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Question 79 of 100

Q.In the Arrhenius equation for a first order reaction, the values of 'A' or 'Ea' are 4×1013 sec−14 \times 10^{13}\ sec^{-1} and 98.6 kJ mol−198.6\ kJ\ mol^{-1} respectively. At what temperature will its half life period be 10 minutes? [R=8.314 J K−1 mol−1R = 8.314\ J\ K^{-1}\ mol^{-1}]

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 3mImportance★★★★★
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Get kk from the given half-life, then solve the Arrhenius equation for TT.

Step 1 — find kk from the half-life (first order, t1/2=10t_{1/2}=10 min =600=600 s):

k=0.693t1/2=0.693600=1.155×10−3 s−1k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{600} = 1.155\times10^{-3}\ s^{-1}

Step 2 — Arrhenius equation:

k=A e−Ea/RT ⇒ ln⁡Ak=EaRT ⇒ T=EaRln⁡(A/k)k = A\,e^{-E_a/RT} \ \Rightarrow\ \ln\dfrac{A}{k} = \dfrac{E_a}{RT} \ \Rightarrow\ T = \dfrac{E_a}{R\ln(A/k)}

Ak=4×10131.155×10−3=3.463×1016\dfrac{A}{k} = \dfrac{4\times10^{13}}{1.155\times10^{-3}} = 3.463\times10^{16}

ln⁡(3.463×1016)=ln⁡(3.463)+16ln⁡(10)=1.242+36.841=38.08\ln(3.463\times10^{16}) = \ln(3.463) + 16\ln(10) = 1.242 + 36.841 = 38.08

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