Question

Please dont use polymath to solve this problem.!. I need Part A only

P7-7A The reaction AB+C was carried out in a constant-volume batch reactor where the following concentration measurements were recorded as a function of time. 0 5 9 15 22 30 40 60 t(min) CA (mol/dm2) 2 1.6 1.35 1.1 0.87 0.70 0.53 0.35 (a) Use nonlinear least squares (i.e., regression) and one other method to determine the reaction order, O., and the specific reaction rate, k. (b) Nicolas Bellini wants to know, if you were to take more data, where would you place the points? Why? (c) Prof. Dr. Sven Köttlov from Jofostan University always asks his students, if you were to repeat this experiment to determine the kinetics, what would you do differently? Would you run at a higher, lower, or the same temperature? Take different data points? Explain what you think he is expecting her to say and why. (d) It is believed that the technician made a dilution error in the concentration measured at 60 min. What do you think? How do your answers compare using regression (Polymath or other software) with those obtained by graphical methods?

Answer 1

the problem is based on the concepts of the reaction engineering and kinetics of reaction.

rate of the reaction is used to determine the reaction speed. rate of the reaction is dependent on the rate constant.

order of the reaction is power of the concentration terms in the rate equations.

zero order reaction is the reaction whose rate is independent of the concentration of the reactant. it' concentration time reaction is given as,

...............................1

A [Ao] – kt

here ,

k = rate constant

t = time

[A0] = initial concentration of the reactant

so , the plot between [A] and t will be a straight line.

first order reaction is the reaction whose rate is directly proportional to the concentration of the reactant. it's concentration time relation is given as,

...............................2

In([4]) = ln([40]) – kt

k = rate constant

t = time

[A0] = initial concentration of the reactant

so , the plot between [A] and t will be a straight line.

so the plot between ln([A]) vs time is straight line with slope of the line equals to the rate constant (k).

hence,

slope = -k

so,

k = -slope..............................3

now to find the activation energy we have to use Arrhenius relation which is given as,

...................................4

Εα In(k) = In(ko) RT

here,

k0 = arrhenius parameter

Ea = activation energy

R = 8.314 J/mol.K

T = temperature in Kelvin

we are given the data of concentration vs time as,

t[min] Ca[mol/L) 0 2 5 1.6 9 1.35 15 1.1 22 30 40 60 0.87 0.7 0.53 0.35

a)

as we don't know the order of the reaction,

we first assume that it is a zero order reaction, so the plot of Ca vs t will be,

Zero order 2.5 2 1.5 Ca (mol/L) 1 in y=-0.026x +1,6508 R = 0.8621 0 0 10 20 30 40 50 60 70 t[min]

we can observe that it is NOT a straight line , hence order is NOT a zero order .

now we assume that order is first order , so the plot of ln(Ca) vs t will be,

First order 1 0.5 0 In(ca) 0 10 20 ..30 40 50 60 70 -0.5 -1 y=-0.0289x + 0.5769. R2 = 0.9821 -1.5 t[min]

from the graph we can see that it is straight line, that means that order of the reaction IS first order.

so, the equation for the first order is given as,

In([4]) = ln([40]) – kt

so,

we know that slope of the ln(A) v/s time will give rate constant,

hence from the graph we get,

slope = -0.0289 1/s

and

slope = - k

so,

we get

rate constant = k = 0.0289 1/s......................[ans.]

hence,

rate constant = 0.0289 1/s

order = 1st order.

b)

if we have to take more data point we will prefer to take the data point in the range of 40 min to 60 min time interval, because the major deviation is seen there. more data points will means more precision.

c)

if the experiments has to be repeated to conform the data points and the students must perform the reaction at the same temperature, but now student should record the data at different times. and if the students want's to determine the activation energy of the reaction, then the experiment must be performed at higher temperature, we get a higher rate constant and then use the Arrhenius equation to determine the activation energy.

d)

the last point that is at t = 60min deviates form the regression the most,

from the regression we get the line equation as,

so , for t = 60 min, we get

taking anti ln we get,

hence, 'the theoretical value of k = 0.3143

and ,

actual value = 0.35 1/s

hence the % error in k = (0.35 - 0.3143)1/s / 0.3143 1/s = 0.1135

so, the % error = 11.35%.

less that 10% error is always acceptable hence the data point needs to be taken again.