Question

Use the graphical analysis for each set of data to determine:

1. Order of the reaction
2. Rate constant (k)
3. Initial concentration, [A]₀
4. Half-life

Time, min	ln[A]t	[A]	1/[A]
0	?
15	0.562
30	0.365
45	0.201
60	0.060
75	-0.063
90	-0.173
105	-0.272
120	-0.362
135	-0.445
150	-0.521

Answer 1

a.

In order to find the order of the reaction, we must plot different graphs trying to obtain a straight line which can be interpreted to determine the kinetics.

For a zero order reaction:

At-kt A

Hence, if our reaction is zero order, a plot of [A] vs t will be a straight line with a slope of -k.

For a first order reaction:

In At kt+In A

Hence, if our reaction is first order, a plot of will be a straight line with a slope of -k

In At vs t US

For a second order reaction:

kt+ [Ao [Alt

We will fill the table by calculating [A] and 1/[A] from ln[A]

| 1/[A] Time(min In[A] [A] 0 15 0.562 1.754177 0.570068 30 0.365 1.440514 0.694197 45 0.201 1.222625 0.817912 60 0.06 1.061837 0.941765 75 -0.063 0.938943 1.065027 90 0.173 0.841138 1.188866 105 0.272 0.761854 1.312587 120 0.362 0.696282 1.436199 135 0.445 0.640824 1.56049 150 -0.521 0.593926 1.683711

Now we will check if this reaction is a zero order reaction by plotting [A] vs t

[A] 2 1.8 1.6 1.4 1.2 0.8 0.6 0.4 0.2 40 60 80 100 120 140 160 time (min) 20

Note that the graph is not a straight line. Hence,our reaction is not a zero order reaction.

Now, we will check for first order reaction by plotting ln[A] vs t.

In[A] 0.8 0.6 0.4 0.2 0 40 80 100 20 60 120 140 160 -0.2 -0.4 -0.6 time (min) In[A]

Note that this graph is also not a straight line. Hence, our reaction is not a first order reaction.

Now, we will check for 2nd order reaction by plotting 1/[A] vs t

1/[A] 1.8 1,6 y 0.0082x+0.4466 R2 1 1.4 1.2 1 0.8 0.6 0.4 0.2 0 C 160 140 120 100 80 40 60 0 20 time (min) 1/[A]

Since the plot of 1/[A] vs time is a straight line, the reaction must be second order with respect to the concentration of A.

Hence, the order of reaction is two.

b.

From the straight line fit of the plot 1/[A] vs t, we obtain the following fit equation

y0.0082 0.4466)

where y = 1/[A] and has a unit of using the the concentration unit is in M

M

Note that concentration unit is not provided in the question, hence we will assume it is M. If it is something different, feel free to change it back as it wont have any effect on the calculated numbers.

x is time in minutes.

The slope of the graph must equal k since

kt+ [Ao [Alt

Hence, value of k is 0.0082.

The unit of rate constant k will be

mol L min Mmin or

Hence, the rate constant of the reaction is
min o.0082 M
.

c.

The initial concentration can be calculated by putting t = 0 in our equation for the second order rate fit equation.

A]

y 0.0082.x 0.4466 1 0.0082 x 00.4466 [Alo 1 Ao 0.4466 2.24 M

Hence, the initial concentration of A is 2.24 M.

d.

Half life is defined as the time it takes for the concentration of the reactant to be half the original value.

To calculate half life of the reaction, we set and put it in the rate law equation.

[A0 At 2

$\frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \\ \Rightarrow \frac{1}{[A]_0/2} = kt_{1/2} + \frac{1}{[A]_0} \\ \Rightarrow kt_{1/2} = \frac{2}{[A]_0} - \frac{1}{[A]_0} \\ \Rightarrow t_{1/2} = \frac{1}{k[A]_0}$

We know that

0.0082 M min

Hence, half life of the reaction is

$t_{1/2} = \frac{1}{k[A]_0} = \frac{1}{0.0082 \ M^{-1}\ min^{-1}\times 2.24 \ M} \approx 54.46 \ min = 54 \ min\ 28 \ sec$

Hence, the half life of the reaction is about 54 min and 28 sec.