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Q.The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea. (R=8.314 JK⁻¹mol⁻¹)

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2025Subjective· 2mImportance★★★★★
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Apply the two-temperature Arrhenius equation with k2/k1 = 2, T1 = 298 K, T2 = 308 K.

Given: T1=298 KT_1 = 298\ K, T2=298+10=308 KT_2 = 298 + 10 = 308\ K, k2k1=2\dfrac{k_2}{k_1} = 2, R=8.314 J K−1mol−1R = 8.314\ J\,K^{-1}mol^{-1}.

Arrhenius equation (two-temperature form):

ln⁡k2k1=EaR(1T1−1T2)\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)

ln⁡2=Ea8.314(1298−1308)\ln 2 = \dfrac{E_a}{8.314}\left(\dfrac{1}{298} - \dfrac{1}{308}\right)

1298−1308=308−298298×308=1091784=1.0896×10−4 K−1\dfrac{1}{298} - \dfrac{1}{308} = \dfrac{308-298}{298\times308} = \dfrac{10}{91784} = 1.0896\times10^{-4}\ K^{-1}

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