1.) For the reaction D A + C when 1/[D] is plotted versus the time in seconds, a
ID: 981530 • Letter: 1
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?
Explanation / Answer
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.
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