Part D ? Item 9 The following reaction was monitored as a function of time: AB A

Question

Part D ?

Item 9 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 5.3x10 2 (M.s) You may want to reference (DA ages 568-575) section 15.5 while completing this problem. What is the half-life when the initial concentration is 0.59 M Express your answer using two significant figures 1/2 32 s Submit My Answers Give Up Correct The half-life of a second order reaction is calculated by the formula 1/2 k IAB From the equation, note that for a second order reaction, the half-life depends on the initial concentrat Part D If the initial concentration of AB is 0.220 and the reaction mixture initially contains no products, wha concentrations of A and B after 70 s Express your answers numerically using two significant figures, separated by a comma A1, DB M. Submit My Answers Give Up

Explanation / Answer

Solution :-

As we know that reaction is second order and the rate constant K = 5.3*10^-2 M-1 s-1

Initial concentration [A]o = 0.220 M

Time = 70 s

Final concetration of the A and B= ?

Lets first calculate the final concentration of the AB using the second order rate equation

1/[A]t = 1/[A]o + K*t

1/[A]t = [1/0.220] + 5.3*10^-2 M-1 s-1 * 70 s

1/[A]t = 8.2555

1/8.2555 = [A]t

0.1211 M= [A]t

So the concentration of the AB remain after time 70 s is 0.1211 M

Now lets calculate the change in the concentration

Change in concetration of AB = initial – final

                                                     = 0.220 M – 0.1211 M

                                                     = 0.0989 M

So the concentration of the A and B after 70 s is A = 0.0989 M and B= 0.0989 M

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