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Chemistry Question on Chemical Kinetics

The rate constant of a reaction is 1.5×10−31.5 \times 10^{-3} at 25∘C25^{\circ} C and 2.1×10−22.1 \times 10^{-2} at 60∘C60^{\circ} C. The activation energy is

A

35333Rlog⁡e2.1×10−21.5×10−2\frac{35}{333} R \log _{e} \frac{2.1 \times 10^{-2}}{1.5 \times 10^{-2}}

B

298×33335Rlog⁡e211.5\frac{298 \times 333}{35} R \log _{e} \frac{21}{1.5}

C

298×33335Rlog⁡e2.1\frac{298 \times 333}{35} R \log _{e} 2.1

D

298×33335Rlog⁡e2.11.5\frac{298 \times 333}{35} R \log _{e} \frac{2.1}{1.5}

Answer

298×33335Rlog⁡e211.5\frac{298 \times 333}{35} R \log _{e} \frac{21}{1.5}

Explanation

Solution

T1=273+25=298 KT_{1}=273+25=298 \,K T2=273+60=333 KT_{2}=273+60=333\, K log⁡k2k1=Ea2.3R(T2−T1T1T2)\log \frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.3 R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right) log⁡ek2k1=EaR(T2−T1T1T2)\log _{e} \frac{k_{2}}{k_{1}}=\frac{E_{a}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right) or log⁡e2.1×10−21.5×10−3=EaR(35333×298)\log _{e} \frac{2.1 \times 10^{-2}}{1.5 \times 10^{-3}}=\frac{E_{a}}{R}\left(\frac{35}{333 \times 298}\right) ∴Ea=298×33335×R×log⁡e211.5\therefore E_{a}=\frac{298 \times 333}{35} \times R \times \log _{e} \frac{21}{1.5}