Question

At $500\, K$, the half life period of a gaseous at an inital pressure of $80\, kPa$ is $350\, \sec$. When the pressure is $40\, kPa$, the half life period is $175\, \sec$, the order of the reaction is

Answer 1

$\frac{\left(t_{1 / 2}\right)_{1}}{\left(t_{1 / 2}\right)_{2}}=\left(\frac{P_{2}}{P_{1}}\right)^{n-1}$
$\frac{350}{175}=\left(\frac{40}{80}\right)^{n-1}$
$\Rightarrow 2=\left(\frac{1}{2}\right)^{n-1} n=1$