Chemistry · Ch 6 — Chemical Kinetics

Graphical representation of the first order reactions

6.5.5

Graphical representation of the first order reactions

i. The differential rate law for the first order reaction $\mathrm{A \longrightarrow P}$ is

\text{rate} = -\frac{\mathrm{d[A]}}{\mathrm{d}t} = k\,\mathrm{[A]_t} + 0

The book marks the terms of this equation with small arrows mapping it onto the straight-line form $y = mx + c$ : the rate is $y$ , $k$ is the slope $m$ , $\mathrm{[A]_t}$ is $x$ and the added zero is the intercept $c$ . A plot of rate versus concentration $\mathrm{[A]_t}$ is therefore a straight line passing through the origin, as shown in Fig. 6.3. The slope of the straight line $= k$ .

Figure 6.3Straight-line plot of reaction rate against initial concentration for a first order reaction, passing through the origin with slope equal to the rate constant k.

Fig. 6.3 — Straight-line plot of reaction rate against initial concentration for a first order reaction, passing through the origin with slope equal to the rate constant k.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. For a first order reaction the differential rate law rate $= k\,\mathrm{[A]_t}$ is a straight line through the origin when rate is plotted against concentration: doubling [A] doubles the rate, and the slope of the line is $k$ . *(The book's printed x-axis label reads "intitial Concerntration" — its own double mispr …

ii. From Eq. (6.7) the integrated rate law is

k = \frac{2.303}{t}\,\log_{10}\frac{\mathrm{[A]_0}}{\mathrm{[A]_t}}

On rearrangement, the equation becomes

\frac{kt}{2.303} = \log_{10}\mathrm{[A]_0} - \log_{10}\mathrm{[A]_t}

\text{Hence,}\quad \log_{10}\mathrm{[A]_t} = -\frac{k}{2.303}\,t + \log_{10}\mathrm{[A]_0}

This too maps term by term onto the straight-line equation: $\log_{10}\mathrm{[A]_t}$ is $y$ , $-k/2.303$ is the slope $m$ , $t$ is $x$ and $\log_{10}\mathrm{[A]_0}$ is the intercept $c$ . A graph of $\log_{10}\mathrm{[A]_t}$ versus $t$ yields a straight line with slope $-k/2.303$ and y-axis intercept $\log_{10}\mathrm{[A]_0}$ , as shown in Fig. 6.4.

Figure 6.4Plot of log10 of reactant concentration versus time for a first order reaction: a descending straight line whose slope is minus k over 2.303 and whose y-intercept is log10 of the initial concentration.

Fig. 6.4 — Plot of log10 of reactant concentration versus time for a first order reaction: a descending straight line whose slope is minus k over 2.303 and whose y-intercept is log10 of the initial concentration.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. Plotting $\log_{10}\mathrm{[A]_t}$ against time t turns the first order integrated rate law into a straight line: $\log_{10}\mathrm{[A]_t} = -\dfrac{k}{2.303}\,t + \log_{10}\mathrm{[A]_0}$ . The line falls with slope $-k/2.303$ and cuts the y-axis at $\log_{10}\mathrm{[A]_0}$ — measuring th …

iii. Eq. (6.7) gives …

Figure 6.5Plot of log10 of the ratio of initial to remaining concentration versus time for a first order reaction: a straight line through the origin with slope k over 2.303.

Fig. 6.5 — Plot of log10 of the ratio of initial to remaining concentration versus time for a first order reaction: a straight line through the origin with slope k over 2.303.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

What this figure shows. The same integrated rate law rearranged as $\log_{10}\dfrac{\mathrm{[A]_0}}{\mathrm{[A]_t}} = \dfrac{k}{2.303}\,t$ has no intercept term — so the plot of $\log_{10}\dfrac{\mathrm{[A]_0}}{\mathrm{[A]_t}}$ versus t is a straight line **passing t …

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