Chemistry · Ch 6 — Chemical Kinetics

Graphical determination of activation energy

6.7.2

Graphical determination of activation energy

Taking the logarithm of both sides of Eq. (6.18) we obtain

\ln k = -\frac{E_a}{RT} + \ln A \qquad \text{...(6.19)}

Converting the natural base to base 10 we write

\log_{10} k = -\frac{E_a}{2.303\,R}\,\frac{1}{T} + \log_{10} A \qquad \text{...(6.20)}

This equation is of the form of a straight line, $y = mx + c$ : the book's term-by-term arrows map $\log_{10} k$ to $y$ , $-E_a/2.303R$ to the slope $m$ , $1/T$ to $x$ and $\log_{10} A$ to the intercept $c$ .

The Arrhenius plot of $\log_{10} k$ versus $1/T$ , giving a straight line, is shown in Fig. (6.8). The slope of the line is $-E_a/2.303R$ , with its intercept being $\log_{10} A$ .

Figure 6.8Arrhenius plot of log10 k against the reciprocal of temperature: a straight line falling with slope minus Ea over 2.303 R, whose y-intercept is log10 A.

Fig. 6.8 — Arrhenius plot of log10 k against the reciprocal of temperature: a straight line falling with slope minus Ea over 2.303 R, whose y-intercept is log10 A.

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. Eq. (6.20) says $\log_{10} k$ depends linearly on $1/T$ — so plotting $\log_{10} k$ against $1/T$ gives a straight line. The line falls (larger $1/T$ = lower temperature = smaller $k$ ); its slope equals $-E_a/2.303R$ , and its intercept on the y-axis equals $\log_{10} A$ . Measuring the slope of …

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