Exam 1
Question1
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the
following data table is obtained:
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
Data Table #2
Time (sec) [N2O5] [O2]
0 0.300 M 0
300 0.272 M 0.014 M
600 0.224 M 0.038 M
900 0.204 M 0.048 M
1200 0.186 M 0.057 M
1800 0.156 M 0.072 M
2400 0.134 M 0.083 M
3000 0.120 M 0.090 M
1. Using the [O2] data from the table, show the calculation of the average rate over
the measured time interval from 0 to 3000 secs.
The average rate over the measured time interval from 0 to 3000 secs is:
rate = ∆[O2] / ∆t = (0.090 - 0) / 3000 - 0 = 3.00 x 10-5 mol/L•s
, 2. Using the [O2] data from the table, show the calculation of the instantaneous
rate late in the reaction (2400 secs to 3000 secs).
The late instantaneous rate over the measured time interval from 2400 to 3000 secs is:
rate = ∆[O2] / ∆t = (0.090 - 0.083) / 3000 - 2400 = 1.17 x 10 -5 mol/L•s
3. Explain the relative values of the average rate and the late instantaneous rate.
The late instantaneous rate is smaller since the concentrations of reactants is lowest
during the late stages of the reaction.
Question 2
The following rate data was obtained for the hypothetical reaction: B + D → X +
Y
Experiment # [B] [D] rate
1 0.50 0.50 2.4
2 1.00 0.50 19.2
3 1.00 1.00 38.4
1. Determine the reaction order with respect to [B].
3rd order with respect to [B] : 19.2/2.4 =k (1.00)n (0.50)m / k (0.50)n (0.50)m : n = 3
2. Determine the reaction order with respect to [D].
1st order with respect to [D] : 38.4/19.2 =k (1.00)n (1.00)m / k (1.00)n (0.50)m : m = 1
3. Write the rate law in the form rate = k [B]n [D]m (filling in the correct exponents).
rate = k [B]3 [D]1
4. Show the calculation of the rate constant, k.
k = 2.4 / (0.50)3 (0.50)1 = 38.4
Question 3
ln [A] - ln [A]0 = - k t 0.693 = k t1/2
Show the calculation of the decay constant (k) and the half-life (t 1/2) of a radioactive
nucleus if 44.3% of the material has decayed in 365 days.
ln [55.7] - ln [100] = - k 365 days 4.0200 - 4.6052 = -k (365)
, -0.5852 = - 365 k k = - 0.5852 / - 365 = 1.60 x 10-3
0.693 = k t1/2 t1/2 = 0..60 x 10-3 = 433.1 yrs.
Question 4
Using the potential energy diagram below, state whether the reaction described by the
diagram is endothermic or exothermic and spontaneous or nonspontaneous, being sure
to explain your answer.
Small Eact = spontaneous ∆H+ = endothermic
Question 5
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.300 mole of
CO and 0.900 mole of H2 in a 3.00 liter container forms an equilibrium mixture containing 0.117
mole of H2O and corresponding amounts of CO, H2, and CH4.
At equilibrium
H2O = 0.117 mole (as stated)
CH4 = 0.117 mole (1 mole of CH4 forms for every mole of H2O that is formed)
CO = 0.300 - 0.117 mole (1 mole of CO reacts for every mole of H 2O that is
formed)
H2 = 0.900 - 3 x 0.117 mole (3 mole of H2 reacts for every mole of H2O that is
formed)
Question1
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below, the
following data table is obtained:
2 N2O5 (g) → 4 NO2 (g) + O2 (g)
Data Table #2
Time (sec) [N2O5] [O2]
0 0.300 M 0
300 0.272 M 0.014 M
600 0.224 M 0.038 M
900 0.204 M 0.048 M
1200 0.186 M 0.057 M
1800 0.156 M 0.072 M
2400 0.134 M 0.083 M
3000 0.120 M 0.090 M
1. Using the [O2] data from the table, show the calculation of the average rate over
the measured time interval from 0 to 3000 secs.
The average rate over the measured time interval from 0 to 3000 secs is:
rate = ∆[O2] / ∆t = (0.090 - 0) / 3000 - 0 = 3.00 x 10-5 mol/L•s
, 2. Using the [O2] data from the table, show the calculation of the instantaneous
rate late in the reaction (2400 secs to 3000 secs).
The late instantaneous rate over the measured time interval from 2400 to 3000 secs is:
rate = ∆[O2] / ∆t = (0.090 - 0.083) / 3000 - 2400 = 1.17 x 10 -5 mol/L•s
3. Explain the relative values of the average rate and the late instantaneous rate.
The late instantaneous rate is smaller since the concentrations of reactants is lowest
during the late stages of the reaction.
Question 2
The following rate data was obtained for the hypothetical reaction: B + D → X +
Y
Experiment # [B] [D] rate
1 0.50 0.50 2.4
2 1.00 0.50 19.2
3 1.00 1.00 38.4
1. Determine the reaction order with respect to [B].
3rd order with respect to [B] : 19.2/2.4 =k (1.00)n (0.50)m / k (0.50)n (0.50)m : n = 3
2. Determine the reaction order with respect to [D].
1st order with respect to [D] : 38.4/19.2 =k (1.00)n (1.00)m / k (1.00)n (0.50)m : m = 1
3. Write the rate law in the form rate = k [B]n [D]m (filling in the correct exponents).
rate = k [B]3 [D]1
4. Show the calculation of the rate constant, k.
k = 2.4 / (0.50)3 (0.50)1 = 38.4
Question 3
ln [A] - ln [A]0 = - k t 0.693 = k t1/2
Show the calculation of the decay constant (k) and the half-life (t 1/2) of a radioactive
nucleus if 44.3% of the material has decayed in 365 days.
ln [55.7] - ln [100] = - k 365 days 4.0200 - 4.6052 = -k (365)
, -0.5852 = - 365 k k = - 0.5852 / - 365 = 1.60 x 10-3
0.693 = k t1/2 t1/2 = 0..60 x 10-3 = 433.1 yrs.
Question 4
Using the potential energy diagram below, state whether the reaction described by the
diagram is endothermic or exothermic and spontaneous or nonspontaneous, being sure
to explain your answer.
Small Eact = spontaneous ∆H+ = endothermic
Question 5
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.300 mole of
CO and 0.900 mole of H2 in a 3.00 liter container forms an equilibrium mixture containing 0.117
mole of H2O and corresponding amounts of CO, H2, and CH4.
At equilibrium
H2O = 0.117 mole (as stated)
CH4 = 0.117 mole (1 mole of CH4 forms for every mole of H2O that is formed)
CO = 0.300 - 0.117 mole (1 mole of CO reacts for every mole of H 2O that is
formed)
H2 = 0.900 - 3 x 0.117 mole (3 mole of H2 reacts for every mole of H2O that is
formed)
