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Q.The rate constants of a reaction at 500 K and 700 K are 0.02 s−10.02\ \text{s}^{-1} and 0.07 s−10.07\ \text{s}^{-1} respectively. Calculate the activation energy of the reaction. (R=8.314 J K−1mol−1R = 8.314\ \text{J K}^{-1}\text{mol}^{-1})

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2024Subjective· 2mImportance★★★★★
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Substituting the two (T, k) pairs into the two-point Arrhenius equation and solving for EaE_a gives an activation energy of about 18.2 kJ mol⁻¹.

The Arrhenius equation in two-temperature (comparative) form is:

log⁡k2k1=Ea2.303 R(1T1−1T2)\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)

Given: k1=0.02 s−1k_1 = 0.02\ s^{-1} at T1=500 KT_1=500\ K; k2=0.07 s−1k_2=0.07\ s^{-1} at T2=700 KT_2=700\ K; R=8.314 J K−1mol−1R=8.314\ J\,K^{-1}mol^{-1}.

Step 1 — Ratio of rate constants

k2k1=0.070.02=3.5,log⁡(3.5)=0.5441\dfrac{k_2}{k_1} = \dfrac{0.07}{0.02} = 3.5, \qquad \log(3.5) = 0.5441

Step 2 — Temperature term

1T1−1T2=1500−1700=0.002−0.0014286=5.714×10−4 K−1\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{500}-\dfrac{1}{700} = 0.002 - 0.0014286 = 5.714\times10^{-4}\ K^{-1}

Step 3 — Solve for EaE_a …

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