Question

27) (14 pts) Two reactions #1 and #2 that can be represented as A 2B were followed as shown in the table below. Time, min 20 80 100 C 10 30 40 50 60 Reaction #1 [A], M 1.6001290 1081 0.930 0.816 0.727 0,656 0.548 0.471 [A], M 1.600 1.345 1131 0951 0.800 0.673 0566 0.400 0283 Reaction #2 A) (4 pts) Determine the reaction order for each reaction. B) (4 pts) Determine the value and units for the rate constant for each reaction. C) (6 pts) If the initial concentration of B was 0.0 M, how many minutes does it take for [B] to reach 2.80 M in reaction #1? Show your work ool(0) poJ.s0or) no (000) 00 E(100 0) gol S 0.0 gol h(t 0) go :TS TASHO ayorTAUO VI0 V enutehomet mo01l D ol x(82)a 82 V Kd belcil Idbyselbclo 8 XIMA W HOL

Answer 1

For reaction 1: A plot of time versus 1 / [A] is made, and I throw a line, it means that the reaction fulfills order 2, where the constants is:

1/[A]1 Time [A]1 2,5 0 1,6 0,625 y 0,015x 0,6255 0,7751938 10 1,29 20 1,081 0,92506938 2 0,93 1,07526882 30 0,816 40 1,2254902 1,5 50 0,727 1,37551582 Series1 0,656 1,52439024 60 - Lineal (Series1) 0,548 1,82481752 80 1 100 0,471 2,12314225 0,5 0 20 60 80 100 120 40 이음88

k = m = 0.015 L / mol * min

For reaction 2: A plot of time versus ln [A] is drawn and a line is drawn, which means that the reaction fulfills order 1, where the constant is:

Time In [A]2 [A]2 0,6 0 1,6 0,47000363 0,4 10 1,345 0,29639401 y 0,0173x+0,4697 20 1,131 0,1231022 0,2 0,951 -0,05024122 30 0,8-0,22314355 40 20 60 80 100 120 0,673 -0,39600995 -0,2 50 - Series1 0,566 0,5691612 -0,4 0,4 -0,91629073 0,6 60 Lineal (Series1) 80 100 0,283 -1,26230838 -0,8 -1 -1,2 -1,4 40

k = - m = - (- 0.0173) = 0.0173 s-1

c) For these concentrations of B to be met, then [Ao] = 1.6 M and [A] = 0.2 M, by means of the reaction equation of order 2, it is calculated:

1 / [A] = k * t + 1 / [Ao]

it is replaced:

1 / 0.2 = 0.015 * t + 1 / 1.6

It clears:

t = 291.67 min

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