Question

1.) For the reaction D → A + C when 1/[D] is plotted versus the time in seconds, a straight line is obtained whose slope is 0.054 M-1s-1. What is the concentration of D (in M) after 36.0 s if [D]o = 0.80 M?

2.) For the reaction A → B + C when [A] is plotted versus the time in minutes a straight line is obtained whose slope is -0.021 M/min. What is the concentration of A (in M) after 10.00 min if [A]o = 0.80 M?

3.) For the reaction A → B + C, when the natural log of [A] is plotted versus the time in seconds a straight line is obtained whose slope is -0.033 s-1. What is the concentration of A (in M) after 58.0 s if [A]o = 0.60 M?

Answer 1

1) At, time, t=0, 1/[D] = 0.8 M -1

=> [D] = 1.25 M

Lets assume at time 36s, 1/[D] = y2

slope, m = (y2-y1)/(x2-x1)

=> 0.054 = (y2-1.25)/(36-0)

=> y2 = (0.054*36+1.25) M-1 = 3.194 M-1

So, at 36 s, the concentration of D will be 1/3.194 M = 0.313

2) At, time, t=0, [A] = 0.8 M

Lets assume at time 10 min, [D] = y2

slope, m = (y2-y1)/(x2-x1)

=> -0.021 = (y2-0.8)/(10-0)

=> y2 = (-0.021*10+0.8) M = 0.59 M

So, after 10mins the concentration of A will be 0.59 M.

3) At, time, t=0, [A]o = 0.6 M

Lets assume at time 10 min, A = A2

slope, m = (logA2-logy1)/(x2-x1)

=> -0.033 = log(A2/0.6)/(58-0)

=> log(A2/0.6) = (-0.033*58) = -1.914

=> A2/0.6 = 0.012189

=> A2 = 0.0073 M

So, after 58s the concentration of A will be 0.0073 M.