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Byju's Answer

Standard XIII

Chemistry

Arrhenius Equation

The first ord...

Question

The first order rate constant for the decomposition of $C a C O_{3}$ at $700 K$ is $6.36 \times 10^{- 3} s^{- 1}$ and activation energy is $209 k J {mol}^{- 1}$ . Its rate constant (in $s^{- 1}$ ) at $600 K is x \times 10^{- 6}$ . The value of is (Nearest integer)
[Given $R = 8.31 J K^{- 1} {mol}^{- 1}; log 6.36 \times 10^{- 3} = - 2.19, 10^{- 4.79} = 1.62 \times 10^{- 5}$

 JEE MAINS 2021

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Solution

$k_{1} = 6.36 \times 10^{- 3} s^{- 1} T_{1} = 700 K$
$E_{a} = 209 k J m o l^{- 1}$
$k_{2} = x \times 10^{- 6} s^{- 1} T_{2} = 600 K$

$ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$log (\frac{x \times 10^{- 6}}{6.36 \times 10^{- 3}}) = \frac{209 \times 10^{3}}{8.31 \times 2.303} (\frac{1}{700} - \frac{1}{600})$

$log (x \times 10^{- 6}) = - 4.79 x \times 10^{- 6} = 10^{- 4.79}$

$x \times 10^{- 6} = 1.62 \times 10^{- 5}$

$x = 16.2 = 16$ (Nearest integer)

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k1=6.36×10−3 s−1T1=700 K Ea=209 kJ mol−1 k2=x×10−6 s−1T2=600 K lnk2k1=EaR(1T1−1T2) log(x×10−66.36×10−3)=209×1038.31×2.303(1700−1600) log (x×10−6)=−4.79x×10−6=10−4.79 x×10−6=1.62×10−5 x=16.2 =16 (Nearest integer)