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

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

((t1 / 2)1/(t1 / 2)2)=((P2/P1))n-1 (350/175)=((40/80))n-1 ⇒ 2=((1/2))n-1 n=1

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

$\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{2}} = (\frac{P _{2}}{P _{1}})^{n - 1}$
$\frac{350}{175} = (\frac{40}{80})^{n - 1}$
$\Rightarrow 2 = (\frac{1}{2})^{n - 1} n = 1$

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

(t1/2​)2​(t1/2​)1​​=(P1​P2​​)n−1 175350​=(8040​)n−1 ⇒2=(21​)n−1n=1

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

$\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{2}} = (\frac{P _{2}}{P _{1}})^{n - 1}$
$\frac{350}{175} = (\frac{40}{80})^{n - 1}$
$\Rightarrow 2 = (\frac{1}{2})^{n - 1} n = 1$