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Answer the following in brief · Q14

Q.xiv. How will you determine activation energy:

(a) graphically using Arrhenius equation
(b) from rate constants at two different temperatures?
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1. (a) Graphical method. Take base-10 logs of the Arrhenius equation: log⁡10k=−Ea2.303R⋅1T+log⁡10A\log_{10}k=-\dfrac{E_a}{2.303R}\cdot\dfrac{1}{T}+\log_{10}A, a straight line (y=mx+cy=mx+c) in log⁡10k\log_{10}k vs 1/T1/T. Measuring kk at several temperatures and plotting gives a line of slope −Ea/(2.303R)-E_a/(2.303R); multiplying the (negative) slope by −2.303R-2.303R gives EaE_a.

Step 2. (b) Two-temperature method. Writing the equation at T1T_1 (rate constant k1k_1) and T2T_2 (rate constant k2k_2) and subtracting eliminates log⁡10A\log_{10}A: log⁡10k2k1=Ea2.303R(T2−T1T1T2)\log_{10}\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\left(\dfrac{T_2-T_1}{T_1T_2}\right). …

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