Question

The first order rate constant for the decomposition of $CaC{O}_3$ at $700K$ is $6.36\times 10^{-3}{s}^{-1}$ and activation energy is $209kJmo{l}^{-1}$. Its rate constant (in $s^{-1}$ ) at $600K$ is $x\times 10^{-6}$. The value of $x$ is _______ (Nearest integer)
$[$ Given $R=8.31J{K}^{-1}mo{l}^{-1};\log 6.36\times 10^{-3}=-2.19$, $10^{-4.79}=1.62\times 10^{-5}]$

Answer 1

Answer: 16

$K_{700}=6.36\times 10^{-3}{s}^{-1}$
$K_{600}=x\times 10^{-6}{s}^{-1}$
$E_a=209kJ/mol$
Applying ;
$\log \left(\frac{K_{T_2}}{K_{T_1}}\right)=\frac{-E_a}{2.303R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$
$\log \left(\frac{K_{700}}{K_{600}}\right)=\frac{-E_a}{2.303R}\left(\frac{1}{700}-\frac{1}{600}\right)$
$\log \left(\frac{6.36\times 10^{-3}}{K_{600}}\right)=\frac{+209\times 1000}{2.303\times 8.31}\left(\frac{100}{700\times 600}\right)$
$\log \left(6.36\times 10^{-3}\right)-\log K_{600}=2.6$
$\Rightarrow \log K_{600}=-2.19-2.6=-4.79$
$\Rightarrow K_{600}=10^{-4.79}=1.62\times 10^{-5}$
$=16.2\times 10^{-6}$
$=x\times 10^{-6}$
$\Rightarrow x=16$