The first order rate constant for the decomposition of CaCO 3 at 700   K is 6 . 36 × 10 - 3  …

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

$K_{600}=x\times {10}^{-6}{ \mathrm{s}}^{-1}$

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

Applying :

$\log \left(\frac{{\mathrm{K}}_{{\mathrm{T}}_2}}{{ \mathrm{K}}_{{\mathrm{T}}_1}}\right)=\frac{-{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left(\frac{1}{{ \mathrm{T}}_2}-\frac{1}{{ \mathrm{T}}_1}\right)$

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

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

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

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

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

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

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

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

$\Rightarrow \mathrm{x}=16$