1. The isomerization of methyl isonitrile (CH3NC) to acetonitrile (CH3CN) was studied in the gas phase at 215EC, and the
following data were obtained.
Time(s) [CH3NC](M)
0 0.0165
2000. 0.0110
5000. 0.00591
8000. 0.00314
Calculate the average rate of reaction, in M/s, for the time interval between each measurement.
(1) First interval
– D[CH3NC]/Dt = – [(0.0110 – 0.0165) M]/[(2000. – 0) s]
= – (- 0.0055 M)/(2000 s)
– D[CH3NC]/Dt = 2.8 x 10 – 6 M/sec
(2) Second interval
– D[CH3NC]/Dt = – [(0.00591 – 0.0110) M]/[(5000. – 2000.) s]
= – (- 0.0051 M)/(3000 s)
– D[CH3NC]/Dt = 1.7 x 10 – 6 M/sec
(3) Third interval
– D[CH3NC]/Dt = – [(0.00314 – 0.00591) M]/[(8000. – 5000.) s]
= – (- 0.00277 M)/(3000. s)
– D[CH3NC]/Dt = 9.23 x 10 – 7 M/sec
2. The following data were collected for the reaction given below at 45EC.
2 N2O5 (g) ® 4 NO2 (g) + O2 (g)
time(min) [N2O5](M) [NO2](M) [O2](M)
0 1.24 x 10 – 2 0 0
10 0.92 x 10 – 2 0.64 x 10 – 2 0.16 x 10 – 2
20 0.68 x 10 – 2 1.12 x 10 – 2 0.28 x 10 – 2
60 0.20 x 10 – 2 2.08 x 10 – 2 0.52 x 10 – 2
(a) Calculate the rate of decomposition of N2O5 during the first 10 minute interval.
– D[N2O5]/Dt = – [(0.92×10 – 2 – 1.24×10 – 2) M]/[(10.– 0) min] = – (- 0.0032 M)/(10. min)
– D[N2O5]/Dt = 3.2 x 10 – 4 M/min
(b) Calculate the rate of decomposition of N2O5 during the second 10 minute interval.
– D[N2O5]/Dt = – [(0.68×10 – 2 – 0.92×10 – 2) M]/[(20.– 10.) min] = – (- 0.0024M)/(10. min)
– D[N2O5]/Dt = 2.4 x 10 – 4 M/min
(c) Calculate the rate of formation of NO2 during the first 10 minute interval.
D[NO2]/Dt = [(0.64×10 – 2 – 0) M]/[(10.– 0) min] = (0.0064 M)/(10. min)
D[NO2]/Dt = 6.4 x 10 – 4 M/min
(d) Calculate the rate of formation of O2 during the first 10 minute interval.
D[O2]/Dt = [(0.16×10 – 2 – 0) M]/[(10.– 0) min] = (0.0016 M)/(10. min)
D[O2]/Dt = 1.6 x 10 – 4 M/min
(e) Determine the relationship between the various rates in the first 10 minute interval.
We have the following three relationships between the rates.
- Since D[NO2]/Dt is four times greater than D[O2]/Dt, NO2 is formed four times as fast as O2.
- Since – D[N2O5]/Dt is twice as large as D[O2]/Dt, N2O5 decomposes twice as fast as O2 is formed.
- Since D[NO2]/Dt is twice as large as – D[N2O5]/Dt, NO2 is formed twice as fast as N2O5 decomposes.
3. At a certain temperature the following data were collected for the reaction
2 ICl + H2 ® I2 + 2 HCl
Initial Concentrations (M) Initial Rate of Formation of I2
[ICl] [H2] (mol/LCs)
0.10 0.10 0.0015
0.20 0.10 0.0030
0.10 0.050 0.00075
(a) Determine the reaction order with respect to each reactant and the overall order of the reaction.
Assume that the rate law expression for this reaction has the following form.
rate = k[ICl]x[H2]y
You need to look at the first two trials In order to determine the order of [ICl].
[ICl] doubles, [H2] constant Þ rate doubles
x = 1
The reaction is first order with respect to [ICl].
You need to look at the first and last trials in order to determine the order of [H2].
[ICl] constant, [H2] halves Þ rate halves
y = 1
The reaction is first order with respect to [H2].
Thus, this reaction is second order overall (x + y = 2).
(b) Write the rate law.
rate = k[ICl][H2]
(c) Calculate the rate constant and express it in appropriate units.
Rearrange the rate law to obtain an expression for k:
k = rate/([ICl][H2])
Substitute into this expression data from any of the trials (for example Trial 1).
k = 0.0015 M s – 1/[(0.10 M)(0.10 M)]
k = 0.15 M – 1s – 1
4. The rate law for the decomposition of aqueous hydrogen peroxide (H2O2) at 70EC is first order in [H2O2] with
k = 0.0347 min-1.
2 H2O2(aq) ® 2 H2O(l) + O2(g)
Since the rate constant has units of time – 1 this is a first-order reaction.
(a) Calculate the half-life (t1/2) for this reaction at 70EC.
t1/2 = (ln 2)/k = 0.693/k = 0.693/0.0347 min – 1 = 20.0 minutes
(b) Given that the initial concentration of H2O2 is 0.300 M, calculate the concentration of H2O2 after 60.0 minutes have
elapsed.
[H2O2]t = [H2O2]0 x e – kt
kt = 0.0347 min – 1 x 60.0 min = 2.08
[H2O2]t = [H2O2]0 x e – 2.08 = 0.300 M x 0.125
[H2O2]t = 0.0375 M
Or you could recognize that for this reaction 60.0 minutes is 3 half-lives.
[H2O2]t = (1/2)(1/2)(1/2)(0.300 M) = (1/8)(0.300 M) = 0.0375 M
5. The rate law for the decomposition of hydrogen iodide at 700 K is second order in [HI] with k = 1.2 x 10-3 M – 1s – 1.
2 HI(g) ® H2(g) + I2(g)
(a) If the initial concentration of HI is 0.56 M, calculate the concentration of HI after 2.00 hr have elapsed.
1/[HI]t = kt + 1/[HI]0
We need to convert 2.00 hr to seconds
2.00 hr x (60 min/1 hr)(60 sec/1 min) = 7.20 x 10 3 seconds
1/[HI]t = (1/0.56 M) + (1.2 x 10-3 M – 1s – 1)(7.20 x 10 3 s)
1/[HI]t = 1.8 M – 1 + 8.6 M – 1 = 10.4 M -1
[HI]t = 1/10.4 M – 1 = 0.096 M
(b) Calculate the elapsed time at which [HI] = 0.28 M.
There are two ways to approach this problem.
(1) (1/[HI]t) – (1/[HI]0) = kt so t = { (1/[HI]t) – (1/[HI]0) } / k
t = [(1/0.28 M) – (1/0.56 M)]/1.2 x 10-3 M – 1s – 1
t = (3.6 M – 1 – 1.8 M – 1)/1.2 x 10-3 M – 1s – 1 = 1.8 M – 1/1.2 x 10-3 M – 1s – 1
t = 1.5 x 10 3 seconds or 25 minutes
(2) Recognize that this is equivalent to calculating the first half-life for this
second-order reaction.
t1/2 = 1/(k[HI]0) = 1/[(1.2 x 10 – 3 M – 1s – 1)(0.56 M – 1)]
t1/2 = 1.5 x 10 3 seconds or 25 minutes
6. Using the plot below determine the order of reaction and the value of the rate constant for the following reaction.
C4H8 ® 2 C2H4
Since a plot of ln[C4H8] vs. time gives a straight line with a negative slope this is a first-order reaction. For a first-order reaction the value of the slope is equal to the negative of the rate constant. Thus, we have the following relationship
k = – slope
k = – (- 0.0112 s – 1)
k = 0.0112 s – 1
