The first order rate constant for the decomposition of CaCO _ 3 at...

${K}_{700} = 6 . 36 \times {10}^{- 3}{s}^{- 1}$

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

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

Applying :

$\log \left(\dfrac{{K}_{{T}_{2}}}{{K}_{{T}_{1}}}\right) = \dfrac{- {E}_{a}}{2 . 303R}\left(\dfrac{1}{{T}_{2}} - \dfrac{1}{{T}_{1}}\right)$

$\log \left(\dfrac{{K}_{700}}{{K}_{600}}\right) = \dfrac{- {E}_{a}}{2 . 303R}\left(\dfrac{1}{700} - \dfrac{1}{600}\right)$

$\log \left(\dfrac{6 . 36 \times {10}^{- 3}}{{K}_{600}}\right) = \dfrac{+ 209 \times 1000}{2 . 303 \times 8 . 31}\left(\dfrac{100}{700 \times 600}\right)$

$\log \left(6 . 36 \times {10}^{- 3}\right) - {logK}_{600} = 2 . 6$

$\log \left(6 . 36 \times {10}^{- 3}\right) - log{K}_{600} = 2 . 6$

$⇒ {logK}_{600} = - 2 . 19 - 2 . 6 = - 4 . 79$

$⇒ {K}_{600} = {10}^{- 4 . 79} = 1 . 62 \times {10}^{- 5}$

$= 16 . 2 \times {10}^{- 6}$

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

$⇒ x = 16$