Question

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

Answer 1

$K _{700}=6.36 \times 10^{-3} s ^{-1}$
$ K _{600}=x \times 10^{-6} s ^{-1}$
$E _{ a }=209 \,kJ / mol$
Applying ;
$\log \left(\frac{ K _{ T _{2}}}{ K _{ T _{1}}}\right)=\frac{- E _{ a }}{2.303 R }\left(\frac{1}{ T _{2}}-\frac{1}{ T _{1}}\right) $
$\log \left(\frac{ K _{700}}{ K _{600}}\right)=\frac{- E _{ a }}{2.303 R }\left(\frac{1}{700}-\frac{1}{600}\right) $
$\log \left(\frac{6.36 \times 10^{-3}}{ 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 K _{600}=2.6 $
$\Rightarrow \log K _{600}=-2.19-2.6=-4.79$
$\Rightarrow K _{600}=10^{-4.79} =1.62 \times 10^{-5} $
$=16.2 \times 10^{-6} $
$=x \times 10^{-6}$
$\Rightarrow x =16 $