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Problems · Problem 6.14

Q.The half life of a first order reaction is 900 min at 820 K. Estimate its half life at 720 K if the activation energy is 250 kJ mol−1^{-1}.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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log10_{10}[(t1/2_{1/2})1_1/(t1/2_{1/2})2_2] = 2.212 →\to antilog 162.7 →\to (t1/2_{1/2})1_1 = 900 ×\times 162.7 = 1.464 ×\times 105^5 min.

Step 1. For a first order reaction t1/2_{1/2} = 0.693/kk, so (t1/2)1(t1/2)2=k2k1\dfrac{(t_{1/2})_1}{(t_{1/2})_2} = \dfrac{k_2}{k_1}. Take T1T_1 = 720 K (half life wanted) and T2T_2 = 820 K, where (t1/2_{1/2})2_2 = 900 min.

Step 2. log⁡10(t1/2)1(t1/2)2=Ea2.303 R(1T1−1T2)=250×10319.15×820−720720×820\log_{10}\dfrac{(t_{1/2})_1}{(t_{1/2})_2} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right) = \dfrac{250 \times 10^3}{19.15} \times \dfrac{820 - 720}{720 \times 820} = 2.212. …

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