Question

part 1 - group activity (45 minutes)
tutorial goals: use collision theory to qualitatively explain differences in reaction rates. use reaction coordinate diagrams to identify key parts of a reaction mechanism. describe the effects of a catalyst on the yield and rate of a reaction. determine the differential and integrated rate laws for a given reaction, including the order and the rate - constant for this reaction, using experimental data.
part 1: revisiting tutorial 1
di - t - butyl peroxide (c8h18o2) decomposes to produce acetone (ch3coch3) and ethane (ch3ch3). a student performed an experiment and measured the concentrations of the different species present as a function of time (shown on the right).

1. write the balanced chemical reaction for this process, assuming the reaction will go to completion.
2. identify which trace (line a, b, or c) on the plot shown here describes the behaviour of each component of this reaction. how do you know?
3. all three components are gases. will the total pressure in the container increase or decrease as the reaction proceeds?

part 2: determining a rate law
a detailed view of the data for decomposition of di - t - butyl peroxide is shown here. use it (and your answers from part 1) to answer the following questions.

1. what is the general expression for the rate law for this reaction?
2. using the graph on the right, determine:

a. the initial concentration of di - t - butyl peroxide in the experiment
b. its concentration at t = 80 min
c. the initial rate
d. the rate at t = 80min
hint line a: slope=-1.2 x 10^(-2) μmol l^(-1) min^(-1); line b: slope=-6.1 x 10^(-3) μmol l^(-1) min^(-1); line c: slope=-5.1 x 10^(-3) μmol l^(-1) min^(-1)

Explanation:

Step1: Write balanced chemical reaction

Di - t - butyl peroxide ($C_8H_{18}O_2$) decomposes to produce acetone ($C_3H_6O$) and ethane ($C_2H_6$). The balanced chemical equation is $C_8H_{18}O_2
ightarrow 2C_3H_6O + C_2H_6$.

Step2: Identify traces for components

The reactant (di - t - butyl peroxide) concentration decreases over time. So, the line with decreasing concentration (Line C) represents di - t - butyl peroxide. As the reaction proceeds, the products (acetone and ethane) are formed. Since more moles of products are formed than moles of reactant consumed (1 mole of reactant gives 3 moles of products), the concentrations of acetone and ethane increase. The steeper increasing line (Line A) represents acetone (because the stoichiometric coefficient of acetone is 2) and the less - steep increasing line (Line B) represents ethane.

Step3: Determine pressure change

According to the ideal gas law $PV = nRT$. Since the reaction produces more moles of gas products (3 moles) than the moles of gas reactant (1 mole) and assuming constant volume and temperature, the total number of moles of gas $n$ increases. So, the total pressure in the container will increase as the reaction proceeds.

Step4: General rate law expression

For the reaction $C_8H_{18}O_2
ightarrow 2C_3H_6O + C_2H_6$, the general rate law expression is rate $=k[C_8H_{18}O_2]^m$, where $k$ is the rate constant and $m$ is the order of the reaction with respect to di - t - butyl peroxide.

Step5: Determine initial concentration

From the graph of di - t - butyl peroxide concentration vs time, the initial concentration (at $t = 0$) is $1.6\ \mu mol/L$.

Step6: Determine concentration at $t = 80$ min

From the graph, the concentration of di - t - butyl peroxide at $t = 80$ min is $0.8\ \mu mol/L$.

Step7: Determine initial rate

The initial rate is the rate at $t = 0$. The slope of the tangent to the concentration - time curve of di - t - butyl peroxide at $t = 0$ gives the initial rate. The slope of Line C at $t = 0$ is $- 5.1\times10^{-3}\ \mu mol\ L^{-1}\ min^{-1}$. The negative sign indicates the decrease in the concentration of the reactant. So, the initial rate is $5.1\times10^{-3}\ \mu mol\ L^{-1}\ min^{-1}$.

Step8: Determine rate at $t = 80$ min

The rate at $t = 80$ min is given by the slope of the tangent to the concentration - time curve of di - t - butyl peroxide at $t = 80$ min. From the hint, the slope of Line C at $t = 80$ min is $- 1.2\times10^{-2}\ \mu mol\ L^{-1}\ min^{-1}$. The negative sign indicates the decrease in the concentration of the reactant. So, the rate at $t = 80$ min is $1.2\times10^{-2}\ \mu mol\ L^{-1}\ min^{-1}$.

Answer:

1. $C_8H_{18}O_2

ightarrow 2C_3H_6O + C_2H_6$

1. Line C: Di - t - butyl peroxide (concentration decreases); Line A: Acetone (steeper increase as $n = 2$ in stoichiometry); Line B: Ethane (less - steep increase as $n = 1$ in stoichiometry)
2. Increase
3. rate $=k[C_8H_{18}O_2]^m$
4. $1.6\ \mu mol/L$
5. $0.8\ \mu mol/L$
6. $5.1\times10^{-3}\ \mu mol\ L^{-1}\ min^{-1}$
7. $1.2\times10^{-2}\ \mu mol\ L^{-1}\ min^{-1}$