Question

At $500$ $K$, the half-life period of a gaseous reaction at an initial 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:

• A. Zero
• B. One
• C. Two
• D. Three

Answer 1

The correct option is A Zero

$\frac{{\left(t_{1/2}\right)}_1}{{\left(t_{1/2}\right)}_2}={\left(\frac{p_2}{p_1}\right)}^{n-1}$
$\frac{350}{175}={\left(\frac{40}{80}\right)}^{n-1}$
$2={\left(\frac{1}{2}\right)}^{n-1}$
$n-1=-1$
$n=0$ (zero order reaction)