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CBSE Class 12 Chemistry 2023 Outside Delhi Set 1 Solved Paper

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Question : 28 of 35
Marks: +1, -0
A first order reaction is 50%50\% complete in 3030 minutes at 300 K300\,\mathrm{K} and in 10 minutes at 320 K320\,\mathrm{K}. Calculate activation energy (Ea)(E_a) for the reaction.
[R=8.314 J K−1 mol−1][R=8.314\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}]
[Given: log⁡2=0.3010,log⁡3=0.4771,[\text{Given: } \log 2=0.3010, \log 3=0.4771, log⁡4=0.6021]\log 4=0.6021]
Solution:     👈: Video Solution
K1K_1 at 27∘C27^{\circ}\mathrm{C} or 300 K=0.69330 min=0.0231 min−1300\,\mathrm{K}=\frac{0.693}{30\,\mathrm{min}}=0.0231\,\mathrm{min}^{-1}
K2K_2 at 47∘C47^{\circ}\mathrm{C} or 320 K1=0.69310 min=0.0693 min−1320\,K_1=\frac{0.693}{10\,\mathrm{min}}=0.0693\,\mathrm{min}^{-1}
Using Arrhenius equation:
log⁡(K2K1)=Ea2.303 R\log\left(\frac{K_2}{K_1}\right)=\frac{Ea}{2.303\,R} (T2−T1T1T2)\left(\frac{T_2-T_1}{T_1 T_2}\right)
log⁡(0.06930.0231)=Ea2.303×8.314×10−3 kj mol−1 K−1(20300×320)\log\left(\frac{0.0693}{0.0231}\right)=\frac{Ea}{2.303\times8.314\times10^{-3}\,\mathrm{kj}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}}\left(\frac{20}{300\times320}\right)
Ea=43.85 kj/molE_a=43.85\,\mathrm{kj}/\mathrm{mol}.
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