At 300 K, half life of a gaseous reactant initially at 58 KPa is 320 min. When the pressure is 29 KPa, the half life is 160 mm. The order of the reaction is

Question

At 300 K, half life of a gaseous reactant initially at 58 KPa is 320 min. When the pressure is 29 KPa, the half life is 160 mm. The order of the reaction is (A) 1 (B) 2 (C) 3 (D) 0

Tardigrade Admin · Accepted Answer

t1 / 2 propto (1/an - 1) (t2/t1)=((a1/a2))n - 1 (160/320)=((58/29))n - 1 n = 0.

Q. At 300 K, half life of a gaseous reactant initially at 58 KPa is 320 min. When the pressure is 29 KPa, the half life is 160 mm. The order of the reaction is

1

2

3

0

$t_{1/2} \propto \frac{1}{a ^{n - 1}}$

$\frac{t _{2}}{t _{1}} = (\frac{a _{1}}{a _{2}})^{n - 1}$

$\frac{160}{320} = (\frac{58}{29})^{n - 1}$

n = 0.

Q. At 300 K, half life of a gaseous reactant initially at 58 KPa is 320 min. When the pressure is 29 KPa, the half life is 160 mm. The order of the reaction is

1

2

3

0

t1/2​∝an−11​ t1​t2​​=(a2​a1​​)n−1 320160​=(2958​)n−1 n = 0.

$t_{1/2} \propto \frac{1}{a ^{n - 1}}$

$\frac{t _{2}}{t _{1}} = (\frac{a _{1}}{a _{2}})^{n - 1}$

$\frac{160}{320} = (\frac{58}{29})^{n - 1}$

n = 0.