Determine which of these data sets represents a zero-order, which represents a f
ID: 512685 • Letter: D
Question
Determine which of these data sets represents a zero-order, which represents a first order, and which represents a second order reaction. For each compound, A, B and C:
Time
(min)
0.0
10.0
20.0
30.0
40.0
50.0
60.0
80.0
100.0
150.0
200.0
[A], M
2.5
2.25
2.07
1.94
1.79
1.69
1.61
1.49
1.28
1.06
[B], M
1.51
1.22
1.06
0.95
0.8
0.73
0.69
0.58
0.48
0.32
0.2
[C], M
3.25
3.05
2.85
2.65
2.45
2.24
2.11
1.78
1.43
0.72
0.05
Write the rate law and integrated rate law.
Determine the value of the rate constant, use appropriate units? Justify.
Calculate the half-life.
Determine the concentration of A, B, and C at 400 minutes.
Time
(min)
0.0
10.0
20.0
30.0
40.0
50.0
60.0
80.0
100.0
150.0
200.0
[A], M
2.5
2.25
2.07
1.94
1.79
1.69
1.61
1.49
1.28
1.06
[B], M
1.51
1.22
1.06
0.95
0.8
0.73
0.69
0.58
0.48
0.32
0.2
[C], M
3.25
3.05
2.85
2.65
2.45
2.24
2.11
1.78
1.43
0.72
0.05
Explanation / Answer
Calculate slope for each A, B and C
For zero-order,
slope = -(final - initial) concentration/(final - initial) time = constant
slope = rate constant (k)
For C,
from Ist and IInd data set, slope = -(3.05 - 3.25)/10 = 0.02
from IInd and IIIrd data set, slope = -(2.85 - 3.05)/10 = 0.02
rate constant = 0.02 M-1.min-1
So,
[C] (M) vs time is a zero-order reaction
rate law,
rate = k = constant
independent of initial concentration
half-life, t1/2 = [A]o/2k
= 3.25/2 x 0.02
= 81.25 min
concentration of C after 400 min,
[C] (at t = 400 min) = 3.25 - 0.02 x 400
= -4.75 M
So technically all of [C] is consumed at this stage.
For a first order reaction,
slope = -(ln(final concentration) - ln(initial concentration))/(final - initial) time
= rate constant
For A,
from Ist and IInd data set, slope = -(ln(2.25) - ln(2.5))/10 = 0.01
from IVth and Vth data set, slope = -(ln(1.79) - ln(1.94))/10 = 0.01
rate constant k = 0.01 min-1
So,
[A] (M) vs time is a first-order reaction
rate law,
rate = k[A]
dependent of initial concentration
half-life, t1/2 = ln(2)/k
= ln(2)/0.01
= 69.31 min
concentration of A after 400 min,
ln[A] (at t = 400 min) = ln(2.5) - 0.01 x 400
[A] = 0.046 M
Fors second-order reaction,
slope = -(1/[A]t - 1/[A]o)/(final - initial) time
From IInd and IIIrd data set,
slope = 0.012
From Vth and VIth data set,
slope = 0.012
rate constant = slope = 0.012 M-1.min-1
So,
[B] is second-order reaction,
rate law = k[B]^2
half-life = 1/0.012 x 1.51
= 55.19 min
after 400 min
1/[B] = 1/1.51 + 0.012 x 400
concentration of [B] = 0.18 M
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