Question

time and concentration data were collected for the reaction

\\(a \
ightarrow \text{products}\\)

\

$$\begin{tabular}{|c|c|} \\hline t (s) & a (m) \\\\ \\hline 0 & 0.52 \\\\ \\hline 20 & 0.43 \\\\ \\hline 40 & 0.35 \\\\ \\hline 60 & 0.29 \\\\ \\hline 80 & 0.23 \\\\ \\hline 100 & 0.19 \\\\ \\hline \\end{tabular}$$

the blue curve is the plot of the data. the straight orange line is tangent to the blue curve at \\(t = 40\text{ s}\\).

approximate the instantaneous rate of this reaction at time \\(t = 40\text{ s}\\).

instantaneous rate: m/s

Explanation:

Identify the target and method

We need to find the instantaneous rate of the reaction \(A
ightarrow \text{products}\) at \(t = 40\text{ s}\).
The instantaneous rate of consumption of \(A\) is given by:

$$\text{Rate} = -\frac{d[A]}{dt}$$

This corresponds to the negative slope of the tangent line (the orange line) to the concentration-time curve at \(t = 40\text{ s}\).
We will use the Reaction Rates concept to relate the slope of the tangent line to the reaction rate.

Find points on the tangent line

Let's identify two clear points on the straight orange tangent line from the graph:

1. At \(t = 0\text{ s}\), the orange line starts at approximately \([A] = 0.48\text{ M}\).
2. At \(t = 95\text{ s}\), the orange line intersects the horizontal axis at \([A] = 0.17\text{ M}\).

Let's check another point: at \(t = 40\text{ s}\), the tangent line touches the curve at \([A] = 0.35\text{ M}\).
Let's calculate the slope using \((0, 0.48)\) and \((40, 0.35)\):

$$\text{Slope} = \frac{0.35 - 0.48}{40 - 0} = \frac{-0.13}{40} = -0.00325\text{ M/s}$$

Let's check using \((40, 0.35)\) and the end of the line at approximately \((95, 0.17)\):

$$\text{Slope} = \frac{0.17 - 0.35}{95 - 40} = \frac{-0.18}{55} \approx -0.00327\text{ M/s}$$

Calculate the instantaneous rate

The instantaneous rate of the reaction is the positive value of this rate of change:

$$\text{Rate} = -\text{Slope} \approx 0.0033\text{ M/s}$$

This can also be written in scientific notation as \(3.3 \times 10^{-3}\text{ M/s}\).

Answer:

Approximate the instantaneous rate of this reaction at time \(t = 40\text{ s}\).

instantaneous rate: <blank>\(0.0033\)</blank> M/s