Question

Time and concentration data were collected for the reaction A ⟶ products t (s) [A] (M) 0 0.52 20 0.43 40 0.34857 60 0.28538 80 0.23365 100 0.19130 The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at t = 40 s. A plot has the concentration of A in molar on the y axis and time in seconds on the x axis. A curve contains the points (0, 0.52), (20, 0.43), (40, 0.35), (60, 0.29), (80, 0.24), and (100, 0.20). A line touches the curve at (40, 0.35) and has a y intercept of (0, 0.48). Approximate the instantaneous rate of this reaction at time t = 40 s.

Answer 1

`\large \boxed{\text{0.003 25 mol$\cdot$L$^(-1)$s}^(-1)}`

Explanation:

A ⟶ Products

`\begin{array}{rc}\textbf{t/s} & \textbf{[A]/mol$\cdot$L}^{\mathbf{-1}} \\0 & 0.52 \\20 & 0.43 \\40 & 0.35 \\60 & 0.29 \\80 & 0.24 \\100 & 0.20 \\\end{array}`

I plotted your blue curve and the orange line in the diagram below.

Calculate the instantaneous rate

The instantaneous rate of reaction is the negative slope of the orange line.

`\begin{array}{rcl}m & = & (y_(2) - y_(1))/(x_(2) - x_(1))\\\\ & = & \frac{\text{(0.35 - 0.48) mol/L}}{\text{(40 - 0) s}}\\\\& = & -\frac{\text{0.13 mol/L}}{\text{40 s}}\\\\& = & -\textbf{0.003 25 mol$\cdot$L$^{\mathbf{-1}}$s}^{\mathbf{-1}} \\\\\end{array}\\\text{The instantaneous rate of reaction is $\large \boxed{\textbf{0.003 25 mol$\cdot$L$^{\mathbf{-1}}$ s}^{\mathbf{-1}}}$}`