Question

QUESTION 8 The following information is available for questions 6 and 7. The concentration of A was found as a function of time during the course of the reaction: AC Time, s [A], mol/L 0 0.9101 10.1893 20.1056 30.07327 4 0.05608 5 0.04542 6 0.03817 70.03291 8 0.02893 The data was graphed several different ways: 0.8 0.6 0.4 A N 15 Time, s

question 8 please, both pics are from same question

Answer 1

Note that for the given data, the plot of 1/[A] vs t is a straight line.

We know that for a second order reaction, the integrate rate equation is given as

= kt +

Hence, a plot of 1/[A] vs t results in a straight line with slope k and intercept 1/[A]o.

Hence, to find the value of k, we just have to calculate the slope of the third plot.

Since all the points in the plot lie in the straight line, we can calculate the slope by taking two points on the line.

Lets first calculate the value of 1/[A] for the points from the data given

t(s) 0 [A] 1/[A] 0.9101 1.09878 0.1893 5.28262 0.1056 9.469697 0.07327 13.64815 0.05608 17.83167 0.04542 22.01673 0.03817 26.19859 0.03291 30.3859 0.02893 34.56619

The second and third row points taken for the plot are

(21,41) = (1,5.28262)

(.22,42) = (2,9.469697)

Hence, the slope can be calculated as

y2- yi .22-21 9.469697 - 5.28262 2-1 4.187

Hence, the correct choice for k is c. 4.18.

Note: we can also plot the data in excel and get the trendline showing the slope as follows:

1/[A] vst y = 4.1834x + 1.0994 1/A] 1 2 3 4 5 6 7 8 9 t(s)

Note that the fit equation has 4.1834 as slope.