 1/[AB] vs time = 5.4*10-2 (M.s)-1

We know that, the slope of the plot of a second order reaction (1/[AB] vs time) is the rate constant of the reaction, k.

So, k = 5.4*10-2 (M.s)-1

Using second order rate law:

1/[AB] = 1/[AB]o + k*t

1/[AB] = (1/0.210) + 5.4*10-2 * 80

= 4.7619 + 4.32

= 9.08

[AB] = 0.110 M

Concentration of AB converted to products = [AB]o - [AB]

= 0.210 - 0.110

= 0.10 M

So,

[A], [B] = 0.10 M , 0.10 M

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