Question

Below is data that a researcher collected on the observed decay of a chemical in water as a function of time. The Researcher used an analytically technique to quantify the chemical that reports relative “peak area” or “response” values. As a result, the researcher analyzed a series of known “standards” of chemical “A” to construct a standard curve, from which the concentrations of “A” in the experimental samples could be calculated. It is known that the chemical decays by a first-order process. (always use the “xy scatter” option under graph type).

Standard Curve:
[A], Concentration of A (Molarity)	Peak Area
0.002	1220
0.004	2520
0.006	3680
0.008	4940
0.01	6120

Experimental Data:
Time (hours)	Peak Area
0	6120
10	4320
20	3060
30	2250
40	1660
50	1170
60	830

Data Table 1)

Find the slope, intercept, and correlation coefficient of the standard curve regressing [A] on the x-axis, and “Peak Area” on the y-axis (Use Excels built-in functions for all three parameters). Plot the data points, and on the same graph, plot the regression line (solve for values on the line on the spreadsheet, including at x=0) (Figure 1).  

Useful table headings:

A(Molarity) Peak Area Line

Data Table 2)

From the standard curve and the peak areas of the samples, calculate the concentrations of “A” in each sample. Find the natural logarithm (ln) of these values. Find the slope, intercept, and correlation coefficient for the least-squares analysis of time (x-axis) versus ln([A]) (y-axis) (use the built-in functions). Using the slope and intercept, calculate points on the regression “line”. Plot the experimental values of ln[A] versus time and the regression line on a graph (Figure 2). From the regression (slope and intercept), calculate [A]_t=0 and k. (note that k is always positive).  

Useful headings:

time(hrs) Peak Area A(Molarity) ln(A)    ln(A)_calculated

Data Table 3)

Use Solver to calculate [A]_t=0 and k without “transforming the data."

Useful table headings:

time(hrs) A_expected (Molarity)         A_calc (Molarity) (A_expected - A_calculated)²

For these data, does the Solver solution agree well with the transformed (i.e., logarithmic) model results? Why or why not.

Table 4) Report the experimental data in units of miliMolar.

Useful table headings:

time(hrs) A(Molarity) A(miliMolar)

Table 5) Calculate the “model" fit at 5 hours or smaller increments, using the values of [A]_t=0 and k calculated both ways, and convert to miliMolar concentrations. Plot the original data (Table 4) and the 2 model lines (solid and dashed) (Table 5) on the same plot (Figure 3).

Useful table headings:

Transformed Solver   

time (hrs) A (Molarity) A(miliMolar) A(M)    A(mM)

Format all Tables and Figures on 2 pages in Excel.

Answer 1

There are many subparts in this questions. I am answering questions for table 1 and 2 (Chegg Policy: 1 questions per post or 4 sub parts in 1 question)

ANSWER

Table 1

a) Table 1

A (mol/L)	Peak Area	Line
0		30
0.002	1220	1252
0.004	2520	2474
0.006	3680	3696
0.008	4940	4918
0.01	6120	6140

b) For the standard curve we plot Peak area (y-axis) vs. [A] (x-axis) we get

c) The regression analysis gives the following values

slope	611000
intercept	30
correlation coefficient	0.9997

and the line equation is:

this equation is used to calculate the values in the "line" column of table 1 (calculated peak area on the regression line)

---------------------------------------------------------------------------------

Table 2

a) Table 2

Time (h)	Peak Area	A (mol/L)	ln (A)	ln (A) calculated
0	6120	9.97E-03	-4.608	-4.617
10	4320	7.02E-03	-4.959	-4.951
20	3060	4.96E-03	-5.307	-5.286
30	2250	3.63E-03	-5.618	-5.620
40	1660	2.67E-03	-5.927	-5.954
50	1170	1.87E-03	-6.284	-6.289
60	830	1.31E-03	-6.638	-6.623

b) The values of A were calculated with the equation line of part 1:

c) And plotting ln (A) (y-axis) vs. Time (x-axis) we get

d) The regression analysis gives the following values

slope	-0.03343
intercept	-4.617
correlation coefficient	0.9995

and the line equation is:

this equation is used to calculate the values in the "ln (A) calculated" column of table 2 (calculated ln (A) on the regression line).

e) As A decay follows a first-order process, this means that integrated rate law for A decay is

It is similar to the line equation second graph. Then, we can calculate the calues of k and [A]0: