Then decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0×10−3S−1 and the activaton energy Ea=11.488KJ mol−1, the rate constant at 200K is ………..×10−5s−1. (Round off to the Nearest Integer). (Given: R=8.314J mol−1K−1 )

T1=200,T2=300K1=? K2=10−3log(10−3K1)=11.488×1032.303×8.314(1200−1300)log(10−3K1)=1148819.147(100200×300)log(10−3K1)=(1148811488)log10−3−logK1=1−3−1=logK1logK1=−4K1=10−4K1=10×10−5

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Then decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is $1.0 \times 10^{- 3} S^{- 1}$ and the activaton energy $E_{a} = 11.488 K J m o l^{- 1},$ the rate constant at 200K is ……….. $\times 10^{- 5} s^{- 1} .$ (Round off to the Nearest Integer). (Given: $R = 8.314 J m o l^{- 1} K^{- 1}$ )

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Detailed Solution

$\begin{array}{l} T_{1} = 200, T_{2} = 300 \\ K_{1} = ? K_{2} = 10^{- 3} \\ \log (\frac{10^{- 3}}{K_{1}}) = \frac{11.488 \times 10^{3}}{2.303 \times 8.314} (\frac{1}{200} - \frac{1}{300}) \end{array}$

$\begin{array}{l} \log (\frac{10^{- 3}}{K_{1}}) = \frac{11488}{19.147} (\frac{100}{200 \times 300}) \\ \log (\frac{10^{- 3}}{K_{1}}) = (\frac{11488}{11488}) \\ \log 10^{- 3} - \log K_{1} = 1 \\ - 3 - 1 = \log K_{1} \\ \log K_{1} = - 4 \\ K_{1} = 10^{- 4} \\ K_{1} = 10 \times 10^{- 5} \end{array}$