Question

For the given gaseous reaction $A⟶2B+C$ pressure after sufficient time was found to be $300\text{mm Hg}$ and after time $2.303\text{sec}$, it was found to be $200\text{mm Hg}$, the half-life is (Given log 2 = 0.3010)

Answer 1

$A⟶2B+C$
$t=0P_o--$

$t=t{P}_o-x2xx$
$t=\infty -2P_o{P}_o$
$P_{\infty }\Rightarrow 300=3P_o\Rightarrow P_o=100\text{mm Hg}$
$P_o+2x=200\text{mm Hg}$
$2x=200-100=100\text{mm Hg}$
$\Rightarrow x=50\text{mm Hg}$
$K=\frac{2.303}{t}log\frac{P_o}{P_o-x}$
$K=\frac{2.303}{2.303}log\frac{100}{50}=0.3010$
$t_{1/2}=\frac{0.6932}{K}=\frac{0.6932}{0.3010}=2.303\text{sec}$