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xiv. How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?

**chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution**

### 3. Answer the following in brief.

xiv. How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?

Answer:-
**chapter 6. CHEMICAL KINETICS class 12 chemistry textbook solution page 137**

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- xiv. How will you determine activation energy: (a) graphically using Arrhenius equation (b) from rate constants at two different temperatures?

The Arrhenius equation is given as:

\[k = A e^{-\frac{E_a}{RT}}\]

Taking the logarithm of both sides of the equation, we obtain:

\[\ln k = -\frac{E_a}{RT} + \ln A\]

Converting the natural base to base 10, we write:

\[\log_{10} k = -\frac{E_a}{RT} + \log_{10} A\]

This equation is in the form of a straight line, \(y = mx + c\).

The Arrhenius plot of \(\log_{10} k\) versus \(\frac{1}{T}\) gives a straight line, as shown in the diagram. The slope of the line is \(-\frac{E_a}{2.303R}\) with its intercept being \(\log_{10} A\).

From the slope of the line, the activation energy can be determined.