Question

Experimental data are listed for the following hypothetical reaction: $$ \mathrm{A}+\mathrm{B} \longrightarrow \text { products } $$ $$ \begin{array}{lcccccc} \hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ {[\mathrm{~B}]} & 0.51 & 0.43 & 0.39 & 0.35 & 0.33 & 0.31 \\ \hline \end{array} $$ (a) Plot these data as in Figure 11.3 . (b) Draw a tangent to the curve to find the instantaneous rate at 60 seconds. (c) Find the average rate over the 20 to 80 second interval. (d) Compare the instantaneous rate at 60 seconds with the average rate over the 60 -second interval.

Short answer

Answer: [Enter the slope of the tangent from step (b) and compare it with the average rate (-1/600) from step (c). State whether these rates are the same or different.]

Step-by-step solution

(a) Plot the given data on a graph

To plot the data, use the given time and concentration values for B and create an XY graph. Time will be on the x-axis (horizontal), and concentration of B will be on the y-axis (vertical). Plot the points and connect them with a smooth curve.

(b) Draw a tangent and calculate the slope at 60 seconds

At 60 seconds, draw a straight line that touches but does not cross the curve (the tangent) on the graph. Calculate the slope of this tangent by finding the change in the concentration of B divided by the change in time (rise over run) over a very small interval around the 60 seconds.

(c) Find the average rate between 20 and 80 seconds

To calculate the average rate over the given interval, use the following formula: $$ \text { Average rate } = \frac{\text { change in concentration of B }}{\text { change in time }} $$ Substitute the values accordingly: $$ \text { Average rate } = \frac{[B]_{80} - [B]_{20}}{80 - 20} $$ Plug in the values for the concentrations at the corresponding times: $$ \text { Average rate } = \frac{0.33 - 0.43}{80 - 20} = -\frac{0.1}{60} $$ Calculate the average rate: $$ \text { Average rate } = -\frac{1}{600} $$

(d) Compare the instantaneous rate and average rate

Now we can compare the instantaneous rate at 60 seconds and the average rate over the 60-second interval. If the slope of the tangent at 60 seconds (calculated in step b) is equal to the average rate (-1/600) calculated in step c, it means the reaction has a constant rate over this time interval, and these rates are the same. If not, the rates are different, and it could indicate that the reaction rate changes over time.