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Q.The rate constant of a first order reaction increases from 2×10−2 s−12\times10^{-2}\,\text{s}^{-1} to 4×10−2 s−14\times10^{-2}\,\text{s}^{-1} when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea)(E_a). [log⁡2=0.3010,log⁡2.5=0.3979,log⁡4=0.6021,R=8.314 JK−1mol−1][\log 2 = 0.3010, \log 2.5 = 0.3979, \log 4 = 0.6021, R = 8.314\,\text{JK}^{-1}\text{mol}^{-1}]

Karnataka PUCKarnataka II PUC Board 2025Subjective· 3mImportance★★★★★
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The Arrhenius two-temperature equation gives an activation energy of about 53.6 kJ mol⁻¹.

Formula (Arrhenius, two temperatures):

log⁡k2k1=Ea2.303 R(T2−T1T1T2)\log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)

so

Ea=2.303 R T1T2T2−T1log⁡k2k1E_a = \frac{2.303\,R\,T_1 T_2}{T_2 - T_1}\log\frac{k_2}{k_1}

Given: k1=2×10−2 s−1k_1 = 2\times10^{-2}\,\text{s}^{-1}, k2=4×10−2 s−1k_2 = 4\times10^{-2}\,\text{s}^{-1}, T1=300 KT_1 = 300\,\text{K}, T2=310 KT_2 = 310\,\text{K}, R=8.314 J K−1mol−1R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}.

k2k1=4×10−22×10−2=2⇒log⁡k2k1=log⁡2=0.3010\frac{k_2}{k_1} = \frac{4\times10^{-2}}{2\times10^{-2}} = 2 \Rightarrow \log\frac{k_2}{k_1} = \log 2 = 0.3010

Substitute: …

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