The first order rate constant for the decomposition

Question:

The first order rate constant for the decomposition

of $\mathrm{CaCO}_{3}$ at $700 \mathrm{~K}$ is $6.36 \times 10^{-3} \mathrm{~s}^{-1}$ and activation energy is $209 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Its rate constant (in $\mathrm{s}^{-1}$ ) at $600 \mathrm{~K}$ is $\mathrm{X} \times 10^{-6}$. The value of $\mathrm{x}$ is_________________(Nearest integer)

$\left[\right.$ Given $\mathrm{R}=8.31 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} ; \log 6.36 \times 10^{-3}=-2.19$, $\left.10^{-4.79}=1.62 \times 10^{-5}\right]$

Solution:

$\mathrm{K}_{700}=6.36 \times 10^{-3} \mathrm{~s}^{-1}$

$\mathrm{K}_{600}=x \times 10^{-6} \mathrm{~s}^{-1}$

$\mathrm{E}_{\mathrm{a}}=209 \mathrm{~kJ} / \mathrm{mol}$

Applying ;

$\log \left(\frac{\mathrm{K}_{\mathrm{T}_{2}}}{\mathrm{~K}_{\mathrm{T}_{1}}}\right)=\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{~T}_{2}}-\frac{1}{\mathrm{~T}_{1}}\right)$

$\log \left(\frac{\mathrm{K}_{700}}{\mathrm{~K}_{600}}\right)=\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{700}-\frac{1}{600}\right)$

$\log \left(\frac{6.36 \times 10^{-3}}{\mathrm{~K}_{600}}\right)=\frac{+209 \times 1000}{2.303 \times 8.31}\left(\frac{100}{700 \times 600}\right)$

$\log \left(6.36 \times 10^{-3}\right)-\log \mathrm{K}_{600}=2.6$

$\Rightarrow \log \mathrm{K}_{600}=-2.19-2.6=-4.79$

$\Rightarrow \mathrm{K}_{600}=10^{-4.79}=1.62 \times 10^{-5}$

$=16.2 \times 10^{-6}$

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

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