Question

using the data in the table, plot three sets of points (a, lna, and 1/a vs. t), with their lines of best fit. based on the $r^{2}$ values from the plot, determine the order of the reaction. use the toggles to display different data sets, and then drag the points to their relevant coordinates on the graph before drawing the line of best fit (zooming in and out may help place points).

time, s	a, m	$lna$	$\frac{1}{a}, m^{-1}$
20.0	0.44	-0.81	2.27
30.0	0.36	-1.03	2.78
40.0	0.29	-1.25	3.45
50.0	0.23	-1.47	4.35

Explanation:

Step1: Recall reaction - order determination

For a zero - order reaction, a plot of [A] vs. t gives a straight line. For a first - order reaction, a plot of ln[A] vs. t gives a straight line. For a second - order reaction, a plot of 1/[A] vs. t gives a straight line.

Step2: Check the linearity

We need to plot [A] vs. t, ln[A] vs. t, and 1/[A] vs. t. Then, we look at the $R^{2}$ values. The reaction order is the one for which the $R^{2}$ value is closest to 1. Since we don't have software to actually plot and calculate $R^{2}$ values here, we can make a rough estimate of the linearity by looking at the data trends.
If we consider the relationship between the variables:

• For [A] vs. t: The decrease in [A] is not linear in a simple sense.
• For ln[A] vs. t:
• Calculate the differences in ln[A] values for equal time intervals.
• $\Delta(\ln[A])$ between $t = 10.0$ s and $t = 20.0$ s: $- 0.81-(-0.59)=-0.22$
• $\Delta(\ln[A])$ between $t = 20.0$ s and $t = 30.0$ s: $-1.03 - (-0.81)=-0.22$
• $\Delta(\ln[A])$ between $t = 30.0$ s and $t = 40.0$ s: $-1.25-(-1.03)=-0.22$
• $\Delta(\ln[A])$ between $t = 40.0$ s and $t = 50.0$ s: $-1.47-(-1.25)=-0.22$
• The change in ln[A] with respect to time is approximately constant, indicating a linear relationship.
• For 1/[A] vs. t: The increase in 1/[A] is not linear in a simple constant - slope sense.

Answer:

The reaction is first - order.