At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.8 kPa, the half life was fund to be 170 s. The order of the reaction is . [integer answer]

Question

At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.8 kPa, the half life was fund to be 170 s. The order of the reaction is . [integer answer] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

t 1 / 2 × (1/[ P 0] n -1) ( t 1/ t 2)=(( P 2) n -1/( P 1) n -1) (340/170)=((27.8/55.5)) n -1 ⇒ 2=(1/(2) n -1) n =0

Q. At $345 K$ , the half life for the decomposition of a sample of a gaseous compound initially at $55.5 k P a$ was $340 s$ . When the pressure was $27.8 k P a$ , the half life was fund to be $170 s$ . The order of the reaction is ____. [integer answer]

$t_{1/2} \times \frac{1}{[ P _{0} ] ^{n - 1}}$
$\frac{t _{1}}{t _{2}} = \frac{( P _{2} ) ^{n - 1}}{( P _{1} ) ^{n - 1}}$
$\frac{340}{170} = (\frac{27.8}{55.5})^{n - 1}$
$\Rightarrow 2 = \frac{1}{( 2 ) ^{n - 1}}$
$n = 0$

Q. At 345K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5kPa was 340s. When the pressure was 27.8kPa, the half life was fund to be 170s. The order of the reaction is ____. [integer answer]

t1/2​×[P0​]n−11​ t2​t1​​=(P1​)n−1(P2​)n−1​ 170340​=(55.527.8​)n−1 ⇒2=(2)n−11​ n=0

Q. At $345 K$ , the half life for the decomposition of a sample of a gaseous compound initially at $55.5 k P a$ was $340 s$ . When the pressure was $27.8 k P a$ , the half life was fund to be $170 s$ . The order of the reaction is ____. [integer answer]

$t_{1/2} \times \frac{1}{[ P _{0} ] ^{n - 1}}$
$\frac{t _{1}}{t _{2}} = \frac{( P _{2} ) ^{n - 1}}{( P _{1} ) ^{n - 1}}$
$\frac{340}{170} = (\frac{27.8}{55.5})^{n - 1}$
$\Rightarrow 2 = \frac{1}{( 2 ) ^{n - 1}}$
$n = 0$