The following data are obtained from the decomposition of a gaseous compound The order of the reaction is

Question

The following data are obtained from the decomposition of a gaseous compound <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/839273327d685b4b13eec9706e497525-.png" /> The order of the reaction is (A) 0.5 (B) 1.0 (C) 1.5 (D) 2.0

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(( t 1 / 2)1/( t 1 / 2)2)=(80/113)=( p 2 / p 1) n -1 Here p 2, p 1 are the initial pressures (80/113)=((0.8/1.6)) n -1 Taking logarithms, log 0.7=( n -1) log 0.5 On solving, n =1.5

Q. The following data are obtained from the decomposition of a gaseous compound

The order of the reaction is

0.5

1.0

1.5

2.0

$\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{2}} = \frac{80}{113} = (p_{2} / p_{1})^{n - 1}$
Here $p_{2}, p_{1}$ are the initial pressures
$\frac{80}{113} = (\frac{0.8}{1.6})^{n - 1}$ Taking logarithms,
$lo g 0.7 = (n - 1) lo g 0.5$
On solving, $n = 1.5$

Q. The following data are obtained from the decomposition of a gaseous compound The order of the reaction is

0.5

1.0

1.5

2.0

(t1/2​)2​(t1/2​)1​​=11380​=(p2​/p1​)n−1 Here p2​,p1​ are the initial pressures 11380​=(1.60.8​)n−1 Taking logarithms, log0.7=(n−1)log0.5 On solving, n=1.5

Q. The following data are obtained from the decomposition of a gaseous compound

The order of the reaction is

$\frac{( t _{1/2} ) _{1}}{( t _{1/2} ) _{2}} = \frac{80}{113} = (p_{2} / p_{1})^{n - 1}$
Here $p_{2}, p_{1}$ are the initial pressures
$\frac{80}{113} = (\frac{0.8}{1.6})^{n - 1}$ Taking logarithms,
$lo g 0.7 = (n - 1) lo g 0.5$
On solving, $n = 1.5$