Question

A first order gaseous reactions has $K=1.5\times {10}^{-6}{\sec }^{-1}$ at ${200}^{\circ }C$. If the reaction is allowed to run for $10$ hour, what percentage of initial concentration would have changed into products. What is the half life period of reaction?

Answer 1

$K=\frac{2.303}{t}log\frac{Ao}{A}$

$\frac{1.5\times {10}^{-6}\times 10\times 60\times 60}{2.303}=log\left(\frac{Ao}{A}\right)$

$Ao=1.05A$

$Ao=1.05(Ao-x)$

$1.05x=1.05Ao-Ao$

$x=\frac{0.05}{1.05}Ao$

$x=0.0477Ao$

$x=4.7$% $Ao$

$t1/2=\frac{0.693}{1.5\times {10}^{-6}}$

$t1/2=4.6\times {10}^{5}sec$