Question

3. The rate law expression for a chemical reaction can be described by the formula r = kan Where r is the rate of the reaction, C is the concentration, k is the rate constant, and n is the order of the reaction. The following data were collected for a certain chemical reaction: C(gmol/L) 4 2.25 1.5 r(gmol/L.s) 0.400 0.305 0.245 a) Write the linear form of the above equation that you would use for non-linear regression b) From a), identify the new variables/parameters that would give a linear relationship c) From a) and b), find the new linear set of data points from the given set d) Find the equation of the line of best fit for the new set of data e) Find the values of k and n

Answer 1

(a).

To get the linear form, take logarithm on both sides if the equation. So the linear form is:

$log(r)=nlog(C)+log(k)$

(b)

The equation is of the form: y = m*x + c

where:

y = log(r)

x = log(C)

slope(m) = n

y-intercept(c) = log(k)

Hence the new variables that would give a linear relationship are : log(C) and log(r)

(c)

The new linear set of data points are:

C(g/mol)	4	2.25	1.5
r(g mol/L.s)	0.400	0.305	0.245
log(C)	0.602	0.352	0.176
log(r)	-0.398	-0.516	-0.611

(d)

The best fit line has been found using spreadsheet software. The graph is attached below as well.

log(r) vs. log(C) log(n) = 0.498*x +-0.696 R2 = 0.999 log(r) 0.2 0.3 0.4 0.5 0.6 log(C)

The equation of best fit line is:

y = 0.498*x + (-0.696)

(e)

We have:

slope(m) = n

y-intercept(c) = log(k)

Hence

n = 0.498 ~ 0.5

log(k) = -0.696

OR

k = antilog(-0.696) = 0.20

Hence we have:

n = 0.5

k = 0.20