You lead an interstellar expedition to a remote planet, inhabited by aliens. Unfortunately, the arri — Physical Chemistry — Kinetics Chemistry Question
THEORETICAL PROBLEM 25
You lead an interstellar expedition to a remote planet, inhabited by aliens. Unfortunately, the arrival of your spaceship initiates a reaction causing the atmosphere to decompose by a first-order rate law with a half-life of 13 hours. Everyone will have to leave by the time when only 13 % of the original atmosphere remains.
How much time do you have until then?
Model Answer
Plugging the data into the integrated rate law gives the equation:
0.13 = (1/2)^(t / 13)
t = –13 hr * ln(0.13) / ln(2) = 38.3 hours.
Ethyl-propionate hydrolyzes in aqueous alkaline solution:
C2H5COOC2H5(aq) + OH–(aq) → C2H5COO–(aq) + C2H5OH(aq)
Initial rate data were collected for different concentrations, as shown in the Table:
[C2H5COOC2H5] [OH–] Initial rate (mmol dm–3 s–1)
0.045 0.300 1.09
0.090 0.300 2.15
0.090 0.150 1.11
[VISUAL]
Determine the partial orders of reaction, its kinetic equation and rate coefficient.
Model Answer
Inspection of the data shows a doubling of the rate upon doubling either of the concentrations. Therefore both partial orders are unity, i.e., the kinetic equation is:
r = k [C2H5COOC2H5] [OH–]
The rate coefficient can then be calculated from any row of the table, e.g.:
k = 0.00109 mol dm–3 s–1 / (0.045 mol dm–3 * 0.300 mol dm–3) = 0.081 dm3 mol–1 s–1
The initial rate of the reaction above doubles when the temperature is raised from 25 °C to 42 °C with the same initial concentrations.
What is the Arrhenius activation energy?
Model Answer
Taking the logarithm of the ratio for two Arrhenius expressions gives the equation:
ln(k2 / k1) = -EA / R * (1/T2 - 1/T1)
Rearranging and plugging in the data yields EA = 32 kJ mol–1.
The reaction 2 NO(g) + O2(g) = 2 NO2(g) obeys the following kinetic equation:
r = k [NO]2 [O2]
Explain how the rate of the reaction changes when the following concentration changes are made:
i. [O2] is quadrupled,
ii. [NO] is quadrupled,
iii. [NO] is halved,
iv. [O2] is halved and [NO] is quadrupled,
v. [NO] is halved and [O2] is quadrupled.
Model Answer
i. The rate will increase by a factor of 4.
ii. The rate will increase by a factor of 4^2 = 16.
iii. The rate will decrease by a factor of 2^2 = 4.
iv. The rate will increase by a factor of 4^2 / 2 = 8.
v. The rate will not change, because (1/2)^2 * 4 = 1.
The initial rate of the above reaction remains the same when the temperature is raised from 460 °C to 600 °C, with all the initial concentrations halved.
What is the Arrhenius activation energy?
Model Answer
Since the overall reaction order is 3, halving all concentrations makes the product of the concentration terms in the kinetic equation smaller by a factor of 2^3 = 8; in order to keep the rate unchanged, k must be increased by the same factor.
ln(k2 / k1) = -EA / R * (1/T2 - 1/T1)
EA = 79 kJ mol–1.
The first-order decay of a compound was followed spectrophotometrically. After loading a solution with an initial concentration 0.015 mol dm–3 into a cuvette with a path-length of 0.99 cm, its absorbance (at a wavelength where only this species absorbs light) was recorded as a function of time.
[VISUAL]
From this plot:
i. estimate the molar absorption coefficient,
ii. estimate the initial rate and the rate constant,
iii. estimate the half-life from the rate constant,
iv. discuss whether the estimated half-life is consistent with the plot,
v. calculate the time required to consume 99 % and 99.99 % of the compound.
Model Answer
i. From the absorbance at t = 0: ε = 0.138 / (0.015 mol dm–3 * 0.99 cm) = 9.29 dm3 mol–1 cm–1
ii. Using, e.g., the change from 0 to 25 s:
r = (0.138 - 0.102) / (0.138 * 25 s) * 0.015 mol dm-3 = 1.57*10^-4 mol dm-3 s-1
k = ln(0.138 / 0.102) / 25 s = 1.2*10^-2 s^-1
iii. t½ = ln(2) / k = 57 s
iv. The estimated half-life is consistent with the plot, from which it can be seen that the absorbance (and concentration) drops roughly by half for each half-life interval.
v. t(99 %) = – ln(1 – 0.99) / k = 384 s
t(99.99 %) = – ln(1 – 0.9999) / k = 768 s